0. Then Ar/Ar -> v and Acp/At
angular velocity vector co, and we have
co r sine
(o x r
\co\\ by definition of the
(5)
Figure 1 .63 A carousel wheel.
It's not difficult to see intuitively that v must be perpendicular to both co and r.
A moment's thought about the right-hand rule should enable you to establish the
vector equation
v = co x r.
(6)
If we apply formula (5) to a bicycle wheel, it tells us that the speed of a point
on the edge of the wheel is equal to the product of the radius of the wheel and
the angular speed (9 is n/2 in this case). Hence, if the rate of rotation is kept
constant, a point on the rim of a large wheel goes faster than a point on the rim
of a small one. In the case of a carousel wheel, this result tells you to sit on an
outside horse if you want a more exciting ride. (See Figure 1.63.)
Summary of Products Involving Vectors
Following is a collection of some basic information concerning scalar multipli-
cation of vectors, the dot product, and the cross product:
Scalar Multiplication: ka
Result is a vector in the direction of a.
Magnitude is ||&a|| = \k\ ||a||.
Zero if k = 0 or a = 0.
Commutative: ka = ak.
Associative: k(la) = (kl)a.
Distributive: Â£(a + b) = ka + kb; (k + l)a = ka + la.
Chapter 1 | Vectors
Dot Product: a â€¢ b
Result is a scalar
Magnitude is a â€¢ b = Hall 1 hi cos (9*
G is the an)?1e between a and b
rVTapnitnde is maximized ifa II Yt
Zero if a _L b, a = 0, or b = 0.
Commutative: a â€¢ b = b â€¢ a.
Associativity is irrelevant, since (a â–
b) â€¢ c doesn't make sense.
Distributive: a-(b + c) = a- b + a
â€¢ c.
Ifa = b, then a- a = ||a|| 2 .
Cross Product: a x b
Result is a vector perpendicular to both a and b.
Magnitude is ||a x b|| = ||a|| ||b|| sin#; 0 is the angle between a and b.
Magnitude is maximized if a _L b.
Zero ifa || b, a = 0, orb = 0.
Anticommutative: axb = â€” bxa.
Not associative: In general, a x (b x c) ^ (a x b) x c.
Distributive: ax(b + c) = axb + axc and
(a + b)xc = axc + bxc.
Ifa _Lb, then ||axb|| = ||a|| ||b||.
Addendum: Proofs of Cross Product Properties
Proof of Property 1 To prove the anticommutativity property, we use the right-
hand rule. Since
||axb|| = ||a|| ||b|| sin0,
we obviously have that ||a x b|| = ||b x a||. Therefore, we need only understand
the relation between the direction of a x b and that of b x a. To determine the
direction of a x b, imagine curling the fingers of your right hand from a toward b.
Then your thumb points in the direction of a x b. If instead you curl your fingers
from b toward a, then your thumb will point in the opposite direction. This is the
direction of b x a, so we conclude that a x b = â€” b x a. (See Figure 1.64.) â–
ax b
Figure 1 .64 The right-hand rule shows why a x b = â€” b x a.
1.4 | The Cross Product 37
Proof of Property 2 First, note the following general fact:
PROPOSITION 4.3 Let a and b be vectors in R 3 . If a â€¢ x = b â€¢ x for all vectors x
in R 3 , then a = b.
To establish Proposition 4.3, write a as aii + ci2] + 03k and b as b\i + +
b^k and set x in turn equal to i, j, and k. Proposition 4.3 is valid for vectors in R 2
as well as R 3 .
To prove the distributive law for cross products (property 2), we show that,
for any x e R 3 ,
(a x (b + c)) â€¢ x = (a x b + a x c) â€¢ x.
By Proposition 4.3, property 2 follows.
From the equations in (4),
(a x (b + c)) â€¢ x = (x x a) â€¢ (b + c)
= (x x a) â€¢ b + (x x a) â€¢ c,
from the distributive law for dot products,
= (a x b) â€¢ x + (a x c) â€¢ x
= (a x b + a x c) â€¢ x,
again using (4) and the distributive law for dot products. â–
Proof of Property 3 Property 3 follows from properties 1 and 2. We leave the
details as an exercise. â–
Proof of Property 4 The second equality in property 4 follows from the first
equality and property 1 :
k(a x b) = â€”k(b x a) by property 1
= â€”(kb) x a by the first equality of property 4
= a x (kb) by property 1 .
Hence, we need only prove the first equality.
If either a or b is the zero vector or if a is parallel to b, then the first equality
clearly holds. Otherwise, we divide into three cases: (1) k = 0, (2) k > 0, and
(3) k < 0. If k = 0, then both ka and k(a x b) are equal to the zero vector and
the desired result holds. If k > 0, the direction of (ka) x b is the same as a x b,
which is also the same as k(a x b). Moreover, the angle between ka and b is the
same as between a and b. Calling this angle 0, we check that
||(*a)xb|| = ||*a|| ||b|| sine
= &||a|| ||b|| sin# by part 1 of Proposition 3.4
= k || a x b || by Definition 4. 1
= \\k(a x b)|| by part 1 of Proposition 3.4.
We conclude (ka) x b = k(a x b) in this case.
38 Chapter 1 | Vectors
a
fca, k < 0
fca, k > 0
Figure 1 .65 If the angle between
a and b is 0, then the angle
between Â£a and b is either 0 (if
k > 0) or?r -0 (ifÂ£ < 0).
If k < 0, then the direction of (ka) x b is the same as that of (â€”a) x b, which
is seen to be the same as that of â€”(a x b) and thus the same as that of k(a x b).
The angle between ka and b is therefore n â€” 9, where 9 is the angle between a
and b. (See Figure 1.65.) Thus,
||(*a) x b|| = ||&a|| ||b|| sin(7r - 9) = \k\ ||a|| ||b|| sin 9 = ||*(a x b)||.
So, again, it follows that (ka) x b = k(a x b). â–
1.4 Exercises
Evaluate the determinants in Exercises 1â€”4.
1.
3.
2.
4.
In Exercises 5â€”7, calculate the indicated cross products, using
both formulas (2) and (3).
5. (1, 3, -2) x (-1,5, 7)
6. (3i-2j + k)x(i + j + k)
7. (i + j)x(-3i + 2j)
8. Prove property 3 of cross products, using properties 1
and 2.
9. If a x b = 3i - 7j - 2k, what is (a + b) x (a - b)?
1 0. Calculate the area of the parallelogram having vertices
(1,1), (3, 2), (1,3), and (-1,2).
1 1 . Calculate the area of the parallelogram having vertices
(1,2, 3), (4, -2, 1), (-3, 1, 0), and (0, -3, -2).
1 2. Find a unit vector that is perpendicular to both 2i +
j â€” 3k and i + k.
13. If (a x b) â€¢ c = 0, what can you say about the geomet-
ric relation between a, b, and c?
Compute the area of the triangles described in Exercises
14-17.
1 4. The triangle determined by the vectors a = i + j and
b = 2i-j
15. The triangle determined by the vectors a = i â€” 2j +
6k and b = 4i + 3j - k
16. The triangle having vertices (1,1), (â€”1,2), and
(-2,-1)
17. The triangle having vertices (1,0, 1), (0,2,3), and
(-1,5,-2)
1 8. Find the volume of the parallelepiped determined by
a = 3i - j, b = -2i + k, and c = i - 2j + 4k.
1 9. What is the volume of the parallelepiped with vertices
(3,0,-1), (4,2,-1), (-1,1,0), (3,1,5), (0,3,0),
(4,3,5), (-1,2, 6), and (0,4,6)?
20. Verify that (a x b) â€¢ c
a i
a 2 a-}
b 2 b 3
C2 c 3
21 . Show that (a x b) â€¢ c = a â€¢ (b x c) using Exercise 20.
22. Use geometry to show that |(axb)-c| =
|b-(axc)|.
23. (a) Show that the area of the triangle with vertices
P\(xi,y\), P 2 (x 2 , yi), and P 3 (x 3 , y 3 ) is given by
the absolute value of the expression
1 1 1
x\ x 2 JC 3
yi yi yi
(b) Use part (a) to find the area of the triangle with
vertices (1,2), (2, 3), and (-4, -4).
24. Suppose that a, b, and c are noncoplanar vectors in R 3 ,
so that they determine a tetrahedron as in Figure 1 .66.
Figure 1.66 The tetrahedron of
Exercise 24.
Give a formula for the surface area of the tetrahedron
in terms of a, b, and c. (Note: More than one formula
is possible.)
1.4 | Exercises 39
25. Suppose that you are given nonzero vectors a, b, and c
in R 3 . Use dot and cross products to give expressions for
vectors satisfying the following geometric descriptions:
(a) A vector orthogonal to a and b
(b) A vector of length 2 orthogonal to a and b
(c) The vector projection of b onto a
(d) A vector with the length of b and the direction of a
(e) A vector orthogonal to a and b x c
(f) A vector in the plane determined by a and b and
perpendicular to c.
26. Suppose a, b, c, and d are vectors in R 3 . Indicate which
of the following expressions are vectors, which are
scalars, and which are nonsense (i.e., neither a vector
nor a scalar).
(a) (a x b) x c
(c) (a â€¢ b) x (c â€¢ d)
(e) (a â€¢ b) x (c x d)
(g) (a x b) â€¢ (c x d)
(b) (a-b)-c
(d) (a x b) â€¢ c
(f) ax[(b-c)d]
(h) (a â€¢ b)c - (a x b)
Exercises 27â€”32 concern several identities for vectors a, b, c,
and d in R 3 . Each of them can be verified by hand by writing
the vectors in terms of their components and by using formula
(2) for the cross product and Definition 3.1 for the dot product.
However, this is quite tedious to do. Instead, use a computer
algebra system to define the vectors a, b, C, and d in general
and to verify the identities.
^27. (a x b) x c = (a â€¢ c)b - (b â€¢ c)a
^ 28. a â€¢ (b x c) = b â€¢ (c x a) = c â€¢ (a x b)
= -a â€¢ (c x b) = -c â€¢ (b x a)
= -b â€¢ (a x c)
^ 29. (a x b) â€¢ (c x d) = (a â€¢ c)(b â€¢ d) - (a â€¢ d)(b â€¢ c)
a â€¢ c a â€¢ d
b-c b-d
^ 30. (a x b) x c + (b x c) x a + (c x a) x b = 0 (this is
known as the Jacobi identity).
^ 31. (a x b) x (c x d) = [a â€¢ (c x d)]b - [b â€¢ (c x d)]a
^ 32. (a x b) â€¢ (b x c) x (c x a) = [a â€¢ (b x c)] 2
33. Establish the identity
(a x b) â€¢ (c x d) = (a â€¢ c)(b â€¢ d) - (a â€¢ d)(b â€¢ c)
of Exercise 29 without resorting to a computer algebra
system by using the results of Exercises 27 and 28.
34. Egbert applies a 20 lb force at the edge of a 4 ft
wide door that is half-open in order to close it. (See
Figure 1.67.) Assume that the direction of force is per-
pendicular to the plane of the doorway. What is the
torque about the hinge on the door?
35. Gertrude is changing a flat tire with a tire iron. The tire
iron is positioned on one of the bolts of the wheel so
Figure 1.67 Figure for Exercise 34.
40 lb
Figure 1.68 The configuration for
Exercise 35.
that it makes an angle of 30Â° with the horizontal. (See
Figure 1.68.) Gertrude exerts 40 lb of force straight
down to turn the bolt.
(a) If the length of the arm of the wrench is 1 ft, how
much torque does Gertrude impart to the bolt?
(b) What if she has a second tire iron whose length is
18 in?
36. Egbert is trying to open a jar of grape jelly. The ra-
dius of the lid of the jar is 2 in. If Egbert imparts 15 lb
of force tangent to the edge of the lid to open the jar,
how many ft-lb, and in what direction, is the resulting
torque?
37. A 50 lb child is sitting on one end of a seesaw, 3 ft
from the center fulcrum. (See Figure 1 .69.) When she is
1.5 ft
Figure 1.69 The seesaw of Exercise 37.
40 Chapter 1 | Vectors
1 .5 ft above the horizontal position, what is the amount
of torque she exerts on the seesaw?
38. For this problem, note that the radius of the earth is
approximately 3960 miles.
(a) Suppose that you are standing at 45Â° north latitude.
Given that the earth spins about its axis, how fast
are you moving?
(b) How fast would you be traveling if, instead, you
were standing at a point on the equator?
39. Archie, the cockroach, and Annie, the ant, are on an
LP record. Archie is at the edge of the record (ap-
proximately 6 in from the center) and Annie is 2 in
closer to the center of the record. How much faster is
Archie traveling than Annie? (Note: A record playing
on a turntable spins at a rate of 33 j revolutions per
minute.)
40. A top is spinning with a constant angular speed of 12
radians/sec. Suppose that the top spins about its axis
of symmetry and we orient things so that this axis is
the z-axis and the top spins counterclockwise about it.
(a) If, at a certain instant, a point P in the top has
coordinates (2,-1, 3), what is the velocity of the
point at that instant?
(b) What are the (approximate) coordinates of P one
second later?
41. There is a difficulty involved with our definition of
the angular velocity vector go, namely, that we cannot
properly consider this vector to be "free" in the sense
of being able to parallel translate it at will. Consider
the rotations of a rigid body about each of two parallel
axes. Then the corresponding angular velocity vectors
go 1 and 0, 0 < * 0 since r
is the radius of a circle. It also makes good sense to require 0 < 9 < 2jt, for then
every point in the plane, except the origin, has a uniquely determined pair of polar
coordinates. Occasionally, however, it is useful not to restrict r to be nonnegative
and 9 to be between 0 and lit. In such a case, no point of R 2 will be described by
a unique pair of polar coordinates: If P has polar coordinates (r, 9), then it also
has (r, 9 + 2nn) and (â€” r, 9 + (2n + \)tt) as coordinates, where n can be any
integer. (To locate the point having coordinates (r, 9), where r < 0, construct the
ray making angle 9 with respect to the positive x-axis, and instead of marching
| r | units away from the origin along this ray, go | r \ units in the opposite direction,
as shown in Figure 1.86.)
EXAMPLE 1 Polar coordinates may already be familiar to you. Nonetheless,
make sure you understand that the points pictured in Figure 1.87 have the coor-
dinates indicated. â™¦
(3V3, kI6)
(2, 5n/6)
(2, kI6)
(5,0)
(-1, 5tt/6) or
(1, 1171/6) or
(3, 3tt/2) <
(l,-7i/6)
Figure 1 .87 Figure for
Example 1.
Figure 1 .88 The graph of
r = 6 cos 6 in Example 2.
EXAMPLE 2 Let's graph the curve given by the polar equation r = 6cos#
(Figure 1.88). We can begin to get a feeling for the graph by compiling values, as
in the adjacent tabulation.
64 Chapter 1 [ Vectors
Thus, r decreases from 6 to 0 as 9 increases from 0 to ji/2; r decreases from
0 to â€”6 (or is not defined if you take r to be nonnegative) as 9 varies from ir/2
to 7r; r increases from â€”6 to 0 as 9 varies from n to 3jt/2; and r increases from
0 to 6 as 9 varies from 3jt/2 to 2ir . To graph the resulting curve, imagine a radar
screen: As 9 moves counterclockwise from 0 to 2n, the point (r, 9) of the graph is
traced as the appropriate "blip" on the radar screen. Note that the curve is actually
traced twice: once as 9 varies from 0 to n and then again as 9 varies from it to
2jt. Alternatively, the curve is traced just once if we allow only 9 values that yield
nonnegative r values. The resulting graph appears to be a circle of radius 3 (not
centered at the origin), and in fact, one can see (as in Example 3) that the graph
is indeed such a circle. â™¦
The basic conversions between polar and Cartesian coordinates are provided
by the following relations:
Polar to Cartesian:
J x = r cos9 ,
1 y = r sin 9 '
Cartesian to polar:
\ r 2 = x 2 + y 2
(tan<9 = y/x - (2)
Note that the equations in (2) do not uniquely determine r and 9 in terms of x
and y. This is quite acceptable, really, since we do not always want to insist that
r be nonnegative and 9 be between 0 and 2n. If we do restrict r and 9, however,
then they are given in terms of x and y by the following formulas:
r = y/x 2 + y 2 ,
tan -1 y/x
if x
>
0,y
>
0
tan -1 y/x + 2jt
ifx
>
0,y
<
0
tan -1 y/x + n
ifx
<
0,y
>
0
tt/2
ifx
0,y
>
0
3tt/2
ifx
0,y
<
0
indeterminate
ifx
y =
0
The complicated formula for 9 arises because we require 0 < 9 < 2n , while the
inverse tangent function returns values between â€” ir/2 and n/2 only. Now you
see why the equations given in (2) are a better bet!
EXAMPLE 3 We can use the formulas in (1) and (2) to prove that the curve in
Example 2 really is a circle. The polar equation r = 6 cos 9 that defines the curve
requires a little ingenuity to convert to the corresponding Cartesian equation. The
trick is to multiply both sides of the equation by r. Doing so, we obtain
r 2 = 6r cosO.
Now (1) and (2) immediately give
x 2 + y 2 = 6x.
1.7 | New Coordinate Systems 65
We complete the square in x to find that this equation can be rewritten as
(x - 3) 2 + y 2 = 9,
which is indeed a circle of radius 3 with center at (3, 0). â™¦
Cylindrical Coordinates
Cylindrical coordinates on R 3 are a "naive" way of generalizing polar coordinates
to three dimensions, in the sense that they are nothing more than polar coordinates
used in place of the x- and y-coordinates. (The z-coordinate is left unchanged.)
The geometry is as follows: Except for the z-axis, fill all of space with infinitely
extended circular cylinders with axes along the z-axis as in Figure 1.89. Then
any point P in R 3 not lying on the z-axis lies on exactly one such cylinder.
Hence, to locate such a point, it's enough to give the radius of the cylinder, the
circumferential angle 9 around the cylinder, and the vertical position z along the
cylinder. The cylindrical coordinates of P are (r, 9, z), as shown in Figure 1.90.
Algebraically, the equations in (1) and (2) can be extended to produce the basic
conversions between Cartesian and cylindrical coordinates.
The basic conversions between cylindrical
and Cartesian coordinates
are
provided by the following relations:
x = r cos 6*
Cylindrical to Cartesian:
y = r sin 9 ;
Z = Z
(3)
r 2 = x 2 + y 2
Cartesian to cylindrical:
tan 9 = y/x .
(4)
z = z
As with polar coordinates, if we make the restrictions r > 0, 0 < 9 < 2n, then
all points of R 3 except the z-axis have a unique set of cylindrical coordinates. A
point on the z-axis with Cartesian coordinates (0,0, zo) has cylindrical coordinates
(0, 9, zo), where 9 can be any angle.
Cylindrical coordinates are useful for studying objects possessing an axis of
symmetry. Before exploring a few examples, let's understand the three "constant
coordinate" surfaces.
â€¢ The r = T(, surface is, of course, just a cylinder of radius ro with axis the
z-axis. (See Figure 1.91.)
â€¢ The 9 = 9q surface is a vertical plane containing the z-axis (or a half-plane
with edge the z-axis if we take r > 0 only). (See Figure 1.92.)
â€¢ The z = zo surface is a horizontal plane. (See Figure 1.93.)
Half-plane only
EXAMPLE 4 Graph the surface having cylindrical equation r = 6 cos 8. (This
equation is identical to the one in Example 2.) In particular, z does not appear
in this equation. What this means is that if the surface is sliced by the horizontal
plane z = c where c is a constant, we will see the circle shown in Example 2, no
matter what c is. If we stack these circular sections, then the entire surface is a
circular cylinder of radius 3 with axis parallel to the z-axis (and through the point
(3, 0, 0) in cylindrical coordinates). This surface is shown in Figure 1.94. â™¦
EXAMPLE 5 Graph the surface having equation z = 2r in cylindrical co-
ordinates.
Here the variable 6 does not appear in the equation, which means that the
surface in question will be circularly symmetric about the z-axis. In other words,
if we slice the surface by any plane of the form 6 = constant (or half-plane, if we
take r > 0), we see the same curve, namely, a line (respectively, a half-line) of
slope 2. As we let the constant-0 plane vary, this line generates a cone, as shown
in Figure 1 .95. The cone consists only of the top half (nappe) when we restrict r
to be nonnegative.
The Cartesian equation of this cone is readily determined. Using the formulas
in (4), we have
Z = 2r => z 2 =4r 2 <^=> z 2 = 4(x 2 + y 2 ).
Since z can be positive as well as negative, this last Cartesian equation describes
the cone with both nappes. If we want the top nappe only, then the equation
z = 2y 'x 2 + y 2 describes it. Similarly, z = â€”2^x 2 + y 2 describes the bottom
nappe. â™¦
Spherical Coordinates
Fill all of space with spheres centered at the origin as in Figure 1 .96. Then every
point P e R 3 , except the origin, lies on a single such sphere. Roughly speaking,
the spherical coordinates of P are given by specifying the radius p of the sphere
containing P and the "latitude and longitude" readings of P along this sphere.
More precisely, the spherical coordinates (p, *

* 0). surface *

* 0) is, of course, a sphere of radius po,
as shown in Figure 1.99. The surface given by 9 = 9o is a half-plane just as in the
cylindrical case. The *

* 0. As with the graph of the cone with cylindrical equation z = 2r,
note that the equation is independent of 9. Thus, all sections of this surface made
by slicing with the half-plane 9 = c must be the same. If we compile values as
in the adjacent table, then the section of the surface in the half-plane 9 = 0 is as
shown in Figure 1.103. Since this section must be identical in all other constant-0
half-planes, we see that this surface appears to be a sphere of radius a tangent to
the xy-plane, which is shown in Figure 1.104.
(2a, 0, 0)
Section of
p = 2a cos q>
Figure 1.103 The cross section
of p = 2a cos 05 in the half-plane
6 = 0.
Figure 1 .1 04 The graph of p =
2a cos*

* between 2n/3 and it, p
is bounded by the surface of Venus (the plane z = â€” | in Cartesian coordinates)
and the spherical surface of the probe (whose equation in spherical coordinates is
p = 5). See Figure 1.107. From the formulas in (7) the equation z = â€” f corre-
sponds to the spherical equation p cos (p = â€” | or, equivalently, to p = â€” | sec *

*2 - - â– sm *

2 â– *3 = pcos^i These relations are the same as those given in (7), so hyperspherical coordinates are indeed the same as spherical coordinates when n = 3. In analogy with (5), it is standard practice to impose the following restrictions on the range of values for the coordinates: P > 0, 0 < (pi < 7T for k = 1, . . . , n â€” 2, 0 < q> n -\ < lit. (1 1) 1.7 | Exercises 73 Then, with these restrictions, we can convert from hypersphencal coordinates to Cartesian coordinates by means of the following formulas: ' p 2 = x \ + x\ H hiâ€ž 2 tanpi = ^2 H hx2_,/xâ€ž tan ^2 = J A H h x*_ 2 /Xn-i n 9 , tan

0.) (a) Sketch the intersection of this surface with the half- plane 8 = jt/2. (b) Sketch the entire surface. 20. (a) Graph the curve in R 2 having polar equation r = 2a sin 9, where a is a positive constant. (b) Graph the surface in R 3 having spherical equation p = 2a sin

2, fo), we have seen in equations (3) and (4) of Â§1.2 how to parametrize the line through Pq and Pi as r(r) = O Pq + t Pq Pi , where t can be any real number. (Recall that r = OP, the position vector of an arbitrary point P on the line.) (a) For what value of / does r(f) = OPb? For what value of t does r(r) = OP\l (b) Explain how to parametrize the line segment join- ing Pq and Pi. (See Figure 1.112.) z y Figure 1.112 The segment joining Po and Pi is a portion of the line containing Po and P|. (See Exercise 4.) (c) Give a set of parametric equations for the line seg- ment joining the points (0, 1, 3) and (2, 5, â€”7). 5. Recall that the perpendicular bisector of a line seg- ment in R 2 is the line through the midpoint of the seg- ment that is orthogonal to the segment. (a) Give a set of parametric equations for the perpen- dicular bisector of the segment joining the points Pi(-l,3)andP 2 (5, -7). (b) Given general points Pi(ai,fl2) and P2(b\,b2), provide a set of parametric equations for the per- pendicular bisector of the segment joining them. 6. If we want to consider a perpendicular bisector of a line segment in R 3 , we will find that the bisector must be a plane. (a) Give an (implicit) equation for the plane that serves as the perpendicular bisector of the segment join- ing the points P(6, 3, -2) and P 2 (-4, 1, 0). (b) Given general points P\{a\, 02, 03) and Pi{b\, bj, ^3), provide an equation for the plane that serves as the perpendicular bisector of the segment joining them. 7. Generalizing Exercises 5 and 6, we may define the per- pendicular bisector of a line segment in R" to be the hyperplane through the midpoint of the segment that is orthogonal to the segment. (a) Give an equation for the hyperplane in R 5 that serves as the perpendicular bisector of the seg- ment joining the points Pi(l, 6, 0, 3, â€” 2) and P 2 (-3,-2,4, 1,0). (b) Given arbitrary points P\(a\ , . . . , aâ€ž) and Pzibi, , . . , b n ) in R", provide an equation for the hyperplane that serves as the perpendicular bisector of the segment joining them. 8. If a and b are unit vectors in R 3 , show that || a x b|| 2 + (a-b) 2 = 1. 9. (a) If a â€¢ b = a â€¢ c, does it follow that b = c? Explain your answer. (b) If a x b = a x c, does it follow that b = c? Explain. 10. Show that the two lines h : x = t-3, y=\-2t, z = 2t + 5 h : x = 4 - It, y = At + 3, z = 6 - At are parallel, and find an equation for the plane that contains them. 1 1 . Consider the two planes x + y = 1 and y + z = 1 . These planes intersect in a straight line. ( a) Find the ( acute) angle of intersection between these planes. (b) Give a set of parametric equations for the line of intersection. 1 2. Which of the following lines whose parametric equa- tions are given below are parallel? Are any the same? (a) x = At + 6, y = 2 - It, z = 8* + 1 Miscellaneous Exercises for Chapter 1 (b) x = 3 - 6t , y = 3f , z = 4 - 9t (c) x = 2 - 2t, y = t + 4, z = -4r - 7 (d) jc = 2f + 4, y = 1 - f , z = 3? - 2 13. Determine which of the planes whose equations are given below are parallel and which are perpendicular. Are any of the planes the same? (a) 2x + 3y - z = 3 (b) -6x + 4y - 2z + 2 = 0 (c) x + y - z = 2 (d) 10* + I5y - 5z = 1 (e) 3x - 2y + z = 1 14. (a) What is the angle between the diagonal of a cube and one of the edges it meets? (Hint: Locate the cube in space in a convenient way.) (b) Find the angle between the diagonal of a cube and the diagonal of one of its faces. 15. Mark each of the following statements with a 1 if you agree, â€” 1 if you disagree: (1) Red is my favorite color. (2) I consider myself to be a good athlete. (3) I like cats more than dogs. (4) I enjoy spicy foods. (5) Mathematics is my favorite subject. Your responses to the preceding "questionnaire" may be considered to form a vector in R 5 . Suppose that you and a friend calculate your respective "response vec- tors" for the questionnaire. Explain the significance of the dot product of your two vectors. 16. The median of a triangle is the line segment that joins a vertex of a triangle to the midpoint of the opposite side. The purpose of this problem is to use vectors to show that the medians of a triangle all meet at a point. (a) Using Figure 1.113, write the vectors B M\ and C Mi in terms of AB and AC. (b) Let P be the point of intersection of BM\ and CMi. Write BP and CP in terms of AB and AC. (c) Use the fact that CB = CP + Jb = CA + AB to show that P must lie two-thirds of the way from B to Mi and two-thirds of the way from C to Mi. (d) Now use part (c) to show why all three medians must meet at P . 17. Suppose that the four vectors a, b, c, and d in R 3 are coplanar (i.e., that they all lie in the same plane). Show that then (a x b) x (c x d) = 0. 18. Show that the area of the triangle, two of whose sides are determined by the vectors a and b (see Figure 1.114), is given by the formula Area = ^7||a|| 2 ||b|| 2 - (a â€¢ b) 2 . Figure 1.1 14 The triangle in Exercise 18. 19. Let A(l,3,-1), 5(4,-1,3), C(2, 5,2), and D(5, 1, 6) be the vertices of a parallelogram. (a) Find the area of the parallelogram. (b) Find the area of the projection of the parallelogram in the xy -plane. 20. (a) For the line / in R 2 given by the equation ax + by = d, find a vector v that is parallel to /. (b) Find a vector n that is normal to / and has first component equal to a. (c) If Po(-*o, Vo) is any point in R 2 , use vectors to de- rive the following formula for the distance from P 0 to /: \axo + bye â€” d\ Distance from Pq to / = Figure 1.113 Two of the three medians of a triangle in Exercise 16. Va 2 + b 2 To do this, you'll find it helpful to use Figure 1.115, where Pi(xu yi) is any point on I. (d) Find the distance between the point (3, 5) and the line 8jc â€” 5y = 2. 21. (a) If PoC^o, yo, Zo) is any point in R 3 , use vectors to derive the following formula for the distance from Po to the plane n having equation Ax + By + Cz = D: \Ax 0 + By 0 + Czo- D\ Distance from Pq to fl VA 2 + B 2 + C 2 78 Chapter 1 [ Vectors y Pa \ n / I: ax + by = Figure 1.115 Geometric construction for Exercise 20. Figure 1.116 should help. (P\{x\, yi, Zi) is any point in n.) Distance ~-*P^J Yl:Ax + By + X Figure 1.116 Geometric construction for Exercise 2 1 . (b) Find the distance between the point (1,5, â€”3) and the plane x â€” 2y + 2z + 12 = 0. 22. (a) Let P be a point in space that is not contained in the plane n that passes through the three noncollinear points A , B , and C . Show that the distance between P and n is given by the expression |p-(bxc)| lib x o || where p = AP, b = AÂ§, and c = AC. (b) Use the result of part (a) to find the distance between (1, 0, â€”1) and the plane containing the points (1, 2, 3), (2, -3, 1), and (2,-1, 0). 23. Let A, B, C, and D denote four distinct points in R 3 . (a) Show that A, B, and C are collinear if and only if Xi x A~t = 0. (b) Show that A, B,C, and D are coplanar if and only if (AT? x At)-ci> = Q. 24. Let x = OP, the position vector of a point P in R 3 . Consider the equation x-k 1 II || ~ V2' Describe the configuration of points P that satisfy the equation. 25. Let a and b be two fixed, nonzero vectors in R 3 , and let c be a fixed constant. Explain how the pair of equations, a â€¢ x = c a x x = b , completely determines the vector x e R 3 . 26. (a) Give examples of vectors a, b, c in R 3 that show that, in general, it is not true that a x (b x c) = (a x b) x c. (That is, the cross product is not as- sociative.) (b) Use the Jacobi identity (see Exercise 30 of Â§1.4) to show that, for any vectors a, b, c in R 3 , a x (b x c) = (a x b) x c if and only if (c x a) x b = 0. 27. (a) Given an arbitrary (i.e., not necessarily regular) tetrahedron, associate to each of its four triangular faces a vector outwardly normal to that face with length equal to the area of that face. (See Fig- ure 1.117.) Show that the sum of these four vec- tors is zero. (Hint: Describe Vi , . . . , V4 in terms of some of the vectors that run along the edges of the tetrahedron.) v 3 Figure 1.117 The tetrahedron of part (a) of Exercise 27. (b) Recall that a polyhedron is a closed surface in R 3 consisting of a finite number of planar faces. Suppose you are given the two tetrahedra shown in Figure 1.118 and that face ABC of one is con- gruent to face A' B'C' of the other. If you glue the tetrahedra together along these congruent faces, then the outer faces give you a six-faced polyhe- dron. Associate to each face of this polyhedron an outward-pointing normal vector with length equal to the area of that face. Show that the sum of these six vectors is zero. Miscellaneous Exercises for Chapter 1 C A B B Figure 1.118 In Exercise 27(b), glue the two tetrahedra shown along congruent faces. 28. (c) Outline a proof of the following: Given an n-faced polyhedron, associate to each face an outward- pointing normal vector with length equal to the area of that face. Show that the sum of these n vectors is zero. Consider a right tetrahedron, that is, a tetrahedron that has a vertex R whose three adjacent faces are pair- wise perpendicular. (See Figure 1.119.) Use the result of Exercise 27 to show the following three-dimensional analogue of the Pythagorean theorem: If a, b, and c denote the areas of the three faces adjacent to R and d denotes the area of the face opposite R, then d 2 . Figure 1.119 The right tetrahedron of Exercise 28. The three faces containing the vertex R are pairwise perpendicular. 29. (a) Use vectors to prove that the sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the four sides. (b) Give an algebraic generalization of part (a) for R" . 30. Show that for any real numbers a\ we have n 2 â– â– , a n ,bi, n 31 . To raise a square (n x n) matrix A to a positive integer power n, one calculates A" as A â– A â– â– â– A (n times), (a) Calculate successive powers A, A 2 , A 3 , A 4 of the 1 1 matrix A = q \ (b) Conjecture the general form of A" for the matrix A of part (a), where n is any positive integer. (c) Prove your conjecture in part (b) using mathemat- ical induction. 32. A square matrix A is called nilpotent if A" some positive power n, 0 1 1 0 for (a) Show that A is nilpotent. 0 0 0 0 0 0 (b) Use a calculator or computer to show that A = 0 0 0 0 0 is nilpotent. ^ 33. The n x n matrix H n whose ijth entry is l/(i + j â€” 1) is called the Hilbert matrix of order n . (a) Write out H2, H], H^, H5, and H^, Use a com- puter to calculate their determinants exactly. What seems to happen to det H n as n gets larger? (b) Now calculate #10 and det H\q. If you use exact arithmetic, you should find that det H\q / 0 and hence that //10 is invertible. (See Exercises 30-38 of Â§ 1 .6 for more about invertible matrices.) (c) Now give a numerical approximation A for H\q. Calculate the inverse matrix B of this approxima- tion, if your computer allows. Then calculate AB and B A . Do you obtain the 10x10 identity matrix ha in both cases? (d) Explain what parts (b) and (c) suggest about the difficulties in using numerical approximations in matrix arithmetic. As a child, you may have played with a popular toy called a SpirographÂ®. With it one could draw some appealing geomet- ric figures. The Spirograph consists of a small toothed disk with several holes in it and a larger ring with teeth on both inside and outside as shown in Figure 1.120. You can draw pictures by meshing the small disk with either the inside or outside circles of the ring and then poking a pen through one of the holes of the disk while turning the disk. (The large ring is held fixed.) 80 Chapter 1 [ Vectors An idealized version of the Spirograph can be obtained by taking a large circle (of radius a) and letting a small circle (of radius b) roll either inside or outside it without slipping. A "Spirograph " pattern is produced by tracking a particu- lar point lying anywhere on (or inside) the small circle. Exer- cises 34-37 concern this set-up. 34. Suppose that the small circle rolls inside the larger circle and that the point P we follow lies on the circum- ference of the small circle. If the initial configuration is such that P is at (a, 0), find parametric equations for the curve traced by P, using angle t from the posi- tive .r-axis to the center B of the moving circle. (This configuration is shown in Figure 1.121.) The result- ing curve is called a hypocycloid. Two examples are shown in Figure 1.122. y a (b. y\ fp ^\ At ( , b\ Figure 1.121 The coordinate configuration for finding parametric equations for a hypocycloid. 35. Now suppose that the small circle rolls on the outside of the larger circle. Derive a set of parametric equa- tions for the resulting curve in this case. Such a curve is called an epicycloid, shown in Figure 1.123. 36. (a) A cusp (or corner) occurs on either the hypocy- cloid or epicycloid every time the point P on the small circle touches the large circle. Equivalently, y Figure 1.123 An epicycloid with a = 4,b= 1. this happens whenever the smaller circle rolls through 2tt. Assuming that a/b is rational, how many cusps does a hypocycloid or epicycloid have? (Your answer should involve a and b in some way.) (b) Describe in words and pictures what happens when a/b is not rational. 37. Consider the original Spirograph set-up again. If we now mark a point P at a distance c from the center of the smaller circle, then the curve traced by P is called a hypotrochoid (if the smaller circle rolls on the inside of the larger circle) or an epitrochoid (if the smaller circle rolls on the outside). Note that we must have b < a, but we can have c either larger or smaller than b. (If c < b, we get a "true" Spirograph pattern in the sense that the point P will be on the inside of the smaller circle. The situation when c > b is like having P mounted on the end of an elongated spoke on the smaller circle.) Give a set of parametric equations for the curves that result in this way. (See Figure 1.124.) Exercises 38-43 are made feasible through the use of appropri- ate software for graphing in polar, cylindrical, and spherical Miscellaneous Exercises for Chapter 1 coordinates. (Note: When using software for graphing in spher- ical coordinates, be sure to check the definitions that are used for the angles

* Y for a function. Such notation indicates all
the ingredients of a particular function, although it does not make the nature of
the rule of assignment explicit. This notation also suggests the "mapping" nature
of a function, indicated by Figure 2.1.
EXAMPLE 1 Abstract definitions are necessary, but it is just as important that
you understand functions as they actually occur. Consider the act of assigning to
each U.S. citizen his or her social security number. This pairing defines a function:
Each citizen is assigned one social security number. The domain is the set of U.S.
citizens and the codomain is the set of all nine-digit strings of numbers.
On the other hand when a university assigns students to dormitory rooms, it
is unlikely that it is creating a function from the set of available rooms to the set of
students. This is because some rooms may have more than one student assigned
to them, so that a particular room does not necessarily determine a unique student
occupant. â™¦
2.1 | Functions of Several Variables; Graphing Surfaces
DEFINITION 1 .1 The range of a function f:X ^ Y is the set of those
elements of Y that are actual values of /. That is, the range of / consists of
those y in Y such that y = f(x) for some x in X.
Using set notation, we find that
Range / = {y e Y \ y = f(x) for some x e X] .
In the social security function of Example 1, the range consists of those nine-
digit numbers actually used as social security numbers. For example, the number
000-00-0000 is not in the range, since no one is actually assigned this number.
Figure 2.2 Every y e Y is "hit"
by at least one x e X.
Figure 2.3 The element b e Y
is not the image of any x Â£ X.
DEFINITION 1 .2 A function /: X -> Y is said to be onto (or surjective)
if every element of Y is the image of some element of X, that is, if range
f=Y.
The social security function is not onto, since 000-00-0000 is in the codomain
but not in the range. Pictorially, an onto function is suggested by Figure 2.2. A
function that is not onto looks instead like Figure 2.3. You may find it helpful to
think of the codomain of a function / as the set of possible (or allowable) values
of /, and the range of / as the set of actual values attained. Then an onto function
is one whose possible and actual values are the same.
DEFINITION 1 .3 A function /: X Y is called one-one (or injective) if
no two distinct elements of the domain have the same image under /. That
is, / is one-one if whenever x\, * 2 Â£ X and x\ ^ x 2 , then f{x\) ^ f(xz).
(See Figure 2.4.)
not one-one
Figure 2.4 The figure on the left depicts a one-one mapping; the one
on the right shows a function that is not one-one.
One would expect the social security function to be one-one, but we have heard
of cases of two people being assigned the same number so that, alas, apparently
it is not.
When you studied single-variable calculus, the functions of interest were
those whose domains and codomains were subsets of R (the real numbers). It
was probably the case that only the rule of assignment was made explicit; it is
generally assumed that the domain is the largest possible subset of R for which
the function makes sense. The codomain is generally taken to be all of R.
84 Chapter 2 I Differentiation in Several Variables
EXAMPLE 2 Suppose /: R â€”> R is given by /(jc) = x 2 . Then the domain and
codomain are, explicitly, all of R, but the range of / is the interval [0, oo). Thus
/ is not onto, since the codomain is strictly larger than the range. Note that / is
not one-one, since f(2) = /(â€” 2) = 4, but 2 â€”2. â™¦
EXAMPLE 3 Suppose g is a function such that g(x) = >Jx â€” 1. Then if we
take the codomain to be all of R, the domain cannot be any larger than [1, oo).
If the domain included any values less than one, the radicand would be negative
and, hence, g would not be real-valued. â™¦
Now we're ready to think about functions of more than one real variable. In
the most general terms, these are the functions whose domains are subsets X of
R" and whose codomains are subsets of R"', for some positive integers n and m.
(For simplicity of notation, we'll take the codomains to be all of R'", except when
specified otherwise.) That is, such a function is a mapping f:XQ R" Râ„¢ that
associates to a vector (or point) x in X a unique vector (point) f(x) in R" ! .
EXAMPLE 4 Let T: R 3 -* R be defined by T(x, y, z) = xy + xz + yz. We
can think of T as a sort of "temperature function." Given a point x = (x, y, z) in
R 3 , T(x) calculates the temperature at that point. â™¦
EXAMPLE 5 Let L: R" -> R be given by L(x) = ||x|| . This is a "length func-
tion" in that it computes the length of any vector x in R". Note that L is not
one-one, since L(e,-) = L(e y ) = 1, where e, and e ; - are any two of the standard
basis vectors for R" . L also fails to be onto, since the length of a vector is always
nonnegative. â™¦
EXAMPLE 6 Consider the function given by N(x) = x/||x|| where x is a vector
in R 3 . Note that N is not defined if x = 0, so the largest possible domain for N is
R 3 â€” {0}. The range of N consists of all unit vectors in R 3 . The function N is the
"normalization function," that is, the function that takes a nonzero vector in R 3
and returns the unit vector that points in the same direction. â™¦
EXAMPLE 7 Sometimes a function may be given numerically by a table.
One such example is the notion of windchill â€” the apparent temperature one
feels when taking into account both the actual air temperature and the speed of
the wind. A standard table of windchill values is shown in Figure 2.5. 1 From it
we see that if the air temperature is 20 Â°F and the windspeed is 25 mph, the wind-
chill temperature ("how cold it feels") is 3 Â°F. Similarly, if the air temperature is
35Â°F and the windspeed is 10 mph, then the windchill is 27Â°F. In other words,
if s denotes windspeed and t air temperature, then the windchill is a function
W(s, t). â™¦
The functions described in Examples 4, 5, and 7 are scalar-valued functions,
that is, functions whose codomains are R or subsets of R. Scalar- valued functions
are our main concern for this chapter. Nonetheless, let's look at a few examples
of functions whose codomains are R" ! where m > 1 .
NOAA, National Weather Service, Office of Climate, Water, and Weather Services, "NWS Wind Chill
Temperature Index." February 26, 2004. (July 31, 2010).
2.1 | Functions of Several Variables; Graphing Surfaces
Air Temp
(deg F)
Windspeed (mph)
5
10
15
20
25
30
35 40
45
50
55
60
40
36
34
32
30
29
28
28 27
26
26
25
25
35
31
27
25
24
23
22
21 20
19
19
18
17
30
25
21
19
17
16
15
14 13
12
12
11
10
25
19
15
13
11
9
8
7 6
5
4
4
3
20
13
9
6
4
3
1
0 -1
-2
-3
-3
-4
15
7
3
0
-2
-4
-5
-7 -8
-9
-10
-11
-11
10
1
-4
-7
-9
-11
-12
-14 -15
-16
-17
-18
-19
5
-5
-10
-13
-15
-17
-19
-21 -22
-23
-24
-25
-26
0
-11
-16
-19
-22
-24
-26
-27 -29
-30
-31
-32
-33
-5
-16
-22
-26
-29
-31
-33
-34 -36
-37
-38
-39
-40
-10
-22
-28
-32
-35
-37
-39
-41 -43
-44
-45
-46
-48
-15
-28
-35
-39
-42
-44
-46
-48 -50
-51
-52
-54
-55
-20
-34
-41
-45
-48
-51
-53
-55 -57
-58
-60
-61
-62
-25
-40
-47
-51
-55
-58
-60
-62 -64
-65
-67
-68
-69
-30
-46
-53
-58
-61
-64
-67
-69 -71
-72
-74
-75
-76
-35
-52
-59
-64
-68
-71
-73
-76 -78
-79
-81
-82
-84
-40
-57
-66
-71
-74
-78
-80
-82 -84
-86
-88
-89
-91
-45
-63
-72
-77
-81
-84
-87
-89 -91
-93
-95
-97
-98
Figure 2.6 The helix of
Example 8. The arrow shows the
direction of increasing t .
Figure 2.5 Table of windchill values in English units.
EXAMPLE 8 Define f: R -> R 3 by f(f) = (cos t, sin?, t). The range off is the
curve in R 3 with parametric equations x = cos t , y = sin t , z = t . If we think of
t as a time parameter, then this function traces out the corkscrew curve (called a
helix) shown in Figure 2.6. â™¦
EXAMPLE 9 We can think of the velocity of a fluid as a vector in R 3 . This
vector depends on (at least) the point at which one measures the velocity and also
the time at which one makes the measurement. In other words, velocity may be
considered to be a function v: X c R 4 â€” >â€¢ R 3 . The domain X is a subset of R 4
because three variables x, y, z are required to describe a point in the fluid and a
fourth variable t is needed to keep track of time. (See Figure 2.7.) For instance,
such a function v might be given by the expression
v(x, y, z, t) = xyzt\ + (x 1
y 2 )l
(3z + /)k.
Figure 2.7 A water
pitcher. The velocity v of
the water is a function
from a subset of R 4 to R 3 .
You may have noted that the expression for v in Example 9 is considerably
more complicated than those for the functions given in Examples 4-8. This is
because all the variables and vector components have been written out explicitly.
In general, if we have a function f: X c R" â€” > R m , then xeX can be written as
x = (xi , X2, â– â– â– , xâ€ž) and f can be written in terms of its component functions
/i. fz< â– â– â– . fm- The component functions are scalar- valued functions of x Â£ X
that define the components of the vector f(x) Â£ R" ! . What results is a morass of
symbols:
f(x) = f(jti, X2, â– . â– , x n ) (emphasizing the variables)
â€” (/i( x )> /2( x )> â– â€¢ â€¢ > /m( x )) (emphasizing the component functions)
= {fi{x\,x 2 , . . .,x n ), f 2 (xi,x 2 , . . . ,X n ), f m (Xl,X 2 , â– â– .,xâ€ž))
(writing out all components).
86 Chapter 2 I Differentiation in Several Variables
For example, the function L of Example 5, when expanded, becomes
L(x) = L(x\,x 2 , . . . , xâ€ž) = ^Jxf + x\
The function N of Example 6 becomes
X (X],X2,X-i)
N(x) = â€”
^ llxll
x 2 x 3
and, hence, the three component functions of N are
X\ X2
Ni(xi,x 2 ,X3) = N 2 (xi, x 2 , xi) =
X^ ~\~ X2 ~\~ X^ J X\ -\- x 2 ~\~ x^
x 3
Ni(xi,X2, x 3 ) =
X\ -\- Xj ~"T~ X-i
Although writing a function in terms of all its variables and components has
the advantage of being explicit, quite a lot of paper and ink are used in the process.
The use of vector notation not only saves space and trees but also helps to make
the meaning of a function clear by emphasizing that a function maps points in
R" to points in R'" . Vector notation makes a function of 300 variables look "just
like" a function of one variable. Try to avoid writing out components as much as
you can (except when you want to impress your friends).
Visualizing Functions
No doubt you have been graphing scalar-valued functions of one variable for so
long that you give the matter little thought. Let's scrutinize what you've been do-
ing, however. A function /:XcR^ R takes a real number and returns another
real number as suggested by Figure 2.8. The graph of / is something that "lives"
in R 2 . (See Figure 2.9.) It consists of points (x, y) such that y = f(x). That is,
Graph / = {{x, f(x)) \ x e X} = {(x, y)\xeX,y = f(x)} .
The important fact is that, in general, the graph of a scalar-valued function
of a single variable is a curve â€” a one-dimensional object â€” sitting inside two-
dimensional space.
y
f
SKA
(x,f(x)) /
X
Figure 2.8 A function /:ICR->R, Figure 2.9 The graph off.
2.1 | Functions of Several Variables; Graphing Surfaces
Now suppose we have a function /:XcR 2 ->R, that is, a function of two
variables. We make essentially the same definition for the graph:
Graph / = {(x, /(x)) xel). (1)
Of course, x = (x, y) is a point of R 2 . Thus, {(x, fix))} may also be written as
l(x, y, fix, y))} , or as {(x, y, z) | (x, y) e X, z = f(x, y)} .
Hence, the graph of a scalar-valued function of two variables is something that
sits in R 3 . Generally speaking, the graph will be a surface.
EXAMPLE 1 0 The graph of the function
9 1,1,7
/:R 2 ^R /(x,y)=-y 3 -y--x 2 + -
is shown in Figure 2.10. For each point x = (x, y) in R 2 , the point in R 3 with
coordinates (x, y, ^y 3 â€” y â€” \x 2 + |) is graphed. â™¦
Figure 2.10 The graph of fix, y) = -j^y 3 â€” y â€” \x 2 + |.
Graphing functions of two variables is a much more difficult task than graph-
ing functions of one variable. Of course, one method is to let a computer do
the work. Nonetheless, if you want to get a feeling for functions of more than
one variable, being able to sketch a rough graph by hand is still a valuable skill.
The trick to putting together a reasonable graph is to find a way to cut down on
the dimensions involved. One way this can be achieved is by drawing certain
special curves that lie on the surface z = fix, y). These special curves, called
contour curves, are the ones obtained by intersecting the surface with horizontal
planes z = c for various values of the constant c. Some contour curves drawn
on the surface of Example 10 are shown in Figure 2.1 1. If we compress all the
contour curves onto the xy-plane (in essence, if we look down along the posi-
tive z-axis), then we create a "topographic map" of the surface that is shown in
Figure 2.12. These curves in the xy-plane are called the level curves of the original
function /.
The point of the preceding discussion is that we can reverse the process in
order to sketch systematically the graph of a function / of two variables: We
88 Chapter 2 I Differentiation in Several Variables
Figure 2.1 1 Some contour curves of the Figure 2.1 2 Some level curves of
function in Example 10. the function in Example 10.
first construct a topographic map in R 2 by finding the level curves of /, then
situate these curves in R 3 as contour curves at the appropriate heights, and finally
complete the graph of the function. Before we give an example, let's restate our
terminology with greater precision.
DEFINITION 1 .4 Let /:XcR 2 ^Rbea scalar-valued function of two
variables. The level curve at height c of / is the curve in R 2 defined by the
equation f(x, y) = c, where c is a constant. In mathematical notation,
Level curve at height c = {(x, y) Â£ R 2 | fix, y) = c] .
The contour curve at height c of / is the curve in R 3 defined by the two
equations z = fix, y) and z = c. Symbolized,
Contour curve at height c = {(x, y, z) Â£ R 3 | z = fix, y) = c} .
In addition to level and contour curves, consideration of the sections of a
surface by the planes where x or y is held constant is also helpful. A section of a
surface by a plane is just the intersection of the surface with that plane. Formally,
we have the following definition:
DEFINITION 1 .5 Let /: X C R 2 -> R be a scalar-valued function of two
variables. The section of the graph of / by the plane x = c (where c
is a constant) is the set of points (x, y, z), where z = f(x, y) and x = c.
Symbolized,
Section by x = c is {(x, y, z) e R 3 I z = f(x, y), x = c}.
Similarly, the section of the graph of / by the plane y = c is the set of
points described as follows:
Section by y = c is {(x, y, z) e R 3 | z = fix, y), y = c].
2.1 | Functions of Several Variables; Graphing Surfaces
EXAMPLE 1 1 We'll use level and contour curves to construct the graph of the
function
/:R 2 ^R, f(x,y) = 4-x 2 -y 2 .
By Definition 1 .4, the level curve at height c is
{(x, y) 6 R 2 | 4 - x 2 - y 2 = c} = {(x, y) | x 2 + y 2 = 4 - c} .
Thus, we see that the level curves for c < 4 are circles centered at the origin of
radius -J4 â€” c. The level "curve" at height c = 4 is not a curve at all but just a
single point (the origin). Finally, there are no level curves at heights larger than 4
since the equation x 2 + y 2 = 4 â€” c has no real solutions in x and y. (Why not?)
These remarks are summarized in the following table:
c
Level curve x 2 + y 2 = 4 â€” c
-5
x 2 + y 2 = 9
-1
x 2 + y 2 = 5
0
x 2 + y 2 = 4
1
x 2 + y 2 = 3
3
x 2 + y 2 = l
4
x 2 + y 2 = 0 x = y = 0
c, where c > 4
empty
Thus, the family of level curves, the "topographic map" of the surface z =
4 â€” x 2 â€” y 2 , is shown in Figure 2.13. Some contour curves, which sit in R 3 ,
are shown in Figure 2.14, where we can get a feeling for the complete graph of
z = 4 â€” x 2 â€” y 2 . It is a surface that looks like an inverted dish and is called a
paraboloid. (See Figure 2. 15.) To make the picture clearer, we have also sketched
in the sections of the surface by the planes x = 0 and y = 0. The section by x = 0
is given analytically by the set
{(x, y, z) e R 3 | z = 4 - x 2 - y 2 , x = 0} = {(0, y, z )\z = 4- y 2 } .
Similarly, the section by y = 0 is
{(x, y, z) e R 3 | z = 4 - x 2 - y 2 , y = 0} = {(x, 0, z) \ z = 4 - x 2 } .
Figure 2.13 The topographic Figure 2.14 Some contour Figure 2.15 The graph of
map of z = 4 â€” x 2 â€” y 2 (i.e., curves of z = 4 â€” x 2 â€” y 2 . f(x, y) = 4 â€” x 2 â€” y 2 .
several of its level curves).
Chapter 2 | Differentiation in Several Variables
Since these sections are parabolas, it is easy to see how this surface obtained its
name. â™¦
EXAMPLE 12 We'll graph the function g: R 2 -> R, g(x, y) = y 2 - x 2 . The
level curves are all hyperbolas, with the exception of the level curve at height 0,
which is a pair of intersecting lines.
c = -4
c
Level curve y 2 â€” x 2 = c
-4
x 2 - y 2 = 4
4
-1
x 2 - y 2 = 1
0
y 2 -X 2 =0
*

~~Ra scalar-valued
function, and a e X . If v e R" is any unit vector, then the directional deriva-
tive of / at a in the direction of v, denoted D v /(a), is
/(a + /*v)-/(a)
D v /(a) = hm
h^O h
(provided that this limit exists).
EXAMPLE 1 Suppose f(x, y) = x 2 - 3xy + 2x - 5y. Then, if v = (v, w) e
R 2 is any unit vector, it follows that
f((0,0) + h(v,w))-f(0, 0)
D y f(0, 0) = lim
h^O h
2 v 2 â€” 3h 2 vw + 2hv â€” 5hw
= lim
h^o h
= lim(7ii; 2 â€” 3h vw + 2v â€” 5w)
= 2v â€” 5w.
Thus, the rate of change of / is 2v â€” 5w if we move from the origin in the
direction given by v. The rate of change is zero if v = (5/V29, 2/V29) or
(-5/V29, -2/V59). â™¦
Consequently, we see that the partial derivatives of a function are just the "tip
of the iceberg." However, it turns out that when / is differentiable, the partial
derivatives actually determine the directional derivatives for all directions v. To
see this rather remarkable result, we begin by defining a new function F of a
single variable by
F(0 = /(a-Mv).
Then, by Definition 6. 1, we have
n f( . Y /(a + rv)-/(a) F(t) - F(0)
D v /(a) = hm = hm â€” = F (0).
t^O t r^O t â€” 0
That is,
O v /(a) = ^/(a + *v)| f=0 . (2)
The significance of equation (2) is that, when / is differentiable at a, we can apply
the chain rule to the right-hand side. Indeed, let x(r) = a + fv. Then, by the chain
rule,
^/(a + fv) = Df(x)Dx(t) = D/(x)v.
dt
Evaluation at t = 0 gives
D v /(a) = Z)/(a)v = V/(a).v. (3)
The purpose of equation (3) is to emphasize the geometry of the situation. The
result above says that the directional derivative is just the dot product of the
2.6 | Directional Derivatives and the Gradient
gradient and the direction vector v. Since the gradient is made up of the partial
derivatives, we see that the more general notion of the directional derivative
depends entirely on just the direction vector and the partial derivatives. To be
more formal, we summarize this discussion with a theorem.
THEOREM 6.2 Let X C R" be open and suppose /: X -> R is differentiable
at a e X. Then the directional derivative D v /(a) exists for all directions (unit
vectors) veR" and moreover, we have
D v /(a) = V/(a).v.
EXAMPLE 2 The function f(x, y) = x 2 â€” 3xy + 2x â€” 5y we considered in
Example 1 has continuous partials and hence, by Theorem 3.5, is differentiable.
Thus, Theorem 6.2 applies to tell us that, for any unit vector v = i>i + w\ 6 R 2 ,
D v /(0, 0) = V/(0, 0) â€¢ v = (f x (0, 0)i + /,(0, 0)j) â€¢ (vi + wj)
= (2i-5j)-(ui+ W j)
= 2v â€” 5w,
as seen earlier. â™¦
EXAMPLE 3 The converse of Theorem 6.2 does not hold. That is, a function
may have directional derivatives in all directions at a point yet fail to be differ-
entiable. To see how this can happen, consider the function /: R 2 -> R denned
by
/(*. y) =
if (jc, y) * (0, 0)
x L + y 4
0 if(je,y) = (0,0)
This function is not continuous at the origin. (Why?) So, by Theorem 3.6, it
fails to be differentiable there; however, we claim that all directional derivatives
exist at the origin. To see this, let the direction vector v be vi + wj. Hence, by
Definition 6.1, we observe that
D v /(0, 0) = lim
f((0,0) + Kvi + wj))-f(0, 0)
h
= lim
h^0
hv(hw)
{hvf + (hw) 4
h 2 vw 2
lim
h^o h 2 (v 2 + h 2 w 4 )
,2 â€ž,â€ž2
= lim
vw
h^0 V 2 + h 2 w 4
vw
w
V
Chapter 2 | Differentiation in Several Variables
Thus, the directional derivative exists whenever v 7^ 0. When v = 0 (in which
case v = j), we, again, must calculate
/((0, 0) + fcj)-/(0, 0)
Dj/(0, 0)=lim
h-*o h
= hm
fc->-0
0-0
= hm = 0.
ft->-o h
Consequently, this directional derivative (which is, in fact, df/dy) exists as well.
â™¦
The reason we have restricted the direction vector v to be of unit length in our
discussion of directional derivatives has to do with the meaning of D v /(a), not
with any technicalities pertaining to Definition 6.1 or Theorem 6.2. Indeed, we
can certainly define the limit in Definition 6. 1 for any vector v, not just one of unit
length. So, suppose w is an arbitrary nonzero vector in R" and / is differentiable.
Then the proof of Theorem 6.2 goes through without change to give
hm = V / (a) â€¢ w.
h-*o h
The problem is as follows: If w = kv for some (nonzero) scalar k, then
hm = V / (a) â€¢ w
h-f0 h
= V/(a)-(*v)
= *(V/(a)-v)
n /(a + Av)-/(a)
That is, the "generalized directional derivative" in the direction of kv is k times
the derivative in the direction of v. But v and kv are parallel vectors, and it is
undesirable to have this sort of ambiguity of terminology. So we avoid the trouble
by insisting upon using unit vectors only (i.e., by allowing k to be Â±1 only) when
working with directional derivatives.
Gradients and Steepest Ascent
Suppose you are traveling in space near the planet Nilrebo and that one of your
spaceship's instruments measures the external atmospheric pressure on your ship
as a function f(x, y, z) of position. Assume, quite reasonably, that this function
is differentiable. Then Theorem 6.2 applies and tells us that if you travel from
point a = (a , b, c) in the direction of the (unit) vector u = ui + uj + 10k, the rate
of change of pressure is given by
D u /(a) = V/(a).u.
Now, we ask the following: In what direction is the pressure increasing the most?
If 9 is the angle between u and the gradient vector V/(a), then we have, by
Theorem 3.3 of Â§1.3, that
D n /(a) = ||V/(a)|| ||u|| costf = ||V/(a)|| cos#,
2.6 | Directional Derivatives and the Gradient 163
since u is a unit vector. Because â€” 1 < cos 6 < 1 , we have
-||V/(a)|| < D u /(a)< ||V/(a)||.
Moreover, cos 9 = 1 when 9 = 0 and cos 9 = â€” 1 when 9 = it . Thus, we have
established the following:
THEOREM 6.3 The directional derivative Dâ€žf(a) is maximized, with respect to
direction, when u points in the same direction as V/(a) and is minimized when u
points in the opposite direction. Furthermore, the maximum and minimum values
of D u /(a) are || V/(a)|| and -|| V/(a)||, respectively.
EXAMPLE 4 If the pressure function on Nilrebo is
f(x, y, z) = 5x 2 + 7y 4 + x 2 z 2 arm,
where the origin is located at the center of Nilrebo and distance units are measured
in thousands of kilometers, then the rate of change of pressure at (1,-1,2)
in the direction of i + j + k may be calculated as V/(l, â€” 1, 2) â€¢ u, where u =
(i + j + k)/\/3. (Note that we normalized the vector i + j + k to obtain a unit
vector.) Using Theorem 6.2, we compute
A./(l,-l,2) = V/(l,-l,2).u
i + j + k
= (18i - 28j + 4k) Vâ€”
18-28 + 4
V3
2\/3 atm/Mm.
Additionally, in view of Theorem 6.3, the pressure will increase most rapidly
in the direction of V/(l , â€”1,2), that is, in the
18i-28j+4k 9i-14j + 2k
||18i-28j+4k|| V28T
direction. Moreover, the rate of this increase is
|| V/(l, -1,2)|| = 2^/281 atm/Mm. â™¦
Theorem 6.3 is stated in a manner that is independent of dimension â€” that is, so
that it applies to functions /: X C R" â€”> R for any n > 2. In the case n = 2, there
is another geometric interpretation of Theorem 6.3: Suppose you are mountain
climbing on the surface z = fix, y). Think of the value of / as the height of the
mountain above (or below) sea level. If you are equipped with a map and compass
(which supply information in the xy-plane only), then if you are at the point on
the mountain with xy-coordinates (map coordinates) (a, b), Theorem 6.3 says
that you should move in the direction parallel to the gradient V/(a , b) in order to
climb the mountain most rapidly. (See Figure 2.69.) Similarly, you should move
in the direction parallel to â€” V f(a , b) in order to descend most rapidly. Moreover,
the slope of your ascent or descent in these cases is || V/(a, b)\\ . Be sure that you
understand that V/(a, b) is a vector in R 2 that gives the optimal north-south,
east-west direction of travel.
1 64 Chapter 2 | Differentiation in Several Variables
Tangent Planes Revisited
In Â§2.1, we indicated that not all surfaces can be described by equations of the
form z = f(x, y). Indeed, a surface as simple and familiar as the sphere is not the
graph of any single function of two variables. Yet the sphere is certainly smooth
enough for us to see intuitively that it must have a tangent plane at every point.
(See Figure 2.70.)
How can we find the equation of the tangent plane? In the case of the unit
sphere x 2 + y 2 + z 2 = 1, we could proceed as follows: First decide whether the
point of tangency is in the top or bottom hemisphere. Then apply equation (4) of
Figure 2.70 A sphere and one of Â§2.3 to the graph of z = â€” x 2 â€” y 2 or z = â€”y/\ â€” x 2 â€” y 2 , as appropriate.
its tangent planes. The calculus is tedious but not conceptually difficult. However, the tangent planes
to points on the equator are all vertical and so equation (4) of Â§2.3 does not apply.
(It is possible to modify this approach to accommodate such points, but we will
not do so.) In general, given a surface described by an equation of the form
F(x, v, z) = c (where c is a constant), it may be entirely impractical to solve
for z even as several functions of x and y. Try solving for z in the equation
xyz + ye" â€” x 2 + yz 2 = 0 and you'll see what we mean. We need some other
way to get our hands on tangent planes to surfaces described as level sets of
functions of three variables.
To get started on our quest, we present the following result, interesting in its
own right:
THEOREM 6.4 Let X C R" be open and /: X -> R be a function of class C 1 .
If x 0 is a point on the level set S = {x e X \ f(x) = c}, then the vector V/(x 0 )
is perpendicular to S.
2.6 | Directional Derivatives and the Gradient
PROOF We need to establish the following: If v is any vector tangent to S at xo,
then V/(xo) is perpendicular to v (i.e., V/(xo) â€¢ v = 0). By a tangent vector to S
at xo, we mean that v is the velocity vector of a curve C that lies in S and passes
through Xq. The situation in R 3 is pictured in Figure 2.71.
Figure 2.71 The level set surface
S = {x | /(x) = c}.
Thus, let C be given parametrically by x(t) = (xi(/), X2(t), . . . , xâ€ž(t)), where
a < t < b and x(fo) = xo for some number to in (a, b). (Then, if v is the velocity
vector at x 0 , we must have x'(f 0 ) = v. See Â§3.1 for more about velocity vectors.)
Since C is contained in S, we have
/(x(0) = /(*i(0. *2(0. â€¢ â– â€¢ . *Â»(0) = c -
Hence,
j t imm = jfc\ = o. (4)
On the other hand the chain rule applied to the composite function fox:
(a,b) -> R tells us
^[/(x(f))] = V/(x(0)-x'(0-
Evaluation at to and equation (4) let us conclude that
V/(x(/ 0 ))-x'(*o) = V/(x 0 )-v = 0,
as desired. â–
Here's how we can use the result of Theorem 6.4 to find the plane tan-
gent to the sphere x 2 + y 2 + z 2 = 1 at the point ^â€” 0, ^J. From Â§1.5, we
know that a plane is determined uniquely from two pieces of information: (i) a
point in the plane and (ii) a vector perpendicular to the plane. We are given a
point in the plane in the form of the point of tangency ^â€” -7^, 0, ^=V As for
a vector normal to the plane, Theorem 6.4 tells us that the gradient of the func-
tion f(x, y, z) = x 2 + y 2 + z 2 that defines the sphere as a level set will do. We
have
V f(x, y, z) = 2xi + 2y\ + 2zk,
so that
v/ (-7!'Â°'7!) = ^' +V5k -
Chapter 2 | Differentiation in Several Variables
Hence, the equation of the tangent plane is
V2f, + -L) + ^--L)=0,
= 0,
or
X = y/2.
In general, if S is a surface in R 3 defined by an equation of the form
f(x,y,z) = c,
then if xo Â£ X, the gradient vector V/(xo) is perpendicular to S and, conse-
quently, if nonzero, is a vector normal to the plane tangent to S at x 0 . Thus,
the equation
V/(x 0 )-(x-x 0 ) = 0
or, equivalently,
f x (x 0 , y 0 , z 0 )(x - x 0 ) + f y (x 0 , y 0 , zo)(y - v 0 )
+ fz(*o, yo, zo)(z - zo) = 0
is an equation for the tangent plane to S at x 0 .
(5)
(6)
Note that formula (5) can be used in R" as well as in R 3 , in which case it
defines the tangent hyperplane to the hypersurface S C R" defined by f{x\ ,
X2, â– â– â– , x n ) = c at the point xo e S.
EXAMPLE 5 Considerthe surface S defined by the equation* 3 y â€” yz 2 + z 5 =
9. We calculate the plane tangent to S at the point (3,-1,2).
To do this, we define f(x, y, z) = x 3 y â€” yz 2 + z 5 - Then
V/(3, -1,2)= (3x 2 yi + (x 3 - z 2 )\ + (5 Z 4 - 2yz)k)\ (3 _ h2)
= -21\ + 23j + 84k
is normal to S at (3, â€”1, 2) by Theorem 6.4. Using formula (6), we see that the
tangent plane has equation
or, equivalently,
-270 - 3) + 23(y + 1) + 84(z - 2) = 0
-27x + 23y + 84z = 64.
EXAMPLE 6 Consider the surface defined by z 4 = x 2 + y 2 . This surface is
the level set (at height 0) of the function
f{x, y, Z ) = x 2 + y 2 - z 4 .
The gradient of / is
Vf(x, y,z) = 2xi + 2yj- 4z 3 k.
2.6 | Directional Derivatives and the Gradient
Figure 2.72 The
surface of Example 6.
Note that the point (0, 0, 0) lies on the surface. However, V/(0, 0, 0) = 0, which
makes the gradient vector unusable as a normal vector to a tangent plane. Thus,
formula (6) doesn't apply. What we conclude from this example is that the surface
fails to have a tangent plane at the origin, a fact that is easy to believe from the
graph. (See Figure 2.72.) â™¦
EXAMPLE 7 The equation x 2 + y 2 + z 2 + w 2 = 4 defines a hypersphere of
radius 2 in R 4 . We use formula (5) to determine the hyperplane tangent to the
hypersphere at (â€” 1 , 1 , 1,-1).
The hypersphere may be considered to be the level set at height 4 of the
function f(x, y, z, w) = x 2 + y 2 + z 2 + w 2 , so that the gradient vector is
V f(x, y, z, w) = (2x, 2y, 2z, 2w),
so that
V/(-l, 1,1,-1) = (-2, 2, 2, -2).
Using formula (5), we obtain an equation for the tangent hyperplane as
(-2,2,2, -2)-(x + l,y- 1, z - 1, w + 1) = 0
or
-2(x + 1) + 2(y - 1) + 2(z - 1) - 2(w + 1) = 0.
Equivalently, we have the equation
x - y - z + w + 4 = 0. â™¦
EXAMPLE 8 We determine the plane tangent to the paraboloid z
- v2
3y
at the point (â€”2, 1, 7) in two ways: (i) by using formula (4) in Â§2.3, and (ii) by
using our new formula (6).
First, the equation z = x 2 + 3y 2 explicitly describes the paraboloid as the
graph of the function f(x, y) = x 2 + 3y 2 , that is, by an equation of the form
z = fix, y). Therefore, formula (4) of Â§2.3 applies to tell us that the tangent
plane at (â€”2, 1, 7) has equation
z = f(-2, 1) + f x (-2, l)(x + 2) + f y (-2, l)(y - 1)
or, equivalently,
z = 7 - 4(jc + 2) + 6(y - 1). (7)
Second, if we write the equation of the paraboloid as x 2 + 3y 2 â€” z = 0,
then we see that it describes the paraboloid as the level set of height 0 of the
three-variable function F(x, y, z) = x 2 + 3y 2 â€” z. Hence, formula (6) applies
and indicates that an equation for the tangent plane at (â€”2, 1, 7) is
F x (-2, 1, 7)(jc + 2) + F y {-2, 1, 7)(y - 1) + F z (-2, 1, 7)(z - 7) = 0
or
-4(x + 2) + 6(y- 1) - l(z - 7)
As can be seen, equation (7) agrees with equation (8).
(8)
â™¦
Example 8 may be viewed in a more general context. If S is the surface in R 3
given by the equation z = f(x, y) (where / is differentiate), then formula (4) of
Â§ 2 . 3 tells us that an equation for the plane tangent to S at the point (a , b , / {a , b)) is
z = f(a, b) + f x (a, b)(x -a)+ Ua, b)(y - b).
Chapter 2 | Differentiation in Several Variables
(-2,2,6)
Figure 2.73 The two-sheeted
hyperboloid z 2 /4 â€” x 2 â€” y 2 = 1.
The point (â€”2, 2, 6) lies on the
sheet given by z = 2^/x 2 + y 2 + 1,
and the point (1,1, â€” 2^/3) lies on
the sheet given by
z = -2jx 2 + y 2 +l.
1.
At the same time, the equation for S may be written as
f(x, y) - z = 0.
Then, if we let F(x, y, z) = f(x, y) â€” z, we see that S is the level set of F at
height 0. Hence, formula (6) tells us that the tangent plane at (a, b, f(a, b)) is
F x (a, b, f(a, b))(x - a) + F y (a, b, f(a, b))(y - b)
+ F e (a, b, f(a, b))(z - f(a, b)) = 0.
By construction of F,
dF _df dF _ df dF
dx dx ' dy dy ' dz
Thus, the tangent plane formula becomes
f x (a, b)(x -a) + f y (a, b)(y - b) - (z - f(a, b)) = 0.
The last equation for the tangent plane is the same as the one given above by
equation (4) of Â§2.3.
The result shows that equations (5) and (6) extend the formula (4) of Â§2.3 to
the more general setting of level sets.
The Implicit Function and Inverse Function
Theorems (optional)
We have previously noted that not all surfaces that are described by equations of
the form F(x ,y,z) = c can be described by an equation of the form z = f(x , y).
We close this section with a brief â€” but theoretically important â€” digression about
when and how the level set {(x, y, z) | F(x, y, z) = c} can also be described as
the graph of a function of two variables, that is, as the graph of z = f(x, y).
We also consider the more general question of when we can solve a system of
equations for some of the variables in terms of the others.
We begin with an example.
EXAMPLE 9 Consider the hyperboloid z 2 /4 - x 2
described as the level set (at height 1) of the function
y = 1, which may be
F{x, y, z) =
y
(See Figure 2.73.) This surface cannot be described as the graph of an equation
of the form z = f(x, y), since particular values for x and y give rise to two values
for z. Indeed, when we solve for z in terms of x and y, we find that there are two
functional solutions:
x 2 + y 2 + 1 and z
= -2/
x 2 + y 2 + 1.
(9)
On the other hand these two solutions show that, given any particular point
(*o, yo, zo) of the hyperboloid, we may solve locally for z in terms of x and y.
That is, we may identify on which sheet of the hyperboloid the point {xq, yo, Zo)
lies and then use the appropriate expression in (9) to describe that sheet. â™¦
Example 9 prompts us to pose the following question: Given a surface S,
described as the level set {(x, y, z) | F(x, y, z) = c], can we always determine at
least a portion of S as the graph of a function z = f(x , y)? The result that follows,
2.6 | Directional Derivatives and the Gradient
a special case of what is known as the implicit function theorem, provides
relatively mild hypotheses under which we can.
THEOREM 6.5 (The implicit function theorem) Let F:XcR"->R be
of class C 1 and let a be a point of the level set S = {x 6 R" | F(x) = c}. If
F Xn (a) 0, then there is a neighborhood U of (a\, ci2, . . . , aâ€ž_i) in R n_1 , a
neighborhood V of a n in R, and a function f:U<^ R" _1 -> V of class C 1
such that if (xi, X2, â– â– â– , x n -\) e U andxâ€ž e V satisfy F(x\, X2, â– â€¢ â– , x n ) = c (i.e.,
(x\,x 2 , xâ€ž) e S), thenxâ€ž = f(xi,x 2 , xâ€ž_i).
The significance of Theorem 6.5 is that it tells us that near a point a â‚¬ S
such that dF/dx n ^ 0, the level set S given by the equation F{x\, . . . , x n ) = c
is locally also the graph of a function xâ€ž = /(xj , . . . , xâ€ž_i). In other words, we
may solve locally for x n in terms of x\, . . . , xâ€ž_i, so that S is, at least locally, a
differentiable hypersurface in R" .
EXAMPLE 10 Returning to Example 9, we recall that the hyperboloid is the
level set (at height 1) of the function F(x, y, z) = z 2 /4 â€” x 2 â€” y 2 . We have
dF
9z"
Note that for any point (xo, yo> Zo) in the hyperboloid, we have |zol > 2. Hence,
9F.(xo, yo, zo) 0. Thus, Theorem 6.5 implies that we may describe a portion
of the hyperboloid near any point as the graph of a function of two variables. This
is consistent with what we observed in Example 9. â™¦
Of course, there is nothing special about solving for the particular vari-
able xâ€ž in terms of Suppose a is a point on the level set S de-
termined by the equation F(x) = c and suppose VF(a) / 0. Then F Xj (a) 0 for
some i . Hence, we can solve locally near a for x,- as a differentiable function of
x\, . . . , x,_i, Xi+i, . . . , xâ€ž. Therefore, S is locally a differentiable hypersurface
inR".
EXAMPLE 11 Let S denote the ellipsoid x 2 /4 + y 2 /36 + z 2 /9
is the level set (at height 1) of the function
1. Then S
n*,y,z)=^ + y - +
At the point (V2, V6, V3), we have
dF
~dz
2z
(V2.V6.V3)
2^3
(V2,V6,V3)
Thus, S may be realized near (V2, V6, V3) as the graph of an equation of the
form z = f(x, y), namely, z = 3^/1 - x 2 /4 â€” y 2 /36. At the point (0, â€”6, 0),
however, we see that dF /dz vanishes. On the other hand,
dF
97
(0,-6,0)
2 Â±
36
(0,-6,0)
^0.
Consequently, near (0, â€”6, 0), the ellipsoid may be described by solving for y as
a function of x and z, namely, y = â€”6^/ 1 â€” x 2 /4 â€” z 2 /9. â™¦
Chapter 2 | Differentiation in Several Variables
EXAMPLE 1 2 Consider the set of points S defined by the equation x 2 z 2 â€” y =
0. Then S is the level set at height 0 of the function F(x, y, z) = x 2 z 2 â€” y. Note
that
VF(x,y,z) = (2xz 2 , -l,2x 2 z).
Since dF /dy never vanishes, we see that we can always solve for y as a function
of x and z. (This is, of course, obvious from the equation.) On the other hand, near
points where x and z are nonzero, both dF/dx and dF/dz are nonzero. Hence,
we can solve for either x or z in this case. For example, near (1 , 1 , â€” 1), we have
x= H and z= -Ii-
As just mentioned, Theorem 6.5 is actually a special case of a more general
result. In Theorem 6.5 we are attempting to solve the equation
F(xi, x 2 , . . . , x n ) = c
for x n in terms of x\, . . . , x n -\. In the general case, we have a system of m
equations
F\(x\, . . . , xâ€ž, yi, . . . , y m ) = c 1
F 2 (xi,...,xâ€ž,yi,...,y m ) = c 2
^mv^l ' â€¢ â– â– ' 3^1) â– â– â– ' y?n)
and we desire to solve the system for yi , . . . , y m in terms of x\ , . . . , xâ€ž. Using vec-
tor notation, we can also write this system as F(x, y) = c, where x = (xi, . . . , x n ),
y = (yi, . . . , y m ), c = (ci , . . . , c m ), and Fi, ... , F m make up the component
functions of F. With this notation, the general result is the following:
THEOREM 6.6 (The implicit function theorem, general case) Suppose
F: A ->â€¢ R m is of class C 1 , where A is open in R" +m . Let (a, b) = (a\, . . . , aâ€ž,
b\, . . . , b m ) Â£ A satisfy F(a, b) = c. If the determinant
r 3Fi
dy
A(a, b) = det
(a,b)
9F,
dy,
-(a, b)
dF m
L dy
-(a, b)
dF m
dy r .
(a, b)
^0,
then there is a neighborhood U of a in R" and a unique function f : U â€”> R m of
class C 1 such that f(a) = b and F(x, f(x)) = c for all x e U. In other words, we
can solve locally for y as a function f(x).
EXAMPLE 13 We show that, near the point (x\, x%, x^, y\, y 2 ) = (â€” 1, 1, 1,
2, 1), we can solve the system
x\y 2 + x 2 y\ = 1
xfx 3 yi + x 2 y{ = 3
for y\ and y 2 in terms of x\ , x 2 , x%.
2.6 | Directional Derivatives and the Gradient
We apply the general implicit function theorem (Theorem 6.6) to the system
IFi(xi, x 2 , x 3 , y\,yi) = x\y 2 + x 2 yi = 1
F 2 (xi,x 2 , x 3 , y\,y 2 ) = xfx 3 yi + x 2 y\ = 3
The relevant determinant is
A(-l, 1, 1,2, 1) = det
= det
= det
3Fi dFi
dyi dy 2
dF 2 dF 2
dyi dy 2
x 2 X\
(X1,.T 2 ,X3,3'1,}' 2 )=(- 1,1, 1,2,1)
xfxj, 3x 2 yj
(xi ,a- 2 , x 3 , vi , y 2 )={- 1 , 1 , 1 ,2, 1)
4/0.
Hence, we may solve locally, at least in principle.
We can also use the equations in (11) to determine, for example,
dy 2
(â€” 1, 1, 1), where we treat x\, x 2 , x 3 as independent variables and y\ and
dx\
y 2 as functions of them.
Differentiating the equations in (1 1) implicitly with respect to X\ and using
the chain rule, we obtain
yi + x x - â€” Vx 2 â€”
ax\ ax\
0
Ixix^yi +x\xt,- r-3x 2 y;
dy 2
= 0
.2â€ž
Now, let (xi , x 2 , Xi, y\ , y 2 ) â€” (â€” 1, 1 , 1,2, 1), so that the system becomes
f*(-l,l,l)-|^(-l,l,l) = -l
3xi 6x\
^l(-l,l,l) + 3^(-l,l,l) = 4
3xi dxi
3y2 5
We may easily solve this last system to find that â€” (â€”1,1,1) = â€”. â™¦
3xi 4
Now, suppose we have a system of n equations that defines the variables
yi , . . . , y n in terms of the variables xi, . . . , xâ€ž, that is,
yi = /i(xi, . . . ,xâ€ž)
yi = fi(xu â– . . ,x n )
yâ€ž = fn(X\,...,X n )
(12)
Note that the system given in (12) can be written in vector form as y = f(x). The
question we ask is, when can we invert this system? In other words, when can we
Chapter 2 | Differentiation in Several Variables
solve for x\ , . . . ,x n in terms of yi ,
function g so that x = g(y)?
The solution is to apply Theorem 6.6 to the system
y n , or, equivalently, when can we find a
F 2 (xi,
,x n ,y\,
xâ€ž,y { ,
.x n ,y\,
yâ€ž) = 0
yâ€ž) = o
Jn) = 0
where Ff(jti, . . . , x n , y\, . . . , yâ€ž) = fi(xi, . . . , x n ) â€” y,-. (In vector form, we are
setting F(x, y) = f(x) â€” y.) Then solvability for x in terms of y near x = a, y = b
is governed by the nonvanishing of the determinant
detDf(a) = det
This determinant is also denoted by
3(/i-
3xi
dxt
(a)
(a)
dx n
dfn
dx n
(a)
(a)
d(x u . . . ,xâ€ž)
and is called the Jacobian of f = (/i, . . . , /â€ž). A more precise and complete
statement of what we are observing is the following:
THEOREM 6.7 (The inverse function theorem) Suppose f = (f\ /â€ž)
is of class C 1 on an open set A c R" . If
then there is an open set U c R" containing a such that f is one-one on U, the
set V = f(U) is also open, and there is a uniquely determined inverse function
g: V -> U to f, which is also of class C 1 . In other words, the system of equations
y = f(x) may be solved uniquely as x = g(y) for x near a and y near b.
det Df(a)
3(/i,..../Â»)
3(xi, . . . , x n )
EXAMPLE 14 Consider the equations that relate polar and Cartesian coordi-
nates:
' x = r cos#
y = r sin 6
These equations define x and y as functions of r and 6. We use Theorem 6.7 to
see near which points of the plane we can invert these equations, that is, solve for
r and 9 in terms of x and y.
To use Theorem 6.7, we compute the Jacobian
d(x, y)
d(r, 9)
cos 6 â€”rsmO
sin 0 r cos 9
r.
Thus, we see that, away from the origin (r = 0), we can solve (locally) for r and 9
uniquely in terms of x and y. At the origin, however, the inverse function theorem
2.6 I Exercises 173
does not apply. Geometrically, this makes perfect sense, since at the origin the
polar angle 9 can have any value. â™¦
2.6 Exercises
1 . Suppose f(x , y , z) is a differentiable function of three
variables.
(a) Explain what the quantity V/(x, y, z) â€¢ (â€” k) rep-
resents.
(b) How does V/(x, y, z) â€¢ (-k) relate to 3//3z?
In Exercises 2â€”8, calculate the directional derivative of the
given function f at the point a in the direction parallel to the
vector u.
2. f(x, y) = e v sinx, a = o), u
31- j
10
3. f(x, y) = x 2 - 2x 3 y + 2y\ a = (2, -1), u =
4. f{x, y) =
1
-,a = (3, -2),u = i-j
(x 2 + y 2 )'
5. /(x, y) = e x - x 2 y, a = (1, 2), u = 2i + j
6. f(x, y, z) = xyz, a = (-1, 0, 2), u
2k- i
~7f
7. /(x, y, z) = e~
8- f(x,y,z)
3k
9. For the function
/(*. y)
3z 2 + l
+z \a = (l,2,3),u = i + j + k
, a = (2, -1,0), u = i-2j +
x\y\
0
if (x, y) ? (0, 0)
if(x,y) = (0, 0)
(a) calculate f x (0, 0) and f y (0, 0). (You will need to
use the definition of the partial derivative.)
(b) use Definition 6.1 to determine for which unit
vectors v = ui + urj the directional derivative
D v /(0, 0) exists.
(c) use a computer to graph the surface z = f(x, y).
1 0. For the function
XV
Jx 2 + .
if(x,y)/(0, 0)
if(x,y) = (0, 0)
(a) calculate / v (0, 0) and /,(0, 0).
(b) use Definition 6.1 to determine for which unit
vectors v = vi + w\ the directional derivative
D v /(0, 0) exists.
(c) use a computer to graph the surface z = f(x, y).
1 1 . The surface of Lake Erehwon can be represented by a
region D in the jcv-plane such that the lake's depth (in
meters) at the point (x, y) is given by the expression
400 â€” 3x 2 y 2 . If your calculus instructor is in the wa-
ter at the point (1, â€”2), in which direction should she
swim
(a) so that the depth increases most rapidly (i.e., so
that she is most likely to drown)?
(b) so that the depth remains constant?
12. A ladybug (who is very sensitive to temperature) is
crawling on graph paper. She is at the point (3,7) and
notices that if she moves in the i-direction, the tem-
perature increases at a rate of 3 deg/cm. If she moves
in the j -direction, she finds that her temperature de-
creases at a rate of 2 deg/cm. In what direction should
the ladybug move if
(a) she wants to warm up most rapidly?
(b) she wants to cool off most rapidly?
(c) she desires her temperature not to change?
13. You are atop Mt. Gradient, 5000 ft above sea level,
equipped with the topographic map shown in Fig-
ure 2.74. A storm suddenly begins to blow, necessitat-
ing your immediate return home. If you begin heading
due east from the top of the mountain, sketch the path
that will take you down to sea level most rapidly.
14. It is raining and rainwater is running off an ellipsoidal
dome with equation Ax 2 + y 2 + 4z 2 = 16, where
z > 0. Given that gravity will cause the raindrops to
slide down the dome as rapidly as possible, describe
the curves whose paths the raindrops must follow.
(Hint: You will need to solve a simple differential
equation.)
15. Igor, the inchworm, is crawling along graph paper in
a magnetic field. The intensity of the field at the point
(x, y) is given by M(x, y) = 3x 2 + y 2 + 5000. If Igor
is at the point (8, 6), describe the curve along which he
should travel if he wishes to reduce the field intensity
as rapidly as possible.
In Exercises 16â€” 19, find an equation for the tangent plane to
the surface given by the equation at the indicated point (xo , yo ,
zo)-
1 74 Chapter 2 | Differentiation in Several Variables
Figure 2.74 The topographic map of Mt. Gradient in Exercise 13.
16. x 1 + y 3 + ; 3 = 7, (xq, y 0 , zo) = (0, -1, 2)
17. ze y cosx = 1, (x 0 , yo, zo) = (x, 0, -1)
18. 2xz + yz- x 2 y + 10 = 0, Oo, yo, zo) = (1, -5, 5)
19. 2xy 2 = 2z 2 - xyz, (x 0 , y 0 , Zo) = (2, -3, 3)
20. Calculate the plane tangent to the surface whose equa-
tion is x 2 â€” 2y 2 + 5xz = 7 at the point (â€”1, 0, â€” |)in
two ways:
(a) by solving for z in terms of x and y and using
formula (4) in Â§2.3
(b) by using formula (6) in this section.
21. Calculate the plane tangent to the surface xsiny +
xz 2 = 2e yz at the point (2, j, 0) in two ways:
(a) by solving for x in terms of y and z and using a
variant of formula (4) in Â§2.3
(b) by using formula (6) in this section.
22. Find the point on the surface x 3 â€” 2y 2 + z 2 = 21
where the tangent plane is perpendicular to the line
given parametrically as x = 3f â€” 5, v = 2t + 7, z =
1 - V2f.
23. Find the points on the hyperboloid 9x 2 â€” 45y 2 +
5z 2 = 45 where the tangent plane is parallel to the
plane x + 5y â€” 2z = 7.
24. Show that the surfaces z = lx 2 â€” \2x â€” 5y 2 and
xyz
2 intersect orthogonally at the point (2, 1 , â€” 1).
25. Suppose that two surfaces are given by the equations
Moreover, suppose that these surfaces intersect at the
point (jco, yo, Zo)- Show that the surfaces are tangent at
(x 0 , yo, Zo) if and only if
VF(x 0 , yo, zo) x VG(i 0 , y 0 , zo) = 0.
26. Let S denote the cone x 2 + Ay 2 = z 2 â–
(a) Find an equation for the plane tangent to S at the
point (3, â€”2, â€”5).
(b) What happens if you try to find an equation for a
tangent plane to S at the origin? Discuss how your
findings relate to the appearance of S.
27. Consider the surface S defined by the equation x 3 â€”
x 2 y 2 + Z 2 = 0.
(a) Find an equation for the plane tangent to S at the
point (2, -3/2, 1).
(b) Does S have a tangent plane at the origin? Why or
why not?
If a curve is given by an equation of the form fix, y) = 0, then
the tangent line to the curve at a given point (xo, yo) Â° n it niay
be found in two ways: (a) by using the technique of implicit
differentiation from single-variable calculus and (b) by using
a formula analogous to formula (6). In Exercises 28-30, use
both of these methods to find the lines tangent to the given
curves at the indicated points.
28. x 2 + y 2 = 4, (xo, y 0 ) = (-V2, V2)
29. y 3 =x 2 +x\(x 0 ,yo) = (1,^/2)
F(x, y, z) = c and G(x, y, z) = k.
30. x 5 + 2xy + y 3 = 16, (x 0 , y 0 ) = (2, -2)
2.6 I Exercises 175
Let C be a curve in R 2 given by an equation of the form
f(x, y) = 0. The normal line to C at a point (xo, yo) on it
is the line that passes through (xo, yo) and is perpendicular
to C (meaning that it is perpendicular to the tangent line to
C at (xo, yo)). In Exercises 31-33, find the normal lines to
the given curves at the indicated points. Give both a set of
parametric equations for the lines and an equation in the form
Ax + By = C. (Hint: Use gradients.)
31. x 2 -y 2 = 9,(x 0 ,y 0 ) = (5, -4)
32. x 2 - x 3 = y 2 , (x 0 , yo) = (-1, 72)
33. x 3 - 2xy + y 5 = 11, (x 0 , y 0 ) = (2, -1)
34. This problem concerns the surface defined by the equa-
tion
x 3 z + x 2 y 2 + sin(yz) = -3.
(a) Find an equation for the plane tangent to this sur-
face at the point (â€”1, 0, 3).
(b) The normal line to a surface S in R 3 at a point
(xo, yo, Zo) on it is the line that passes through
(xo, yo, Zo) and is perpendicular to S. Find a set
of parametric equations for the line normal to the
surface given above at the point (â€”1,0, 3).
35. Give a set of parametric equations for the normal line to
the surface defined by the equation e xy + e xz â€” 2e yz =
0 at the point (-1,-1,-1). (See Exercise 34.)
36. Give a general formula for parametric equations for
the normal line to a surface given by the equation
F(x, y, z) = 0 at the point (xo, yo, zo) on the surface.
(See Exercise 34.)
37. Generalizing upon the techniques of this section,
find an equation for the hyperplane tangent to
the hypersurface sinxi + cosx2 + sinx3 + cos X4 +
sinxs = â€” 1 at the point (jr, jt, 3jt/2, 2jt, 2jt) e R 5 .
38. Find an equation for the hyperplane tangent to the
(Â« â€” l)-dimensional ellipsoid
2, 0 2,0 2 , , 2 Â»(" + 1)
X[ + 2x 2 + 3x 3 H h nx n =
at the point (-1, -1,..., -1) e R".
39. Find an equation for the tangent hyperplane to the (n â€”
l)-dimensional sphere x\ + x\ + â– â– â– + x 2 = 1 in R"
at the point (1 j+Jn, \/-Jn, . . . , l/Vn, â€” l/*/n).
Exercises 40â€”49 concern the implicit function theorems and
the inverse function theorem (Theorems 6.5, 6.6, and 6. 7).
40. Let S be described by z 2 y 3 + x 2 y = 2.
(a) Use the implicit function theorem to determine
near which points S can be described locally as
the graph of a C 1 function z = /(x, y).
(b) Near which points can S be described (locally) as
the graph of a function x = g(y, z)?
(c) Near which points can S be described (locally) as
the graph of a function y = h(x, z)?
41 . Let S be the set of points described by the equation
sinxy + e xz + x 3 y = 1.
(a) Near which points can we describe S as the graph
of a C 1 function z = f(x, y)? What is f{x, y) in
this case?
(b) Describe the set of "bad" points of S, that is, the
points (xo, yo, zo) â‚¬ S where we cannot describe
S as the graph of a function z = f(x, y).
(c) Use a computer to help give a complete picture of
S.
42. Let F(x, y) = c define a curve C in R 2 . Suppose
(xo, yo) is a point ofC such that VF(xo, yo) / 0. Show
that the curve can be represented near (xo, yo) as either
the graph of a function y = f(x) or the graph of a
function x = g(y).
43. Let F(x, y) = x 2 â€” y 3 , and consider the curve C de-
fined by the equation F(x, y) = 0.
(a) Show that (0, 0) lies on C and that F y (0, 0) = 0.
(b) Can we describe C as the graph of a function
y = /(x)? Graph C.
(c) Comment on the results of parts (a) and (b) in light
of the implicit function theorem (Theorem 6.5).
44. (a) Consider the family of level sets of the function
F(x, y) = xy + 1. Use the implicit function theo-
rem to identify which level sets of this family are
actually unions of smooth curves in R 2 (i.e., locally
graphs of C 1 functions of a single variable).
(b) Now consider the family of level sets of
F(x, y, z) = xyz + 1. Which level sets of this
family are unions of smooth surfaces in R 3 ?
45. Suppose that F(u, v) is of class C 1 and is such that
F(-2, 1) = 0 and Fâ€ž(-2, 1) = 7, Fâ€ž(-2, 1) = 5. Let
G(x, y, z) = F(x 3 - 2y 2 + z 5 , xy - x 2 z + 3).
(a) Check that G(-l, 1, 1) = 0.
(b) Show that we can solve the equation G(x, y, z) =
0 for z in terms of x and y (i.e., as z = g(x, y), for
(x,y)near(-l, 1) so that g(-l, 1)= 1).
46. Can you solve
X2j2 â€” x \ cos yi = 5
X2 sin yi + x\yi = 2
for yi, y2 as functions of x\, xi near the point
(x\, X2, yu y2) = (2, 3, n, 1)? What about near the
point (xi , X2, yi , ya) = (0, 2, n/2, 5/2)?
Chapter 2 | Differentiation in Several Variables
x\y\
47. Consider the system
2x 2 y 3 = 1
xiy{ + x 2 y 2 - 4y 2 yi = -9 .
x 2 y\ + 3xiyj = 12
(a) Show that, near the point (x\, x%, y\, y%, y{) =
(1, 0, â€”1, 1, 2), it is possible to solve for yi, yj,
ys in terms of x\, x%.
(b) From the result of part (a), we may consider y\,y%,
j3 to be functions of.ti and.*^. Use implicit differ-
dy\
entiation and the chain rule to evaluate (1, 0),
dx\
-i(l,0),and-^(l,0).
9xi dx\
48. Consider the equations that relate cylindrical and
Cartesian coordinates in R 3 :
x = r cos 8
y = r sin 8 .
z = z
(a) Near which points of R 3 can we solve for r, 8, and
z in terms of the Cartesian coordinates?
(b) Explain the geometry behind your answer in
part (a).
49. Recall that the equations relating spherical and Carte-
sian coordinates in R 3 are
' x = p sin ~~

~~ R is a differ-
entiable function. Here's a reminder of how the method works.
We wish to find a number r such that f(r ) = 0. To approximate r, we make
an initial guess xq for r and, in general, we expect to find that f(xo) ^ 0. So next
we look at the tangent line to the graph of / at (x 0 , f(x 0 )). (See Figure 2.75.)
Since the tangent line approximates the graph of / near (xo, f(xo)), we can
find where the tangent line crosses the x-axis. The crossing point (jci, 0) will
generally be closer to (r, 0) than (xq, 0) is, so we take x\ as a revised and improved
approximation to the root r of f{x) = 0.
To find x\ , we begin with the equation of the tangent line
y = f(*o) + /'C*o)(* - *o),
then set y = 0 to find where this line crosses the x -axis. Thus, we solve the equation
/(*o) + f'(x 0 )(xi - x 0 ) = 0
for x\ to find that
f(xo)
X] â€” Xr> .
f'(x 0 )
Once we have x\, we can start the process again using x\ in place of x 0 and
produce what we hope will be an even better approximation x^ via the formula
fix,)
Xi = X\ .
f'(xi)
Indeed, we may iterate this process and define Xk recursively by
f{Xk-\)
xt = Xk-i k = 1,2,...
and thereby produce a sequence of numbers xq, x\, . . . , Xk,
(1)
2.7 | Newton's Method (optional) 177
It is not always the case that the sequence {xk\ converges. However, when
it does, it must converge to a root of the equation f(x) = 0. To see this, let
L = lim^oo Xk. Then we also have lim^oo jc&_i = L. Taking limits in formula
(1), we find
L=L-m,
/'(!)'
which immediately implies that f(L) = 0. Hence, L is a root of the equation.
Now that we have some understanding of derivatives in the multivariable
case, we turn to the generalization of Newton's method for solving systems of n
equations in n unknowns. We may write such a system as
fi(xi, ...,xÂ») = 0
fi(x\, xâ€ž) = 0
. â– (2)
fâ€ž(x\, ...,*â€ž) = 0
We consider the map f: X c R" -> R" defined as f(x) = (/i(x), . . . , / n (x)) (i.e.,
f is the map whose component functions come from the equations in (2). The
domain X of f may be taken to be the set where all the component functions are
denned.) Then to solve system (2) means to find a vector r = (rj , . . . , râ€ž) such
that f(r) = 0. To approximate such a vector r, we may, as in the single- variable
case, make an initial guess xo for what r might be. If f is differentiable, then we
know that y = f(x) is approximated by the equation
y = f(x<,) + Of(x 0 )(x - xo).
(Here we think of f(xo) and the vectors x and xo as n x 1 matrices.) Then we set
y equal to 0 to find where this approximating function is zero. Thus, we solve the
matrix equation
f(xâ€ž) + Â£>f(x 0 )(x 1 - x 0 ) = 0 (3)
for xi to give a revised approximation to the root r. Evidently (3) is equivalent to
Df(xo)( Xl - x 0 ) = -f(x 0 ). (4)
To continue our argument, suppose that Df(xo) is an invertible n x n matrix,
meaning that there is a second n x n matrix [Df(xo)] _1 with the property that
[Df^r'Dftxo) = J Df(x 0 )[Df(x 0 )]" 1 = /â€ž, the n x n identity matrix. (See Ex-
ercises 20 and 30-38 in Â§1.6.) Then we may multiply equation (4) on the left by
[Df(x 0 )] _1 to obtain
7â€ž(X! -x 0 )=-[Df(x 0 )r 1 f(x 0 ).
Since Iâ€žA = A for any n x k matrix A, this last equation implies that
Xl =x 0 -[Â£>f(x 0 )r 1 f(x 0 ). (5)
As we did in the one -variable case of Newton's method we may iterate formula
(5) to define recursively a sequence {x^} of vectors by
x t = xn-[Df( X n)r'%-i) (6)
Chapter 2 | Differentiation in Several Variables
-3 +
Figure 2.76 Finding
the intersection
points of the circle
4 and
the hyperbola
Ax 2 - y 2 = A in
Example 1.
Note the similarity between formulas (1) and (6). Moreover, just as in the case
of formula (1), although the sequence {xo, Xi, . . . , x k , . . .} may not converge, if
it does, it must converge to a root of f(x) = 0. (See Exercise 4.)
EXAMPLE 1 Consider the problem of finding the intersection points of the cir-
cle x 2 + y 2 = 4 and the hyperbola 4x 2 â€” y 2 = 4. (See Figure 2.76.) Analytically,
we seek simultaneous solutions to the two equations
x 2 + y 2 = 4 and Ax 2 - y 2 = 4,
or, equivalently, solutions to the system
x 2 + y 2 - 4 = 0
Ax 2 - y 2
0
(7)
To use Newton's method, we define a function f : R -> R by f(x , y) = (x +
y
A Ax 1
A) and try to approximate solutions to the vector equation
f(x, y) = (0, 0). We may begin with any initial guess, say,
x 0
Xo
T
l
and then produce successive approximations Xi, x 2 ,
mula (6). In particular, we have
to a solution using for-
NotethatdetDf(x, y)
2x
8.v
2y
-2y
Df(x,y) =
-20xy . You may verify (see Exercise 36 in Â§ 1 .6) that
[Df(x,y)T l =
1
-20xy
-2y -2y
-8x 2x
1
2
57
1
1
Thus,
~x k ~
~Xk-\
yk-\_
~x k -\
yk-i
[Df(x k _ uyk-i )] f(*k- 1 . yk-i)
l l
2
L 5y k _
1
10%-i J
4-i + yLi-4
^U-yU-A
J x k -\
L I0y k -i J
Xk-l
yk-\ -
JX k -\
lOxjt i
10 W -i J
2.7 | Newton's Method (optional) 179
Beginning with xo = yo = 1 , we have
5 â– l 2 â€” 8 5 â€¢ l 2 â€” 12
x x = \ = 1.3 yi = 1 = 1.7
10-1 10-1
5(1.3) 2 -8 5(1.7) 2 - 12
x 2 = 1.3 = 1.265385 y 2 = 1.7
10(1.3) y 10(1.7)
= 1.555882, etc.
It is also easy to hand off the details of the computation to a calculator or a
computer. One finds the following results:
k
x k
y*
0
1
l
1
1.3
1.7
2
1.26538462
1.55588235
3
1.26491115
1.54920772
4
1.26491106
1.54919334
5
1.26491106
1.54919334
Thus, it appears that, to eight decimal places, an intersection point of the curves
is (1.26491106, 1.54919334).
In this particular example, it is not difficult to find the solutions to (7) exactly.
We add the two equations in (7) to obtain
5x z - 8 = 0
x 2 =
Thus, x = Â±V8/5. If we substitute these values for x into the first equation of
(7), we obtain
I + r ~ 4 = 0 y
Hence, y = Â±y/l2/5. Therefore, the four intersection points are
11
5 â–
Since 78/5 Â« 1.264911064 and ^/T2/5 Â« 1.54919334, we see that Newton's
method provided us with an accurate approximate solution very quickly. â™¦
EXAMPLE 2 We use Newton's method to find solutions to the system
x 3 - 5x 2 + 2x - y + 13 = 0
x 3 +x 2 - 14x - y - 19 = 0
(8)
As in the previous example, we define f : R 2
y + 13, x 3 + x 2 - 14x - y - 19). Then
R 2 byf(x, y) = (x 3 - 5x 2 + 2x
Df(x,y) =
3x 2 - lOx + 2
3x 2 + 2x - 14
1 80 Chapter 2 | Differentiation in Several Variables
so that det Di(x, y) = \2x â€” 16 and
1
1
[Df(x,y)Y
Thus, formula (6) becomes
1
I2x - 16
-3x 2 - 2x + 14
12x - 16
x k
~x k -\
\2x - 16
-3x 2 -lOx + 2
\2x - 16
1
12x t .
-3x
16
2*fc_i + 14
-3x
12x t .
2
i - 16
lOjCjt-i +2
12x^-1 â€” 16
5x?_, + 2xk-\ -
12xt_
~~** R 3 given by
x(f) = b + ta
defines the path along the straight line parallel to a and passing through the end-
point of the position vector of b as in Figure 3.1. (See formula ( 1 ) of Â§ 1 .2.) â™¦
EXAMPLE 2 The path y: [0, 2tt) -> R 2 given by
y(f) = (3 cosf, 3 sinf)
can be thought of as the path of a particle that travels once, counterclockwise,
around a circle of radius 3 (Figure 3.2). â™¦
190 Chapter 3 I Vector-Valued Functions
Figure 3.4 The path x and its
velocity vector v.
EXAMPLE 3 The map z: R ->â– R 3 defined by
z(f) = (a cos t, a sin?, bt), a, b constants {a > 0)
is called a circular helix, so named because its projection in the xy-plane is a
circle of radius a. The helix itself lies in the right circular cylinder x 2 + y 2 = a 2
(Figure 3.3). The value of b determines how tightly the helix twists. â™¦
We distinguish between a path x and its range or image set x(7), the latter
being a curve in R". By definition, a path is a function, a dynamic object (at least
when we imagine the independent variable / to represent time), whereas a curve
is a static figure in space. With such a point of view, it is natural for us to consider
the derivative Dx(t ), which we also write as x'(/ ) or v(/), to be the velocity vector
of the path. We can readily justify such terminology. Since
x(t) = (x 1 (t),x 2 (t),...,x n (t))
is a function of just one variable,
v(/) = x'(t) = lim
x(r + At) - x(t)
At '
Thus, v(/) is the instantaneous rate of change of position x(t) with respect to t
(time), so it can appropriately be called velocity. Figure 3.4 provides an indication
as to why we draw v(f) as a vector tangent to the path at x(f). Continuing in this
vein, we introduce the following terminology:
DEFINITION 1 .2 Let x: / ->â– R" be a differentiable path. Then the velocity
\(t) = x'(t) exists, and we define the speed of x to be the magnitude of
velocity; that is,
Speed = ||v(OI|.
If v is itself differentiable, then we call V(t) = x"(t) the acceleration of x
and denote it by m(t).
EXAMPLE 4 The helix x(t ) = (a cos t , a sin t , bt) has
v(/) = â€” a sinf i + a cost j + fok and a(f) = â€” a cos/ i â€” a sin/ j.
3.1 ; Parametrized Curves and Kepler's Laws
Thus, the acceleration vector is parallel to the xy-plane (i.e., is horizontal). The
speed of this helical path is
||v(OII = -vA-a sin t) 2 + (a cos tf + b 2 = y/a 2 + b 2 ,
which is constant. â™¦
The velocity vector v is important for another reason, namely, for finding equations
of tangent lines to paths. The tangent line to a differentiate path x, at the point
x 0 = x(?o), is the line through xo that is parallel to any (nonzero) tangent vector
to x at xo. Since v(t), when nonzero, is always tangent to x(t), we may use equa-
tion ( 1 ) of Â§ 1 .2 to obtain the following vector parametric equation for the tangent
line:
l(s) = Xo + SV(). (1)
Here vo = v(to) and s may be any real number.
In equation (1), we have 1(0) = Xo. To relate the new parameter s to the
original parameter t for the path, we set s = t â€” to and establish the following
result:
PROPOSITION 1 .3 Let x be a differentiate path and assume that v 0 =
v(?o) ^ 0. Then a vector parametric equation for the line tangent to x at x 0 = x(t 0 )
is either
l(s) = x 0 + s\ 0 (2)
or
1(0 = x 0 + (t - f 0 )v 0 . (3)
(See Figure 3.5.)
EXAMPLE 5 If x(?) = (it + 2, t 2 - 7, t - t 2 ), we find parametric equations
for the line tangent to x at (5, â€” 6, 0) = x(l).
For this path, v(0 = x'(t) = 3i + 2/j + (1 - 2t)k, so that
Thus, by formula (3),
1(0
v 0 = v(l) = 3i + 2j-k.
(5i-6j) + (f- l)(3i + 2j-k).
Taking components, we read off the parametric equations for the coordinates
of the tangent line as
x = 3t +2
y = 2t - 8 .
z=\-t
The physical significance of the tangent line is this: Suppose a particle of
mass m travels along a path x. If, suddenly, at t = to, all forces cease to act on the
particle (so that, by Newton's second law of motion F = ma, we have a(/) = 0
for t > to), then the particle will follow the tangent line path of equation (3).
192 Chapter 3 I Vector-Valued Functions
EXAMPLE 6 If Roger Ramjet is fired from a camion, then we can use vectors
to describe his trajectory. (See Figure 3.6.)
y
Roger's path
x
Figure 3.6 Roger Ramjet's path.
We'll assume that Roger is given an initial velocity vector vo by virtue of the
firing of the cannon and that thereafter the only force acting on Roger is due to
gravity (so, in particular, we neglect any air resistance). Let us choose coordinates
so that Roger is initially at the origin, and throughout our calculations we '11 neglect
the height of the cannon. Let x(r) = (x(f), y(t)) denote Roger's path. Then the
information we have is
a (t) = x"(t)=-g\
(i.e., the acceleration due to gravity is constant and points downward); hence,
v(0) = x'(0) = v 0
and
x(0) = 0.
Since a(/) = V(t), we simply integrate the expression for acceleration compo-
nentwise to find the velocity:
v(0
= j a(t)dt = j -g]dr = -gtj +
Here c is an arbitrary constant vector (the "constant of integration"). Since v(0) =
vo, we must have c = vo, so that
v(0 = -gtj + v 0 .
Integrating again to find the path,
x(t) = j v(t)dt = j (-gt\ + \ 0 )dt = -^gt 2 ] + t\ 0 + d,
where d is another arbitrary constant vector. From the remaining fact that x(0) = 0,
we conclude that
1 7
x(0 = -^gn + ty 0 (4)
describes Roger's path.
To understand equation (4) better, we write vo in terms of its components:
vo = vo cos 9 i + vo sin 9 j.
Here vo = ||vo|| is the initial speed. (We're really doing nothing more than
expressing the rectangular components of vo in terms of polar coordinates.
3.1 : Parametrized Curves and Kepler's Laws
See Figure 3.7.) Thus,
x(r) = â€” \gt 2 \ + f(i> 0 cos#i + vosin# j)
= (v 0 cos 9)t\ + I (u 0 sin 9)t â€” -gt 2
1
2
From this, we may read off the parametric equations:
x = (vo cos6)t
1 , -
y = (v 0 sinfl)/ - -gt z
from which it is not difficult to check that Roger's path traces a parabola. â™¦
Here are two practical questions concerning the set-up of Example 6: First, for
a given initial velocity, how far does Roger travel horizontally? Second for a given
initial speed how should the cannon be aimed so that Roger travels (horizontally)
as far as possible? To find the range of the cannon shot and thereby answer the
first question, we need to know when y = 0 (i.e., when Roger hits the ground).
Thus, we solve
(u 0 sin 6>)r - \gt 2 = t(v 0 sm9 - \gt) = 0
for t. Hence, y = 0 when t = 0 (which is when Roger blasts off) and when
t = (2vq sin9)/g. At this later time,
(2vo sin#\ sin 29
â€” = â– (5)
8 J S
Formula (5) is Roger's horizontal range for a given initial velocity. To maximize
the range for a given initial speed vo, we must choose 9 so that (v 2 sin29)/g is
as large as possible. Clearly, this happens when sin 29 = 1 (i.e., when 9 = jt/4).
Kepler's Laws of Planetary Motion (optional)
Since classical antiquity, individuals have sought to understand the motions of
the planets and stars. The majority of the ancient astronomers, using a combina-
tion of crude observation and faith, believed all heavenly bodies revolved around
the earth. Fortunately, the heliocentric (or "sun-centered") theory of Nicholas
Copernicus (1473-1543) did eventually gain favor as observational techniques
improved. However, it was still believed that the planets traveled in circular or-
bits around the sun. This circular orbit theory did not correctly predict planetary
positions, so astronomers postulated the existence of epicycles, smaller circular
orbits traveling along the major circular arc, an example of which is shown in
Figure 3.8. Although positional calculations with epicycles yielded results closer
to the observed data, they still were not correct. Attempts at further improvements
were made using second- and third-order epicycles, but any gains in predictive
power were made at a cost of considerable calculational complexity. A new idea
was needed. Such inspiration came from Johannes Kepler (1571-1630), son of a
saloonkeeper and assistant to the Danish astronomer Tycho Brahe. The classical
astronomers were "stuck on circles" for they believed the circle to be a perfect
form and that God would use only such perfect figures for planetary motion.
Kepler, however, considered the other conic sections to be as elegant as the cir-
cle and so hypothesized the simple theory that planetary orbits are elliptical.
Empirical evidence bore out this theory.
194 Chapter 3 | Vector-Valued Functions
Figure 3.9 Kepler's second law
of planetary motion: If
t 2 â€” t\ = t 4 â€” f 3 , then A\ = A 2 ,
where A ; and A 2 are the areas of
the shaded regions.
Kepler's three laws of planetary motion are
1. The orbit of a planet is elliptical, with the sun at a focus of the ellipse.
2. During equal periods of time, a planet sweeps through equal areas with respect
to the sun. (See Figure 3.9.)
3. The square of the period of one elliptical orbit is proportional to the cube of
the length of the semimajor axis of the ellipse.
Kepler's laws changed the face of astronomy. We emphasize, however, that
they were discovered empirically, not analytically derived from general physical
laws. The first analytic derivation is frequently credited to Newton, who claimed
to have established Kepler's laws (at least the first and third laws) in Book I of
his Philosophiae Naturalis Principia Mathematica (1687). However, a number of
scientists and historians of science now consider Newton's proof of Kepler's first
law to be flawed and that Johann Bernoulli (1667-1748) offered the first rigorous
derivation in 1710. 1 In the discussion that follows, Newton's law of universal
gravitation is used to prove all three of Kepler's laws.
In our work below, we assume that the only physical effects are those be-
tween the sun and a single planet â€” the so-called two-body problem. (The n-body
problem, where n > 3 is, by contrast, an important area of current mathematical
research.) To set the stage for our calculations, we take the sun to be fixed at the
origin O in R 3 and the planet to be at the moving position P. We also need the
following two "vector product rules," whose proofs we leave to you:
PROPOSITION 1.4
1. If x and y are differentiable paths in R", then
d dx dy
â€” (x â€¢ y) = v â€¢ h x â€¢ â€” .
dt y J dt dt
2. If x and y are differentiable paths in R 3 , then
d dx dy
â€” (x xy)= â€” xy + xx â€” .
dt y J dt J dt
First, we establish the following preliminary result:
PROPOSITION 1 .5 The motion of the planet is planar, and the sun lies in the
planet's plane of motion.
PROOF Let r = OP. Then r is a vector whose representative arrow has its tail
fixed at O. (Note that r = r(t); that is, r is a function of time.) If v = r'(f), we
will show that r x v is a constant vector c. This result, in turn, implies that r must
always be perpendicular to c and, hence, that r always lies in a plane with c as
normal vector.
To show that r x v is constant, we show that its derivative is zero. By part 2
of Proposition 1.4,
d dr dy
â€” (r xv)= â€” xv + rx â€” = vxv+rxa,
dt dt dt
For an indication of the more recent controversy surrounding Newton's mathematical accomplishments,
see R. Weinstock, "Isaac Newton: Credit where credit won't do," The College Mathematics Journal, 25
(1994), no. 3, 179-192, andC. Wilson, "Newton's orbit problem: A historian's response," Ibid., 193-200,
and related papers.
3.1 i Parametrized Curves and Kepler's Laws
by the definitions of velocity and acceleration. We know that vxv = 0 (why?), so
d
â€” (r x v) = r x a. (6)
dt
Now we use Newton's laws. Newton's law of gravitation tells us that the planet
is attracted to the sun with a force
GMm
F = â€” u, (7)
where G is Newton's gravitational constant (= 6.6720 x 10~ n Nm 2 /kg 2 ), M
is the mass of the sun, m is the mass of the planet (in kilograms), r = ||r||, and
u = r/ 1| i* || (distances in meters). On the other hand Newton's second law of
motion states that, for the planet,
F = ma.
Thus,
GMm
ma = r â€” u,
r L
or
GM
a = -r. (8)
Therefore, a is just a scalar multiple of r and hence is always parallel to r. In
view of equations (6) and (8), we conclude that
d
â€” (r xv) = rxa = 0
dt
(i.e., that r x v is constant).
THEOREM 1 .6 (Kepler's first law) In a two-body system consisting of one
sun and one planet, the planet's orbit is an ellipse and the sun lies at one focus of
that ellipse.
PROOF We will eventually find a polar equation for the planet's orbit and see
that this equation defines an ellipse as described. We retain the notation from
the proof of Proposition 1.5 and take coordinates for R 3 so that the sun is at the
origin, and the path of the planet lies in the xy-plane. Then the constant vector
c = r x v used in the proof of Proposition 1.5 may be written as ck, where c is
some nonzero real number. This set-up is shown in Figure 3.10.
z
c = r x v
Figure 3.10 Establishing Kepler's laws.
196 Chapter 3 I Vector-Valued Functions
Step 1. We find another expression for c. By definition of u in formula (7),
r = ru, so that, by the product rule,
d du dr
v = â€” (ru) = râ€” + â€” u.
dt dt dt
Hence,
/ du dr \ 2 / du\ dr
c = r x v = (ru) x r 1 u =r ux â€” + r â€” (u x u).
\ dt dt J \ dt J dt
Since uxu must be zero, we conclude that
c = r'(uxf). ,9,
Step 2. We derive the polar equation for the orbit. Before doing so, however,
note the following result, whose proof is left to you as an exercise:
PROPOSITION 1.7 If x(r) has constant length (i.e., ||x(f)|| is constant for
all r), then x is perpendicular to its derivative dx/dt.
Continuing now with the main argument, note that the vector r(r) is denned
so that its magnitude is precisely the polar coordinate r of the planet's position.
Using equations (8) and (9), we find that
axe
GM
u x r u x
du
dt
= -GM
= GM
duX
U X | U X
dt )
du\
U X X u
dt J
= GM
= GM
(u-u)
du
dt
du
u- â€” | u
dt
du
1 Ou
dt
(see Exercise 27 of Â§ 1 .4)
(by Proposition 1.7)
= - (GMu),
dt
since G and M are constant. On the other hand, we can "reverse" the product rule
to find that
d\
a x c = â€” x c
dt
dy dc
= â€” xc + vxâ€” (since c is constant)
dt dt
= â€” (v X c).
dt
3.1 ; Parametrized Curves and Kepler's Laws
Figure 3.1 1 The angle 9 is
the angle between r and d.
Thus,
and, hence,
d d
a x c = â€” (GMu) = â€” (v x c),
dt dt
v x c = GMu + d,
(10)
where d is an arbitrary constant vector. Because both v x c and u lie in the xy-
plane, so must d.
Let us adjust coordinates, if necessary, so that d points in the i-direction (i.e.,
so that d = di for some d e R). This can be accomplished by rotating the whole
set-up about the z-axis, which does not lift anything lying in the xy-plane out of
that plane. Then the angle between r (and hence u) and d is the polar angle 6 as
shown in Figure 3.11.
By Theorem 3.3 of Chapter 1,
u-d= ||u|
Since c = ||c||,
2
Idll cos# = dcosi
(11)
c â€¢ c
= (r x v) â€¢ c
= r â€¢ (v x c)
= ru-(GMu + d)
Hence,
(Why? See formula (4) of Â§1.4.)
by equation (10).
GMr + rd cos6Â»
by equation (11). We can readily solve this equation for r to obtain
c 2
r =
(12)
GM + d cosO
the polar equation for the planet's orbit.
Step 3. We now check that equation (12) really does define an ellipse by
converting to Cartesian coordinates. First, we'll rewrite the equation as
c 2 (c 2 /GM)
GM + d cosO ~ l + (d/GM)cose'
and then let p = c 2 / GM, e = d/ GM for convenience. (Note that p > 0.) Hence,
equation (12) becomes
r = â– (13)
1 + e cos e
A little algebra provides the equivalent equation,
r = p â€” er cos6. (14)
Now r cose = x (x being the usual Cartesian coordinate), so that equation (14)
is equivalent to
r = p â€” ex.
To complete the conversion, we square both sides and find, by virtue of the fact
that r 2 = x 2 + y 2 ,
x 2 + y 2 = p 2
2pex + e 2 x 2 .
198 Chapter 3 I Vector-Valued Functions
A little more algebra reveals that
(1- e 2 )x 2 + 2pex + y 2 = p 2 . (15)
Therefore, the curve described by the preceding equation is an ellipse if 0 <
\e\ < 1, a parabola if e = Â±1, and a hyperbola if \e \ > 1 . Analytically, there is no
way to eliminate the last two possibilities. Indeed, "uncaptured" objects such as
comets or expendable deep space probes can have hyperbolic or parabolic orbits.
However, to have a closed orbit (so that the planet repeats its transit across the
sky), we are forced to conclude that the orbit must be elliptical.
More can be said about the elliptical orbit. Dividing equation (15) by 1 â€” e 2
and completing the square in x, we have
\ 2 2 2
x +
1 -e 2 (1 -e 2 ) 2 '
This is equivalent to the rather awkward-looking equation
(x + pe/(l-e 2 )f y 2
p 2 /(l - e 2 ) 2 p 2 /{\ - e 2 ) y J
From equation (16), we see that the ellipse is centered at the point (â€” pe/(l â€” e 2 ),
0), that its semimajor axis has length a = p/(l â€” e 2 ), and that its semiminor axis
has length b = p/^/l â€” e 2 . The foci of the ellipse are at a distance
yV - b 2
P 2 P 2 P\e\
(1 - e 2 ) 2 1 - e 2
from the center. (See Figure 3.12.) Hence, we see that one focus must be at the
origin, the location of the sun. Our proof is, therefore, complete. â–
Fortunately, all the toil involved in proving the first law will pay off in proofs
of the second and third laws, which are considerably shorter. Again, we retain all
the notation we already introduced.
THEOREM 1 .8 (Kepler's second law) During equal intervals of time, aplanet
sweeps through equal areas with respect to the sun.
Figure 3.1 2 The ellipse of equation (16).
3.1 ; Parametrized Curves and Kepler's Laws
Po(r 0 , B 0 )
P(r, 6)
Figure 3.1 3 The shaded area A(9) is
given by \r 2 dip.
PROOF Fix one point Pq on the planet's orbit. Then the area A swept between
Po and a second (moving) point P on the orbit is given by the polar area integral
A{6)
f 6 1 2
L 2 r
dip.
(See Figure 3.13.) Thus, we may reformulate Kepler's law to say that dA/dt is
constant. We establish this reformulation by relating dA/dt to a known constant,
namely, the vector c = r x v.
By the chain rule (in one variable),
dA _ dA d9
dt d6 dt '
By the fundamental theorem of calculus,
dA d f 9 1 , , 1 r ,~
Hence,
dA _ 1 2 dO
dt ~ 2 r dt'
(17)
Now, we relate c to dO/dt by means of equation (9). Therefore, we compute
u x du/dt in terms of 0. Recall that u = - r and r = r cos 0 i + r sin 6* j. Thus,
r
cos 0 i + sin 0 j
u
du
dt dt ' dt
Hence, it follows by direct calculation of the cross product that
dO dO
smtf â€” i + cosO â€” j.
c = r u x
du
dt
2 d0
= r â€” k,
dt
so c = ||c|| = r 2 d0/dt, and equation (17) implies that
dA _ 1
~dt ~ 2 C '
a constant.
(18)
THEOREM 1 .9 (Kepler's third law) If T is the length of time for one plane-
tary orbit, and a is the length of the semimajor axis of this orbit, then T 2 = Ka 3
for some constant K.
Chapter 3 | Vector-Valued Functions
3.1 Exercises
PROOF We focus on the total area enclosed by the elliptical orbit. The area of an
ellipse whose semimajor and semiminor axes have lengths a and b, respectively,
is nab. This area must also be that swept by the planet in the time interval [0, T].
Thus, we have
' T dA
â€” dt
dt
T 1
-cdt by equation ( 1 8)
nab = I
Jo
-L
Hence,
1
= 2 CT -
T = 2 -^, so r 2 = 4 " W . (19)
c c 2
Now, b and c are related to a, so these quantities must be replaced before we are
done. In particular, from equation (16), b 2 = p 2 /(l â€” e 2 ), so
b 2 = pa.
Also
P
c 2
GM
(See equations (12) and (13).) With these substitutions, the result in (19) becomes
t2 = An 2 a 2 {pa) = /4^r 2 \ ^
pGM \Gm)
This last equation shows that T 2 is proportional to a 3 , but it says even more:
The constant of proportionality 4n 2 /GM depends entirely on the mass of the
sun â€” the constant is the same for any planet that might revolve around the sun.
In Exercises 1-6, sketch the images of the following paths, us-
ing arrows to indicate the direction in which the parameter
increases:
. \x = It - 1
1. { , , , -1 < t < 1
[y = 3 - t
2. x(t) = e' i + i + e 3 ' j, t = 0
16. x(t) = 4cos?iâ€” 3sinr j + 5tk, t = tt/3
17. x(t) = (t 2 ,t\t 5 ), t = 2
18. x(r) = (cos(e'), 3 - t 2 , t), t = 1
19. (a) Sketch the path x(f) = (t, f 3 - It + 1).
(b) Calculate the line tangent to x when t = 2.
(c) Describe the image of x by an equation of the form
y = f(x) by eliminating t .
(d) Verify your answer in part (b) by recalculating the
tangent line, using your result in part (c).
Exercises 20-23 concern Roger Ramjet and his trajectory when
he is shot from a cannon as in Example 6 of this section.
20. Verify that Roger Ramjet's path in Example 6 is indeed
a parabola.
21 . Suppose that Roger is fired from the cannon with an
angle of inclination 6 of 60Â° and an initial speed Vq of
100 ft/sec. What is the maximum height Roger attains?
22. Suppose that Roger is fired from the cannon with an an-
gle of inclination 6 of 60Â° and that he hits the ground
1/2 mile from the cannon. What, then, was Roger's
initial speed?
23. If Roger is fired from the cannon with an initial speed of
250 ft/sec, what angle of inclination 6 should be used
so that Roger hits the ground 1 500 ft from the cannon?
24. Gertrude is aiming a Super Drencher water pistol at
Egbert, who is 1.6 m tall and is standing 5 m away.
Gertrude holds the water gun 1 m above ground at an
angle a of elevation. (See Figure 3.14.)
(a) If the water pistol fires with an initial speed of
7 m/sec and an elevation angle of 45Â°, does Egbert
get wet?
(b) If the water pistol fires with an initial speed of
8 m/sec, what possible angles of elevation will
cause Egbert to get wet? (Note: You will want to
use a computer algebra system or a graphics cal-
culator for this part.)
25. A malfunctioning rocket is traveling according to the
pathx(r) = (e 2 ', 3f 3 â€” 2t, t - j) in the hope of reach-
ing a repair station at the point (7e 4 , 35, 5). (Here
t represents time in minutes and spatial coordinates
are measured in miles.) Alt = 2, the rocket's engines
suddenly cease. Will the rocket coast into the repair
station?
26. Two billiard balls are moving on a (coordina-
tized) pool table according to the respective paths
x(f) = (t
!, j - 1J and y(f) = (t, 5 - r), where
t represents time measured in seconds.
(a) When and where do the balls collide?
(b) What is the angle formed by the paths of the balls
at the collision point?
27. Establish part 1 of Proposition 1.4 in this section: If x
and y are differentiable paths in R" , show that
dt
(x-y):
dx
dy
dt '
28. Establish part 2 of Proposition 1.4 in this section: If x
and y are differentiable paths in R 3 , show that
dt
dx dy
(xxy) = â€” xy + xx â€” .
dt at
29. Prove Proposition 1.7.
30. (a) Show that the path x(f) = (cos t, cos t sin t, sin 2 t)
lies on a unit sphere.
(b) Verify that x(r) is always perpendicular to the ve-
locity vector v(f ).
(c) Use Proposition 1 .7 to show that if a differentiable
path lies on a sphere centered at the origin, then
its position vector is always perpendicular to its
velocity vector.
Figure 3.14 Figure for Exercise 24.
Chapter 3 | Vector-Valued Functions
31. Consider the path
x = (a + b cos cot ) cos t
â– y = (a + b cos cot) sin t ,
z = b sin cot
where a, b, and co are positive constants and a > b.
(a) Use a computer to plot this path when
i. a = 3, b = 1, and co = 15.
ii. a = 5, b = 1, and co = 15.
iii. a = 5, b = 1, and w = 25.
Comment on how the values of a, b, and R 2 , x(f ) =
(cosf, sinf) may be the most familiar way to give a
parametric description of a unit circle, in this problem
you will develop a different set of parametric equations
that gives the x- and v-coordinates of a point on the
circle in terms of rational functions of the parameter.
(This particular parametrization turns out to be useful
in the branch of mathematics known as number theory.)
To set things up, begin with the unit circle x 1 +
y 2 = 1 and consider all lines through the point (â€”1,0).
(See Figure 3.15.) Note that every line other than the
vertical line x = â€” 1 intersects the circle at a point
(x, y) other than (â€”1, 0). Let the parameter t be the
slope of the line joining (â€”1,0) and a point (x, y) on
the circle.
y
(-1,0)1
2" Slope?
X
Figure 3.15 Figure for Exercise 34.
(a) Give an equation for the line of slope t joining
(â€”1,0) and (x, y). (Your answer should involve
x, y, and t.)
(b) Use your answer in part (a) to write y in terms of
x and t. Then substitute this expression for y into
the equation for the unit circle. Solve the resulting
equations for x in terms of /. Your answer(s) for x
will give the points of intersection of the line and
the circle.
(c) Use your result in part (b) to give a set of paramet-
ric equations for points (x, y) on the unit circle.
(d) Does your parametrization in part (c) cover the
entire circle? Which, if any, points are missed?
35. Let x(f) be a path of class C 1 that does not pass through
the origin in R 3 . If x(fo) is the point on the image of x
closest to the origin and x'(fo) / 0, show that the po-
sition vector x(?o) is orthogonal to the velocity vector
x'(fo).
3.2 Arclength and Differential Geometry
In this section, we continue our general study of parametrized curves in R 3 ,
considering how to measure such geometric properties as length and curvature.
This can be done by defining three mutually perpendicular unit vectors that form
the so-called moving frame specially adapted to a path x. Our study takes us
briefly into the branch of mathematics called differential geometry, an area where
calculus and analysis are used to understand the geometry of curves, surfaces,
and certain higher-dimensional objects (called manifolds).
3.2 | Arclength and Differential Geometry
Length of a Path
For now, let x: [a, b] â€”> R 3 be a C 1 path in R 3 . Then we can approximate the
length L of x as follows: First, partition the interval [a, b] into n subintervals.
That is, choose numbers to, t\, ...,/â€ž such that a = to < t\ < â– â– â– < tâ€ž = b. If,
for i = 1, . . . , n, we let As, denote the distance between the points x(r,_i) and
x(ti) on the path, then
n
1=1
(See Figure 3.16.) We have x(f) = (x(t), y(t), z(t)), so that the distance formula
(i.e., the Pythagorean theorem) implies
As t = J Ax 2 + Ay, 2 + Az 2 ,
where Axj = x(t { ) - x(ti-\), Ay t = y(u) - y(ti-\), and Az,- = z{U) - zft-i). It
is entirely reasonable to hope that the approximation in (1) improves as the At t 's
become closer to zero. Hence, we define the length L of x to be
n
L= lim Y^A Xi 2 + Ay,- 2 + Az,- 2 . (2)
max Ar,-^0 ~â€”f
i=l
Now, we find a way to rewrite equation (2) as an integral. On each subinterval
fj], apply the mean value theorem (three times) to conclude the following:
1. There must be some number t* in [Â£,-_i, ?, ] such that
x(tt) - x(Â»,_i) = x'(ff)fe - fj_i);
that is, Ax, = x'{t*)At t .
2. There must be another number t** in [/, i , r,] such that
Ay i =y'(t**)At t .
3. There must be a third number t*** in [/,_!, f, ] such that
Az, =z'(t***)At h
Therefore, with a little algebra, equation (2) becomes
L= lim V Jx'{tff + y'(t**) 2 + z'(t***) 2 At h (3)
max Ar,^0 z â€” â– f V
( = 1
When the limit appearing in equation (3) is finite, it gives the value of the definite
integral
b
Jx>(t) 2 + y>(t) 2 + z'(t) 2 dt.
Note that the integrand is precisely ||x'(f)||, the speed of the path. (This makes
perfect sense, of course. Speed measures the rate of distance traveled per unit
time, so integrating the speed over the elapsed time interval should give the total
distance traveled.) Moreover, it's not hard to see how we should go about defining
the length of a path in R" for arbitrary n.
L
204 Chapter 3 | Vector- Valued Functions
DEFINITION 2.1 The length L(x) of a C 1 path x: [a, b] -> R" is found by
integrating its speed:
= f llx'
J a
L(x) = / \\x'(t)\\dt.
Figure 3.17 AC 1 path.
x(a)
Figure 3.1
x: [a, b]
x(b)
8 Apiecewise C 1 path
R 3 .
EXAMPLE 1 To check our definition in a well-known situation, we compute
the length of the path
x: [0, 2n] â€” > R 2 , x(/) = (a cost, a sin t), a > 0.
We have
so
x'(r) = â€” a sinr i + a cos? j,
||x'(f)ll = V o 2 sin 2 1 + a 2 cos 2 f = a.
Thus, Definition 2.1 gives
L(x)
2,t
: dt = lit a.
Since the path traces a circle of radius a once, the length integral works out to be
the circumference of the circle, as it should. â™¦
EXAMPLE 2 For the helix x(t) = (a cost, a sint, bt), 0 < t < 2ir, we have
x'(t) = â€”a sin t i + a cos t j + b k,
so that ||x'(0|| = Va 2 + Â£ 2 , and
L(x)
= / V**

~~â€¢ R is a differentiable function (scalar field),
the gradient of / may be considered to be the result of multiplying the vector V
by the scalar /, except that when we "multiply" each component of V by /, we
actually compute the appropriate partial derivative:
/ 3 3 3 \ df df df
Vf(x,y,z)= iâ€” +jâ€” +k- )f(x,y,z)= -M+/j + /k.
\ dx ay dz ) dx dy dz
The del operator can also be defined in R", for arbitrary n. If we take
x\, X2, â– â– â– , x n to be coordinates for R", then del is simply
where e, â– = (0, . . . , 1, . . . , 0), i = 1, . . . , n, is the standard basis vector for R".
The Divergence of a Vector Field
Whereas taking the gradient of a scalar field yields a vector field the process of
taking the divergence does just the opposite: It turns a vector field into a scalar
field.
DEFINITION 4.1 Let F: X C R" R" be a differentiable vector field.
Then the divergence of F, denoted div F or V â€¢ F (the latter read "del dot
F"), is the scalar field
dFi dF 2 dF n
div F = V â€¢ F = â€” L + â€” L + ... + â€” 1,
3xi dxi dxâ€ž
where Cartesian coordinates for R" and F\ , . . . , Fâ€ž are the
component functions of F.
It is essential that Cartesian coordinates be used in the formula of Definition 4.1.
(Later in this section we shall see what div F looks like in cylindrical and spherical
coordinates for R 3 .)
EXAMPLE 1 IfF = jr 2 vi + x Z j+xyzk,then
3 3d
div F = â€” (x 2 y) + â€” (jcz) + â€” (xyz) = 2xy + 0 + xy = 3xy. â™¦
dx dy dz
3.4 | Gradient, Divergence, Curl, and the Del Operator 229
The notation for the divergence involving the dot product and the del operator
is especially apt: If we write
F = Fid + F 2 e 2 H h F n e n ,
then,
/ 3 3 3 \ ,
V-F= eiâ€” +e 2 â€” + --- + eâ€žâ€” â– (Fid + F 2 e 2 + â– â– â– + Fâ€žeâ€ž)
\ dXl 0X2 ox n J
_ dFi dF 2 dF n
dx\ 9x2 3xâ€ž '
where, once again, we interpret "multiplying" a function by a partial differential
operator as performing that partial differentiation on the given function.
Intuitively, the value of the divergence of a vector field at a particular point
gives a measure of the "net mass flow" or "flux density" of the vector field in
or out of that point. To understand what such a statement means, imagine that
the vector field F represents velocity of a fluid. If V â€¢ F is zero at a point, then
the rate at which fluid is flowing into that point is equal to the rate at which
fluid is flowing out. Positive divergence at a point signifies more fluid flowing out
than in, while negative divergence signifies just the opposite. We will make these
assertions more precise, even prove them, when we have some integral vector
calculus at our disposal. For now, however, we remark that a vector field F such
that V â€¢ F = 0 everywhere is called incompressible or solenoidal.
EXAMPLE 2 The vector field F = xi + yj has
V-F= A (je )+ ^(y) = 2.
dx ay
This vector field is shown in Figure 3.38. At any point in R 2 , the arrow whose
tail is at that point is longer than the arrow whose head is there. Hence, there is
greater flow away from each point than into it; that is, F is "diverging" at every
point. (Thus, we see the origin of the term "divergence.")
The vector field G = â€” xi â€” yj points in the direction opposite to the vector
field F of Figure 3.38 (see Figure 3.39), and it should be clear how G's divergence
of â€”2 is reflected in the diagram. â™¦
EXAMPLE 3 The constant vector field F(x, y, z) = a shown in Figure 3.40
is incompressible. Intuitively, we can see that each point of R 3 has an arrow
representing a with its tail at that point and another arrow, also representing a,
with its head there.
The vector field G = yi â€” xj has
v-g= A (v) + A ( _ x) = o.
ax ay
A sketch of G reveals that it looks like the velocity field of a rotating fluid, without
either a source or a sink. (See Figure 3.41 .) â™¦
The Curl of a Vector Field
If the gradient is the result of performing "scalar multiplication" with the del
operator and a scalar field, and the divergence is the result of performing the
"dot product" of del with a vector field, then there seems to be only one simple
Chapter 3 | Vector-Valued Functions
Figure 3.40 The constant vector
field F = a.
Figure 3.41 The vector field
G = yi â€” x'] resembles the
velocity field of a rotating fluid.
differential operation left to be built from del. We call it the curl of a vector field
and define it as follows:
DEFINITION 4.2 Let F: X c R 3 -> R 3 be a differentiable vector field on
R 3 only. The curl of F, denoted curl F or V x F (the latter read "del cross
F"), is the vector field
(3 3 3 \
V + V + V ) x + F ^ + F3k >
dx ay dz )
i j k
d/dx d/dy d/dz
F\ F 2 F 3
^EL - i + ( ?H _ ^fl\ â– + ( ^fl _ |
dy dz J \ dz dx J \ dx dy
There is no good reason to remember the formula for the components of the
curl â€” instead simply compute the cross product explicitly.
EXAMPLE 4 If F = x 2 yi - 2xzj + (x + y - z)k, then
V x F
i j k
d/dx d/dy d/dz
â€” 2xz x + y â€” z
x 2 y
3 3 \ ( 3 3
â€” (x + y-z)- ^~ 2xz) ) 1 + \dz~ (x2y) ~ 9^ (X + y ~ Z -
3
9
= (1 + 2x)i - j - (x 2 + 2z)k.
3.4 | Gradient, Divergence, Curl, and the Del Operator 231
Figure 3.42 A twig in a pond where water moves with velocity given by a vector field F. In the left figure, the twig does
not rotate as it travels, so curl F = 0. In the right figure, curl F / 0, since the twig rotates.
One would think that, with a name like "curl," V x F should measure how
much a vector field curls. Indeed the curl does measure, in a sense, the twisting
or circulation of a vector field but in a subtle way: Imagine that F represents the
velocity of a stream or lake. Drop a small twig in the lake and watch it travel.
The twig may perhaps be pushed by the current so that it travels in a large circle,
but the curl will not detect this. What curl F measures is how quickly and in what
orientation the twig itself rotates as it moves. (See Figure 3.42.) We prove this
assertion much later, when we know something about line and surface integrals.
For now, we simply point out some terminology: A vector field F is said to be
irrotational if V x F = 0 everywhere.
EXAMPLE 5 Let F = (3x 2 z + y 2 ) i + 2xy j + (x 3 - 2z) k. Then
V x F
l
d/dx
3x 2 z + y 2
a
j
d/dy
2xy
k
d/dz
2z
dy
(x> - 2z)
^(2xy)W(^(3x 2 z + y 2 )
dx
(x 3 -2z)
dx
(2xy)
3
9y
(3* 2 z + y 2 ) I k
= (0 - 0)i + (3x 2 - 3x 2 )\ + (2y - 2y)k = 0.
Thus, F is irrotational.
Two Vector-analytic Results
It turns out that the vector field F in Example 5 is also a gradient field. Indeed
F = V/, where f(x, y, z) = x 3 z + xy 2 â€” z 2 . (We'll leave it to you to verify
this.) In fact, this is not mere coincidence but an illustration of a basic result
about scalar-valued functions and the del operator:
THEOREM 4.3 Let /: X C R 3 -> R be of class C 2 . Then curl (grad/) = 0.
That is, gradient fields are irrotational.
Chapter 3 | Vector-Valued Functions
PROOF Using the del operator, we rewrite the conclusion as
V x (V/) = 0,
which might lead you to think that the proof involves nothing more than noting
that V/ is a "scalar" times V, hence, "parallel" to V, so thatthe cross productmust
be the zero vector. However, V is not an ordinary vector, and the multiplications
involved are not the usual ones. A real proof is needed.
Such a proof is not hard to produce: We need only start calculating V x (V/).
We have
df df df
dx dy dz
Therefore,
V x (V/) =
i j k
d/dx d/dy d/dz
df/dx df/dy df/dz
d 2 f
dydz dzdy
dzdy) \
d 2 f d 2 f
dzdx dxdz
j +
d 2 f d 2 f
dxdy dydx
Since / is of class C 2 , we know that the mixed second partials don't depend
on the order of differentiation. Hence, each component of V x (V/) is zero, as
desired. â–
There is another result concerning vector fields and the del operator that is
similar to Theorem 4.3:
THEOREM 4.4 Let F:XcR 3 ^R 3 be a vector field of class C 2 . Then
div (curl F) = 0. That is, curl F is an incompressible vector field.
The proof is left to you.
EXAMPLE 6 If F = (xz â€” e 2x cos z) i - yz j + e 2l (sin y + 2 sin z) k, then
3 3 3
V â€¢ F = â€” (xz â€” e 2x cosz) + â€” (-yz) + â€” (e 2x (smy + 2sinz))
dx dy dz
= z - 2e 2x cos z - z + 2e 2x cos z = 0
for all (ij,z)Â£R 5 . Hence, F is incompressible. We'll leave it to you to check
that F = V x G, where G(x, y, z) = e 2x cos y i + e 2x sin z j + xyz k, so that, in
view of Theorem 4.4 the incompressibility of F is not really a surprise. â™¦
Other Coordinate Formulations (optional)
We have introduced the gradient, divergence, and curl by formulas in Cartesian
coordinates and have, at least briefly, discussed their geometric significance. Since
certain situations may necessitate the use of cylindrical or spherical coordinates,
we next list the formulas for the gradient, divergence, and curl in these coordinate
systems. Before we do, however, a remark about notation is in order. Recall that
in cylindrical coordinates, there are three unit vectors e r , e$, and e z that point in
the directions of increasing r, 6, and z coordinates, respectively. Thus, a vector
3.4 | Gradient, Divergence, Curl, and the Del Operator 233
field F on R 3 may be written as
F = F r e r + F g e e + F-e..
In general, the component functions F r , Fg, and F z are each functions of the
three coordinates r, 9, and z; the subscripts serve only to indicate to which of
the vectors e, , eg, and e ; that particular component function should be attached.
Similar comments apply to spherical coordinates, of course: There are three unit
vectors e p , e 9 , and eg, and any vector field F can be written as
F = F p e p + F v e v + F e e g .
THEOREM 4.5 Let /: X c R 3 -> R and F: Y C R 3
scalar and vector fields, respectively. Then
3/ 13/ , 3/
â€” e r H e e H e,;
9r r 30 3z
divF =
curl F =
3 3F 0 3
or 39 dz
R be differentiate
(3)
(4)
e r rt 0
3/3r 3/36*
F r rFg
e
3/3z
(5)
PROOF We'll prove formula (4) only, since the argument should be sufficiently
clear so that it can be modified to give proofs of formulas (3) and (5). The idea is
simply to rewrite all rectangular symbols in terms of cylindrical ones.
From the equations in (8) of Â§ 1 .7, we have
e,- = cos 6 i + sin 9 j
eg = â€” sin 9 i + cos 6 j .
e- = k
(6)
From the chain rule, we have the following relations between rectangular and
cylindrical differential operators:
3 3 3
â€” = cos 9 h sin 6 â€”
3r dx 3y
3 3 3
â€” = â€”rsm9 h r cos 9 â€” .
89 dx 3y
3 _ 3
dz dz
These relations can be solved algebraically for d/dx, d/dy, and 3/3z to yield
a
= cos
3
9
sin 9
3
37
dr
r
39
3
= sin 6
3
Vr +
cos 9
8
3v"
r
39
3
3
dz
~ d~Z
(7)
234 Chapter 3 I Vector- Valued Functions
Hence, we can use (6) and (7) to rewrite the expression for the divergence of a
vector field on R 3 :
V-F
3 3 3
IT 1 + IT J + IT k ) * ( F '" e '- + FsCs + F ^
dx dy dz
1 cos
,1
dr
sin# 3
~r r ~d~9
j sin (9
3 cos9 3
8r
r 3(9
dz
[(F r cos 9 - F e sin 9) i + (F r sin <9 + F e cos 0) j + F z k].
(We used the equations in (7) to rewrite the partial operators d/dx, 3/3 y, and
3/3z appearing in del and the equations in (6) to replace the cylindrical basis
vectors e r , eg, and e z by expressions involving i, j, and k.) Performing the dot
product and using the product rule yields
V-F
cost
3 sva9 3
i
dr r d9
3 cos# 3
+ |sin6> 1
dr r d9
(F r cos 9 â€” F g sin#)
(F r sin 9 + F e cos 0) H F-
dz
dF r 3
= cost/ [cos 9 h F r â€” (cos#)
dr dr
cos v sin
3r
dF r
dr
sin9
r
sin
+ sm9 \ sm9 â€” - + F r â€” (sm.9)
dr dr
dF 0 3
+ sin# | cosf h F g â€” (cos$)
dr dr
dF r 3
cos 9 h F r â€” (cos 9)
3(9 3(9
r
cos 9
r
cos 9
sin 9-
dF e
d9
3
F e â€”(sin9)
3(9 v
dF r d
sm9 â€” - + F r â€” (sine)
36> 89
cos9-
d_F^
89
3
Fa â€” (cos 9)
d9
+
dz
After some additional algebra, we find that
3 F
V â€¢ F = (cos 2 9 + sin 2 9) â€” - +
dr
sin 2 9 + cos 2 (
\F r
+
dF r
sin 2 6> + cos 2 9 \ dF e
36
dF z
IF
1 1 dF e 8F,
dr r r d9 dz
(
\
r h F r H rF,
3r 3(9 3z V
as desired.
In spherical coordinates, the story for the gradient, divergence, and curl is
more complicated algebraically, although the ideas behind the proof are essentially
3.4 I Exercises 235
the same. We state the relevant results and leave to you the rather tedious task of
verifying them.
THEOREM 4.6 Let /: X C R 3 -> R and F: Y C R 3 -> R 3 be differentiable
scalar and vector fields, respectively. Then the following formulas hold:
v/ =
1 df
e
p d~~

~~
(sin~~

~~i, V2 v n ), show that _D v F(a) = v.
True/False Exercises for Chapter 3
1 . If a path x remains a constant distance from the origin,
then the velocity of x is perpendicular to x.
2. If a path is parametrized by arclength, then its velocity
vector is constant.
3. If a path is parametrized by arclength, then its velocity
and acceleration are orthogonal.
' â– x(f)|[ = ||x'(f)[|.
4.
dt
d
5. â€” (x x y)
dt K 3
6. K
dT
dt
dy dx
xx h y x â€” .
dt J dt
7. |T|
dB
ds
8. The curvature k is always nonnegative.
9. The torsion r is always nonnegative.
10. N
dT
ds
11. If a path x has zero curvature, then its acceleration is
always parallel to its velocity.
12. If a path x has a constant binormal vector B, then r =0.
|a(r)[| 2
14. grad / is a scalar field.
15. div F is a vector field.
16. curl F is a vector field.
1 7. grad(div F) is a vector field.
18. div(curl(grad /)) is a vector field.
1 9. grad/ x div F is a vector field.
20. The path x(f) = (2 cos t, 4 sinr, t) is a flow line of the
y
vector field F(x ,y, z) = â€” â€” i + 2x j + z k.
21. The path x(?) = (e' cosf , e'(cos t + sinf), e' sin?) is a
flow line of the vector field F(jc, y, z) = (x â€” z)i +
2jc j + y k.
22. The vector field F = 2xy cos z i â€” y 2 cos z j + e xy k is
incompressible.
23. The vector field F = 2xy cos z i â€” y 2 cos z j + e xy k is
irrotational.
24. V x (V/) = 0 for all functions /: R 3 â€” Â» R.
25. If V â€¢ F = 0 and V x F = 0, then F = 0.
26. V â€¢ (F x G) = F â€¢ (V x G) + G â€¢ (V x F).
27. If F = curl G, then F is solenoidal.
28. The vector field F = 2x sin y cos z i + x 2 cos y cos z
j + x 2 sin y sin z k is the gradient of a function / of
class C 2 .
29. There is a vector field F of class C 2 on R 3 such that
V x F = x cos 2 y i + 3y j â€” xyz 2 k.
30. If F and G are gradient fields, then F x G is incom-
pressible.
Miscellaneous Exercises for Chapter 3
1 . Figure 3.47 shows the plots of six paths x in the plane.
Match each parametric description with the correct
graph.
(a) x(/) = (s'mlt, sin 3i)
(b) x(f) = (t + sin5f, t 2 + cos 6?)
(c) x(r) = (f 2 + l,f 3 -?)
(d) x(f) = (2? + sin4f, t - sin5f)
(e) x(t) = {t -t 2 ,t 3 -t)
(f) x(t) = (sin(? + sin 30, cosf)
2. Figure 3.48 shows the plots of six paths x in R 3 . Match
each parametric description with the correct graph.
(a) x(f) = (t + cos3f, t 2 + sin5f, sin At)
(b) x(f) = (2 cos 3 1, 3 sin 3 1, cos2f)
(c) x(0 = (15 cos /, 23 sin t , At)
(d) x(0 = (cos3r, cos5f, sin4f)
(e) x(f) = (2f cosf, 2f sin?, 4f)
(f) x(f) = (r 2 + l,? 3 -f,? 4 -r 2 )
238 Chapter 3 | Vector-Valued Functions
Miscellaneous Exercises for Chapter 3 239
3. Suppose that x is a C 2 path with nonzero velocity. Show
that x has constant speed if and only if its velocity and
acceleration vectors are always perpendicular to one
another.
4. You are at Vertigo Amusement Park riding the new
Vector roller coaster. The path of your car is given by
x(?) = (x(t), y{t)), where
x(f)
t/60 1Tt t/60 â– Jtt
e ' cos â€” ,e ' sm â€” ,
30 30
2r(10 - t)(t - 90) 2
80 +
10 6
where / = 0 corresponds to the beginning of your
three-minute ride, measured in seconds, and spatial
dimensions are measured in feet. It is a calm day, but
after 90 sec of your ride your glasses suddenly fly off
your face.
(a) Neglecting the effect of gravity, where will your
glasses be 2 sec later?
(b) What if gravity is taken into account?
5. Show that the curve traced parametrically by
1
x(f) = ^cos(f - 1), t - 1,
istangenttothesurface.ii; 3 + y 3 + z 3
t = 1.
xyz = 0 when
6. Gregor, the cockroach, is on the edge of a Ferris wheel
that is rotating at a rate of 2 rev/min (counterclock-
wise as you observe him). Gregor is crawling along
a spoke toward the center of the wheel at a rate of
3 in/min.
(a) Using polar coordinates with the center of the
wheel as origin, assume that Gregor starts (at time
t = 0) at the point r = 20 ft, 6 = 0. Give paramet-
ric equations for Gregor's polar coordinates r and
6 at time t (in minutes).
(b) Give parametric equations for Gregor's Cartesian
coordinates at time t.
(c) Determine the distance Gregor has traveled once
he reaches the center of the wheel. Express your
answer as an integral and evaluate it numerically.
If you have used a drawing program on a computer, you have
probably worked with a curve known as a Bezier curve. 1 Such
a curve is defined parametrically by using several control
points in the plane to shape the curve. In Exercises 7-12,
we discuss various aspects of quadratic Bezier curves. These
curves are defined by using three fixed control points {x\ , y\),
( x 2, yi), and(xi, yi) and a nonnegative constant w. The Bezier
curve defined by this information is given by x: [0, 1] â€” > R 2 ,
X(t):
y(t) :
(1 - tfx\ + 2wt(l - t)x 2 + t 2 X }
(\-tf + 2wt(\-t) + t 2
(1 -tf yi +2wtQ_ -r)y 2 + f 2 y 3
(1 -t) 2 + 2wt{\ -t) + t 2
0 < t < 1.
(1)
^ 7. Let the control points be (1,0), (0, 1), and (1, 1).
Use a computer to graph the Bezier curve for w =
0, 1/2, 1, 2, 5. What happens as w increases?
8. Repeat Exercise 7 for the control points (â€”1, â€” 1),
(1,3), and (4, 1).
9. (a) Show that the Bezier curve given by the paramet-
ric equations in ( 1 ) has (x\ , y\ ) as initial point and
(x3, V3) as terminal point.
(b) Show that x(i) lies on the line segment joining
(*2, yi) to the midpoint of the line segment joining
(x\,y\) to (x 3 , y 3 ).
10. In general the control points (xi,y{), (x2,y2), and
(*3 , )?3 ) will form a triangle, known as the control poly-
gon for the curve. Assume in this problem that w > 0.
By calculating x'(0) and x'(l), show that the tangent
lines to the curve at x(0) and x(l) intersect at (X2, V2).
Hence, the control triangle has two of its sides tangent
to the curve.
11. In this problem, you will establish the geometric sig-
nificance of the constant w appearing in the equations
in(l).
(a) Calculate the distance a between x( j ) and (x2 , ^2).
(b) Calculate the distance b between x(^) and the
midpoint of the line segment joining (xi, y\) and
C*3, yj)-
(c) Show that w = b/a. By part (b) of Exercise 9,
x(|) divides the line segment joining (X2, 3^) to
the midpoint of the line segment joining (x\, y\)
to (*3, V3) into two pieces, and w represents the
ratio of the lengths of the two pieces.
12. Determine the Bezier parametrization for the portion
of the parabola y = x 2 between the points (â€”2, 4) and
(2, 4) as follows:
(a) Two of the three control points must be (â€”2, 4) and
(2, 4). Find the third control point using the result
of Exercise 10.
(b) Using part (a) and Exercise 9, we must have that
x(i) lies on the y-axis and, hence, at the point
2 P. Bezier was an automobile design engineer for Renault. See D. Cox, J. Little, and D. O'Shea, Ide-
als, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative
Algebra, 3rd ed. (Springer- Verlag, New York, 2007), pp. 28-29. Exercises 7-1 1 adapted with permission.
240 Chapter 3 | Vector- Valued Functions
(0, 0). Use the result of Exercise 11 to determine
the constant w.
(c) Now write the Bezier parametrization. You should
be able to check that your answer is correct.
1 3. Let x: (0, 7t) -*â– R 2 be the path given by
x(?) = (sin?, cos? + In tan |) ,
where ? is the angle that the y-axis makes with the
vector x(?). The image of x is called the tractrix. (See
Figure 3.49.)
(a) Show that x has nonzero speed except when ? =
n/2.
(b) Show that the length of the segment of the tangent
to the tractrix between the point of tangency and
the y-axis is always equal to 1. This means that
the image curve has the following description: Let
a horse pull a heavy load by a rope of length 1 .
of class C 2 . Use equation (17) in Â§3.2 to derive the
curvature formula
Figure 3.49 The tractrix
of Exercise 13.
Suppose that the horse initially is at (0, 0), the load
at (1, 0), and let the horse walk along the v-axis.
The load follows the image of the tractrix.
14. Another way to parametrize the tractrix path given in
Exercise 13 is
y: (-oo, O)-*- R 2 ,
where y(r)
O-P
dp
(a) Show that y satisfies the property described in part
(b) of Exercise 13.
(b) In fact, y is actually a reparametrization of part of
the path x of Exercise 13. Without proving this fact
in detail, indicate what portion of the image of x
the image of y covers.
15. Suppose that a plane curve is given in polar coordi-
nates by the equation r = f{6), where / is a function
K(9)
' + 2r' :
( r 2 _|_ r /2)3/2
(Hint: First give parametric equations for the curve in
Cartesian coordinates using 9 as the parameter.)
16. Use the result of Exercise 15 to calculate the curvature
of the lemniscate r 2 = cos 26.
Let x: / â€” > R 2 be a path of class C 2 that is not a straight line
and such that x'(?) ^ 0. Choose some to â‚¬ I and let
y(?) = x(?) - s(f)T(f),
where s(t) = f' ||x'(r)|| dr is the arclength function and T is
the unit tangent vector. The path y: / â€”> R 2 is called the invo-
lute of x. Exercises 1 7â€”19 concern involutes of paths.
17. (a) Calculate the involute of the circular path of radius
a, that is, x(?) = (a cos ?, a sin ?). (Take ?o to be 0.)
(b) Let a = 1 and use a computer to graph the path x
and the involute path y on the same set of axes.
1 8. Show that the unit tangent vector to the involute at ?
is the opposite of the unit normal vector N(?) to the
original path x. (Hint: Use the Frenet-Serret formulas
and the fact that a plane curve has torsion equal to zero
everywhere.)
19. Show that the involute y of the path x is formed by
unwinding a taut string that has been wrapped around
x as follows:
(a) Show that the distance in R 2 between a point x(?)
on the original path and the corresponding point
y(?) on the involute is equal to the distance traveled
from x(?o) to x(?) along the underlying curve of x.
(b) Show that the distance between a point x(?) on the
path and the corresponding point y(?) on the in-
volute is equal to the distance from x(?) to y(?)
measured along the tangent emanating from x(?).
Then finish the argument.
Let x: / â€” > R 2 be a path of class C 2 that is not a straight line
and such that x'(?) / 0. Let
e(?) = x(?) + -N(?).
K
This is the path traced by the center of the osculating circle of
the path x. The quantity p = \/k is the radius of the osculat-
ing circle and is called the radius of curvature of the path x.
The path e is called the evolute of the path x. Exercises 20â€”25
involve evolutes of paths.
20. Letx(?) = (?, ? 2 )beaparabolicpath.(SeeFigure3.50.)
(a) Find the unit tangent vector T, the unit normal
vector N, and the curvature k as functions of ?.
(b) Calculate the evolute of x.
Miscellaneous Exercises for Chapter 3 241
(c) Use a computer to plot x(f) and e(?) on the same
set of axes.
y
Figure 3.50 The parabola and its
osculating circle at a point. The centers
of the osculating circles at all points of
the parabola trace the evolute of the
parabola as described in Exercise 20.
21 . Show that the evolute of a circular path is a point.
22. (a) Use a computer algebra system to calculate the for-
mula for the evolute of the elliptical path x(f) =
(a cost, b sinr).
(b) Use a computer to plot x(f ) and the evolute e(f ) on
the same set of axes for various values of the con-
stants a and b. What happens to the evolute when
a becomes close in value to bl
23. Use a computer algebra system to calculate the formula
for the evolute of the cycloid x(f ) = (at â€” a sin t, a â€”
a cost). What do you find?
24. Use a computer algebra system to calculate the formula
for the evolute of the cardioid x(f) = (2a cosf(l +
a cos t), 2a sin t (1 + a cos t )).
25. Assuming ic'(t) ^ 0, show that the unit tangent vector
to the evolute e(f ) is parallel to the unit normal vector
N(f ) to the original path x(f).
26. Suppose that a C 1 path x(f ) is such that both its veloc-
ity and acceleration are unit vectors for all t . Show that
k = 1 for all t.
27. Consider the plane curve parametrized by
x ( s ) = / cos g(t)dt, y(s) = I sin g(t)dt,
Jo Jo
where g is a differentiable function.
(a) Show that the parameter s is the arclength param-
eter.
(b) Calculate the curvature k(s).
(c) Use part (b) to explain how you can create a
parametrized plane curve with any specified con-
tinuous, nonnegative curvature function k(s).
(d) Give a set of parametric equations for a curve
whose curvature k(s) = \s\. (Your answer should
involve integrals.)
(e) Use a computer to graph the curve you found in
part (d), known as a clothoid or a spiral of Cornu.
(Note: The integrals involved are known as Fres-
nel integrals and arise in the study of optics. You
must evaluate these integrals numerically in order
to graph the curve.)
28. Suppose that x is a C 3 path in R 3 with torsion r always
equal to 0.
(a) Explain why x must have a constant binormal vec-
tor (i.e., one whose direction must remain fixed for
all f).
(b) Suppose we have chosen coordinates so that x(0) =
0 and that v(0) and a(0) lie in the .ty-plane (i.e.,
have no k-component). Then what must the binor-
mal vector B be?
(c) Using the coordinate assumptions in part (b), show
that x(f) must lie in the jc;y-plane for all t . (Hint:
Begin by explaining why v(f) â€¢ k = a(r) â€¢ k = Ofor
all t. Then show that if
x(t) = x(t)i + y(t)j + z(t)k,
we must have z(t ) = 0 for all t .)
(d) Now explain how we may conclude that curves
with zero torsion must lie in a plane.
29. Suppose that x is a C 3 path in R 3 , parametrized by arc-
length, with k / 0. Suppose that the image of x lies in
the xy -plane.
(a) Explain why x must have a constant binormal
vector.
(b) Show that the torsion r must always be zero.
Note that there is really nothing special about the im-
age of x lying in the Jfj-plane, so that this exercise,
combined with the results of Exercise 28, shows that
the image of x is a plane curve if and only if r is always
zero and if and only if B is a constant vector.
30. In Example 7 of Â§3.2 we saw that if x is a straight-line
path, then x has zero curvature. Demonstrate the con-
verse; that is, if x is a C 2 path parametrized by arc-
length s and has zero curvature for all s, then x traces
a straight line.
31 . A large piece of cylindrical metal pipe is to be manu-
factured to include a strake, which is a spiraling strip
of metal that offers structural support for the pipe. (See
Figure 3.51.) The pieces of the strake are to be made
from flat pieces of flexible metal whose curved sides
are arcs of circles as shown in Figure 3.52. Assume that
242 Chapter 3 | Vector- Valued Functions
the pipe has a radius of a ft and that the strake makes
one complete revolution around the pipe every h ft. 3
Figure 3.51 A cylindrical Figure 3.52 A section of
pipe with strake attached. the strake. (See Exercise 3 1 .)
(a) In terms of a and h, what should the inner radius r
be so that the strake will fit snugly against the pipe?
(b) Suppose a = 3 ft and h = 25 ft. What is r?
Suppose that x: / â€” > R 3 is a path of class C 3 parametrized by
arclength. Then the unit tangent vector T(s) defines a vector-
valued function T: / â€”> R 3 that may also be considered to be a
path (although not necessarily one parametrized by arclength,
nor necessarily one with nonvanishing velocity). Since T is a
unit vector, the image of the path T must lie on a sphere of
radius 1 centered at the origin. This image curve is called the
tangent spherical image of x. Likewise, we may consider the
functions defined by the normal and binormal vectors N and B
to give paths called, respectively, the normal spherical image
and binormal spherical image of x. Exercises 32-35 concern
these notions.
32. Find the tangent spherical image, normal spherical
image, and binormal spherical image of the circular
helix x(/) = (a cosf , a sinf , bt). (Note: The path x is
not parametrized by arclength.)
33. Suppose that x is parametrized by arclength. Show
that x is a straight-line path if and only if its tangent
spherical image is a constant path. (See Example 7 of
Â§3.2 and Exercise 30.)
34. Suppose that x is parametrized by arclength. Show that
the image of x lies in a plane if and only if its binormal
spherical image is constant. (See Exercises 28 and 29.)
35. Suppose that x is parametrized by arclength. Show
that the normal spherical image of x can never be
constant.
36. In this problem, we will find expressions for velocity
and acceleration in cylindrical coordinates. We begin
with the expression
x(t) = x(t)i + y(t)j + z(t)k
for the path in Cartesian coordinates.
(a) Recall that the standard basis vectors for cylindri-
cal coordinates are
e,- = cosf? i + s'md j,
ee = â€” s'md i + cos# j,
e, = k.
Use the facts that x = r cos 0 and y = r sin 0 to
show that we may write x(r ) as
x(t) = r(t)e r +z(t)e z .
(b) Use the definitions of e r , ee, and e z just given and
the chain rule to find de r /dt, deg/dt, and de./dt
in terms of e r , eg , and e z .
(c) Now use the product rule to give expressions for v
and a in terms of the standard basis for cylindrical
coordinates.
37. Suppose that the path
x(f) = (sin2f, Vlcoslt, sin2f â€” 2)
describes the position of the Starship Inertia at time t .
(a) Lt. Commander Agnes notices that the ship is trac-
ing a closed loop. What is the length of this loop?
(b) Ensign Egbert reports that the Inertia's path is
actually a flow line of the Martian vector field
F(jc, y, z) = yi â€” 2x\ + yk, but he omitted a con-
stant factor when he entered this information in
his log. Help him set things right by finding the
correct vector field.
38. Suppose that the temperature at points inside a room is
given by a differentiable function T(x, y, z). Livinia,
the housefly (who is recovering from a head cold), is in
the room and desires to warm up as rapidly as possible.
(a) Show that Livinia 's path x(?) must be a flow line
of kVT, where & is a positive constant.
(b) If T(x, y, z) = x 2 â€” 2y 2 + 3z 2 and Livinia is ini-
tially at the point (2,3,-1), describe her path
explicitly.
39. Let F = u(x, y)i â€” v(x, y)j be an incompressible,
irrotational vector field of class C 2 .
(a) Show that the functions u and v (which deter-
mine the component functions of F) satisfy the
Cauchy-Riemann equations
9m 3d du dv
â€” = â€” , and â€” = .
dx dv dy ox
3 See F. Morgan, Riemannian Geometry: A Beginner's Guide, 2nd ed. (A K Peters, Wellesley, 1998),
pp. 7-10. Figures 3.51 and 3.52 adapted with permission.
Miscellaneous Exercises for Chapter 3 243
(b) Show that u and v are harmonic, that is, that
(Also see Â§1.4 concerning the notion of torque.) Show
that
d 2 u d 2 U
9^ + 9^2
40. Suppose that a particle of mass m travels along a path
x according to Newton's second law F = ma, where
F is a gradient vector field. If the particle is also con-
strained to lie on an equipotential surface of F, show
that then it must have constant speed.
41 . Let a particle of mass m travel along a differentiable
path x in a Newtonian vector field F (i.e., one that
satisfies Newton's second law F = ma, where a is the
acceleration of x). We define the angular momen-
tum 1(f) of the particle to be the cross product of the
position vector and the linear momentum mv, that is,
l(r) = x(f) x m\(t).
(Here v denotes the velocity of x.) The torque about
the origin of the coordinate system due to the force F
is the cross product of position and force:
M(f ) = x(f) x F(f) = x(f) x ma(f).
dt
M.
Thus, we see that the rate of change of angular mo-
mentum is equal to the torque imparted to the particle
by the vector field F.
42. Consider the situation in Exercise 41 and suppose that
F is a central force (i.e., a force that always points
directly toward or away from the origin). Show that in
this case the angular momentum is conserved, that is,
that it must remain constant.
43. Can the vector field
F = (e x cos y + e~ x sin z) i â€” e x sin y j + e~ x cos z k
be the gradient of a function f(x, y, z) of class C 2 ?
Why or why not?
44. Can the vector field
F = *(y 2 + \)i + (ye x - e z ) j + x 2 e z k
be the curl of another vector field G(x, y, z) of class
C 2 ? Why or why not?
4 Maxima and Minima
in Several Variables
4.1 Differentials and Taylor's
Theorem
4.2 Extrema of Functions
4.3 Lagrange Multipliers
4.4 Some Applications of
Extrema
True/False Exercises for
Chapter 4
Miscellaneous Exercises
for Chapter 4
4.1 Differentials and Taylor's Theorem
Among all classes of functions of one or several variables, polynomials are without
a doubt the nicest in that they are continuous and differentiable everywhere and
display intricate and interesting behavior. Our goal in this section is to provide
a means of approximating any scalar-valued function by a polynomial of given
degree, known as the Taylor polynomial. Because of the relative ease with which
one can calculate with them, Taylor polynomials are useful for work in computer
graphics and computer-aided design, to name just two areas.
Taylor's Theorem in One Variable: A Review
Suppose you have a function /:lcR->R that is differentiable at a point a in
X. Then the equation for the tangent line gives the best linear approximation for
/ near a. That is, when we define p\ by
p\{x) = f(a) + f'(a)(x â€” a), we have pi(x) & fix) if x % a.
(See Figure 4.1.) As explained in Â§2.3, the phrase "best linear approximation"
means that if we take R\(x, a) to be f(x) â€” p\(x), then
lim
Ri(x, a)
0.
Note that, in particular, we have pi(a) = f(a) and p[(a) = f'(a).
Generally, tangent lines approximate graphs of functions only over very small
neighborhoods containing the point of tangency. For a better approximation, we
might try to fit a parabola that hugs the function's graph more closely as in
Figure 4.2. In this case, we want p 2 to be the quadratic function such that
p 2 (a) = f(a), p' 2 (a) = f'(a), and p' 2 \a) = f"(a).
The only quadratic polynomial that satisfies these three conditions is
P2 (x) = f(a) + f(a){x -a)+ ^(x - a) 2 .
It can be proved that, if / is of class C 2 , then
fix) = p 2 (x)+ R 2 (x, a),
4.1 | Differentials and Taylor's Theorem 245
where
Figure 4.3
fix) = In*
y = 2-
Approximations to
Riix, a)
lim = 0.
*-Â»-a ix â€” a) 1
EXAMPLE 1 If fix) = In x, then, for a = 1 , we have
/(l) = lnl=0,
1
I
1
12
/'(I)
/"(I) =
1.
1.
Hence,
piix) = 0 + 1(*
p 2 (x) = 0+ 1(jc
1) = *- 1,
1) - I(* - l) 2 =
2x
The approximating polynomials p\ and p2 are shown in Figure 4.3. â™¦
There is no reason to stop with quadratic polynomials. Suppose we want to
approximate / by a polynomial of degree k, where k is a positive integer.
Analogous to the work above, we require that p k and its first k derivatives agree
with / and its first k derivatives at the point a. Thus, we demand that
Pk(a) =
f{a),
Pk( a ) =
f'(a),
P'l(a) =
pf (Â«) = / w (a).
Given these requirements, we have only one choice for p k , stated in the following
theorem:
- f(k)(
THEOREM 1.1 (Taylor's theorem in one variable) Let X be open in
R and suppose /:XcR^R is differentiable up to (at least) order k.
246 Chapter 4 I Maxima and Minima in Several Variables
Given a e X, let
f"(a) f( k Xa)
p k (x) = f{a) + f'(a)(x -a)+ J -^-(x - af + â€¢ â€¢ â€¢ + J â€”^-{x - af. (1)
2 k\
Then
f(x) = Pk(x)+ Rk{x,a),
where the remainder term R k is such that R k (x, a)/(x â€” af -Â¥ 0 as x â€”* a.
Figure 4.4 The graphs of
(1) y =x -a,
(2) y = (x â€” a) 2 , and
(3) y = (x- af.
Note how much more closely the
graph of (3) hugs the jc-axis than
thatof(l)or(2).
The polynomial defined by formula (1) is called the &th-order Taylor poly-
nomial of / at a. The essence of Taylor's theorem is this: For x near a, the Taylor
polynomial p k approximates / in the sense that the error R k involved in making
this approximation tends to zero even faster than (x â€” af does. When k is large,
this is very fast indeed, as we see graphically in Figure 4.4.
EXAMPLE 2 Consider In x with a = 1 again. We calculate
/(l) = lnl = 0,
/'(l) = y = l>
/"(I)
1
12
1*
i^l = (-l) k+ \k - 1)!.
Therefore,
Pk(x) = (x
lf + - 3 (x
If
+
(-1)
~~* 0 as
x â€” > a.
To obtain the second-order formula, the case k = 2 of (1), we focus on
Ri(x, a) = f*(x â€” t)f"{t)dt and integrate by parts again, this time with u =
f"(t) and v = (x - tf/2, so that dv = -(x - t)dt. We obtain
f\x
J a
t)f"(t)dt =
f"(t)(x - tf
f"(a)(x - a) 2
a Ja
f
f"'(t)dt
(x - tf
f"'(t)dt.
Hence (13) becomes
f"(a)
f(x) = f(a) + f'(a)(x -a)+ J â€”^-{x
2 r* (x-t)
a) + L â€”
Therefore, we have shown, when / is differentiable up to (at least) third order,
that
-Jlf"\t)dt.
R 2 (x,a)= f
J a
We can continue to argue in this manner or use mathematical induction to show
that formula (1) holds in general with
R k (x,a) = j"^t^\t)dt, (14)
assuming that / is differentiable up to order (at least) k+\.
It remains to see that Rk(x, a)/(x â€” a) k â€” > 0 as x â€” > a. In formula (14) we
are only considering t between a and x, so that \x â€” t \ < \x â€” a\. Moreover, since
we are assuming that / is of class C k+1 , we have that f (k+x \t) is continuous and
therefore, bounded for t between a and x (i.e., that \f {k+l \t)\ < M for some
constant M). Thus,
\R k (x,a)\ <
r (x - tf
Ja k\
f (k+ \t)dt
< Â±
f
(x - tf
k\
f ik+l) (t)
dt,
where the plus sign applies if x > a and the negative sign if x < a,
C x M
x - a\ k dt
M
\k+\
Thus,
Rk(x, a)
(x â€” a) k
M
as x â€”> a, as desired.
260 Chapter 4 I Maxima and Minima in Several Variables
Proof of Proposition 1.2 We establish Proposition 1.2 by means of a general
version of the mean value theorem for integrals. This theorem states that for
continuous functions g and h such that h does not change sign on [a, b] (i.e.,
either h(t) > 0 on [a, b] or h(t) < 0 on [a, b]), there is some number z between
a and b such that
f g(t)h(t)dt = g(z) f h(t)dt.
(We omit the proof but remark that this theorem is a consequence of the interme-
diate value theorem.) Applying this result to formula (14) with g(t) = f (k+ ^(t)
and h(t) = (x â€” t) k /k\, we find that there must exist some z between a and x
such that
i) = f (k+ \z)\ X{ ^-^dt=f^\z)
k\
(x - t)
(k+iy.
(k + iy.
(x â€” a)
k+l
Proof of Theorem 1.5 As in the proof of Theorem 1 . 1 , we establish Theorem 1 .5
under the stronger assumption that / is of class C 3 . Begin by setting h = x â€” a,
so that x = a + h, and consider a and h to be fixed. We define the one-variable
function F by F(t) = /(a + th). Since / is assumed to be of class C 3 on an
open set X, if we take x sufficiently close to a, then F is of class C 3 on an open
interval containing [0, 1]. Thus, Theorem 1.1 with k = 2, a = 0, and x = 1 may
be applied to give
F(l) = F(0) + F'(0)(1 - 0) + ^9(1 - 0) 2 + R 2 (l, 0)
F"(0)
= F(0) + F'(0) + â€” ^ + R 2 (l, 0),
(15)
where R 2 (l, 0) = /J ^-F"'(0^.Now we use the chain rule to calculate deriva-
tives of F in terms of partial derivatives of /:
.7 = 1
hi = ^ f XiXj (a + th)hihj;
F'(t) = D/(a + th)h = J2 U â– (Â» + fh )^
F"(r) = Â£
F"'(t) =
k=\
Thus, (15) becomes
n j n
/(a + h) = /(a) + fx,Wi + ~ E UM) h i h J
J2 /u,,, ; (a â€¢ /h)/;,/;,
.W=i
fyfc = X! fxiXjxM + t^hihjhk
i,j,k=\
+
n 1
(l - o 2
f XiXjXk (a + th)hihjh k dt,
4.1 | Differentials and Taylor's Theorem 261
or, equivalently,
n \ n
/(x) = /(a) + /*,(Â»)(*; " ad +~Y1 fxitjW&i ~ a i)( x i ~ a i)
+ # 2 (x, a),
where the multivariable remainder is
R 2 (x,a)= V / y â€”â€” L f XiX]Xl (a + th)h i hjh k dt.
>,j,k=\ Jo l
(16)
We must still show that |i?2(x, a)|/||x â€” a|| 2 -> 0 as x -> a, or, equivalently,
that |^?2(x, a)|/|h|| 2 -> 0 as h -> 0. To demonstrate this, note that, for a and h
fixed the expression (1 â€” t) 2 f XjXjXk (st + fh) is continuous for f in [0, 1] (since /
is assumed to be of class C 3 ), hence bounded. In addition, for i = 1, . . . , n, we
have that |/z,-| < ||h||.
Hence,
|tf 2 (x,a)| =
<
;i 1
<
n Â«1
(i - o 2
2
(1 " tf
fxtXjxM + th)hihjh k dt
f XiXjXt (a + th)hjhjh k dt
M||/i|| 3 dt = n 3 M||h|| 3 = n 3 M||x - a|
Thus,
|i? 2 (x, a)| ,
v '1 <Â« 3 M||x-a|
x- a
0
as x -> a.
Finally, we remark that entirely similar arguments may be given to establish
results for Taylor polynomials of orders higher than two. â–
Lagrange's formula for the remainder (see page 257) Using the function
F(t) = /(a + fh) defined in the proof of Theorem 1.5, Proposition 1.2 implies
that there must be some number c between 0 and 1 such that the one-variable
remainder is
F"'(c) -
tf 2 (l,0)=^-Al-0) 3 .
Now, the remainder term Rz(I, 0) from Proposition 1.2 is precisely ^(x, a) in
Theorem 1.5 and
n n
F"'(c)= f XiXiXt {* + c\v)hihjh k = f XiXjXk (z)hihjh k ,
i,j,k=\ i,j,k=l
where z = a + ch. Since c is between 0 and 1 , the point z lies on the line segment
joining a and x = a + h, and so
1
^2(x, a)=â€” f XiXjXk (z)hihjh k
i,j,k=l
which is the result we desire. The derivation of the formula for R k (x, a) for k > 2
is analogous. â–
262 Chapter 4 I Maxima and Minima in Several Variables
4.1 Exercises
In Exercises 1â€”7, find the Taylor polynomials of given order
k at the indicated point a.
1. f( x ) = e 2x ,a = 0,k = A
2. fix) = ln(l + x), a = 0,k = 3
3. f(x)= l/x 2 ,a = l,k = 4
4. f(x) = Jx,a = l,k = 3
5. f(x) = +2x > + - + " x Â».
(a) Calculate D/(0, 0, . . . , 0) and Hf(0, 0, . . . , 0).
(b) Determine the first- and second-order Taylor poly-
nomials of / at 0.
(c) Use formulas (3) and ( 10) to write the Taylor poly-
nomials in terms of the derivative and Hessian
matrices.
26. Find the third-order Taylor polynomial pi(x, y, z) of
f(x,y,z) = e x+2y+3z
at (0, 0, 0).
27. Find the third-order Taylor polynomial of
f{x, y, z) = x 4 + x 3 y + 2y 3 - xz 2 + x 2 y + 3xy - z + 2
(a) at (0, 0, 0).
(b) at (1,-1,0).
Determine the total differential of the functions given in
Exercises 28â€”32.
28. /(x,y) = x 2 y 3
29. f{x, y, z) = x 2 + 3y 2 - 2z 3
30. fix, y, z) = cos (xyz)
31. fix, y, z) = e x cosy + e y sin z
32. fix, y, z) = l/^Jxyz
33. Use the fact that the total differential df approximates
the incremental change A/ to provide estimates of the
following quantities:
(a) (7.07) 2 (1.98) 3
(b) l/y(4.1)(1.96)(2.05)
(c) (1. 1) cos ((jr -0.03X0.12))
34. Near the point (1, â€”2, 1), is the function g(x, y, z) =
x 3 â€” 2xy + x 2 z + 7z most sensitive to changes in x,
y, or z?
35. To which entry in the matrix is the value of the
determinant
2 3
-1 5
most sensitive?
4.2 | Extrema of Functions 263
36. If you measure the radius of a cylinder to be 2 in, with
a possible error of Â±0. 1 in, and the height to be 3 in,
with a possible error of Â±0.05 in, use differentials to
determine the approximate error in
(a) the calculated volume of the cylinder.
(b) the calculated surface area.
37. A can of mushrooms is currently manufactured to have
a diameter of 5 cm and a height of 12 cm. The man-
ufacturer plans to reduce the diameter by 0.5 cm. Use
differentials to estimate how much the height of the
can would need to be increased in order to keep the
volume of the can the same.
38. Consider a triangle with sides of lengths a and b that
make an interior angle 0 .
(a) If a = 3, b = 4, and 0 = tt/3, to changes in which
of these measurements is the area of the triangle
most sensitive?
(b) If the length measurements in part (a) are in error
by as much as 5% and the angle measurement is
in error by as much as 2%, estimate the resulting
maximum percentage error in calculated area.
39. To estimate the volume of a cone of radius approx-
imately 2 m and height approximately 6 m, how ac-
curately should the radius and height be measured so
that the error in the calculated volume estimate does
not exceed 0.2 m 3 ? Assume that the possible errors in
measuring the radius and height are the same.
40. Suppose that you measure the dimensions of a block
of tofu to be (approximately) 3 in by 4 in by 2 in.
Assuming that the possible errors in each of your mea-
surements are the same, about how accurate must your
measurements be so that the error in the calculated
volume of the tofu is not more than 0.2 in 3 ? What per-
centage error in volume does this represent?
41 . (a) Calculate the second-order Taylor polynomial for
f(x, y) = cosx siny at the point (0, ?r/2).
(b) If h = (h u h 2 ) = (x, y) - (0, tt/2) is such that
\hi \ and \h 2 \ are no more than 0.3, estimate how
accurate your Taylor approximation is.
42. (a) Determine the second-order Taylor polynomial of
f(x, y) = e x+2y at the origin.
(b) Estimate the accuracy of the approximation if \x\
and \ y\ are no more than 0.1.
43. (a) Determine the second-order Taylor polynomial of
f(x, y) = e 2x cos y at the point (0, rr/2).
(b) If h = (h u h 2 ) = (x, y) - (0, tt/2) is such that
\h\\ < 0.2 and \h 2 \ < 0.1, estimate the accuracy
of the approximation to / given by your Taylor
polynomial in part (a).
4.2 Extrema of Functions
The power of calculus resides at least in part in its role in helping to solve a wide
variety of optimization problems. With any quantity that changes, it is natural to
ask when, if ever, does that quantity reach its largest, its smallest, its fastest or
slowest? You have already learned how to find maxima and minima of a function
of a single variable, and no doubt you have applied your techniques to a number of
situations. However, many phenomena are not appropriately modeled by functions
of only one variable. Thus, there is a genuine need to adapt and extend optimization
methods to the case of functions of more than one variable. We develop the
necessary theory in this section and the next and explore a few applications in Â§4.4.
Critical Points of Functions
Let X be open in R" and /:XcR"^Ra scalar- valued function.
DEFINITION 2.1 We say that / has a local minimum at the point a in
X if there is some neighborhood U of a such that f(x) > /(a) for all x
in U. Similarly, we say that / has a local maximum at a if there is some
neighborhood U of a such that /(x) < /(a) for all x in U.
Figure 4.1 2 The graph of When n = 2, local extrema of f(x, y) are precisely the pits and peaks of the
z = f(x, y). surface given by the graph of z = f(x, y), as suggested by Figure 4.12.
Max.
264 Chapter 4 I Maxima and Minima in Several Variables
We emphasize our use of the adjective "local." When a local maximum of
a function / occurs at a point a, this means that the values of / at points near
a can be no larger, not that all values of / are no larger. Indeed, / may have
local maxima and no global (or absolute) maximum. Consider the graphs in
Figure 4.13. (Of course, analogous comments apply to local and global minima.)
Figure 4.1 3 Examples of local and global maxima.
Recall that, if a differentiable function of one variable has a local extremum
at a point, then the derivative vanishes there (i.e., the tangent line to the graph
of the function is horizontal). Figures 4.12 and 4.13 suggest strongly that, if a
function of two variables has a local maximum or minimum at a point in the
domain, then the tangent plane at the corresponding point of the graph must be
horizontal. Such is indeed the case, as the following general result (plus formula
(4) of Â§2.3) implies.
THEOREM 2.2 Let X be open in R" and let /:XcR"^R be differentiable.
If / has a local extremum at a 6 X, then Df(a) = 0.
PROOF Suppose, for argument's sake, that / has a local maximum at a. Then the
one-variable function F defined by F(t) = /(a + 1 h) must have a local maximum
at t = 0 for any h. (Geometrically, the function F is just the restriction of / to the
line through a parallel to h as shown in Figure 4. 14.) From one-variable calculus,
we must therefore have F'(0) = 0. By the chain rule
F'(t) = -[/(a + th)] = D/(a + th)h = V/(a + th) â€¢ h.
at
1 \
1 k
1 / \
1 1
1 / 1
1
1 J
Graph of /
restricted to line
Figure 4.1 4 The graph of / restricted to a line.
4.2 | Extrema of Functions 265
Hence,
/-o
X/ 0 \
\ f>0/
\ 1
\ /
\ /
\ /
\ /
\ /
Figure 4.1 5 The function / is
strictly positive on the shaded
region, strictly negative on the
unshaded region, and zero along
the lines y = Â±x.
0 = F'(0) = D/(a)h = f x ^)h x + / a (a)A 2 + â€¢ â€¢ â€¢ + Aâ€ž(a)fcâ€ž.
Since this last result must hold for all h e R", we find that by setting h in turn
equal to (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1), we have
f Xl {*)= = -â€¢â€¢ = /*â€ž (a) = 0.
Therefore, Df(a) = 0, as desired. â–
A point a in the domain of / where Df(a) is either zero or undefined is called
a critical point of /. Theorem 2.2 says that any extremum of / must occur at a
critical point. However, it is by no means the case that every critical point must
be the site of an extremum.
EXAMPLE 1 If f(x, y) = x 2 - y 2 , then Df(x, y) = [ 2x -2y ] so that,
clearly, (0, 0) is the only critical point. However, neither a maximum nor a mini-
mum occurs at (0, 0). Indeed, inside every open disk centered at (0, 0), no matter
how small, there are points for which f(x, y) > f(0, 0) = 0 and also points where
f(x, y) < /(O, 0). (See Figure 4.15.) â™¦
This type of critical point is called a saddle point. Its name derives from the
fact that the graph of z = f(x , y) looks somewhat like a saddle. (See Figure 4.16.)
Figure 4.1 6 A saddle point.
EXAMPLE 2 Let f(x, y) = ^x 1 + y 2 . The domain off is all of R 2 . We com-
2x 2y
pute that Df(x, y)
; note that Df is unde-
3( x 2 + y 2)2/3 3( x 2 + _ y 2)2/3 _
fined at (0, 0) and nonzero at all other (x, y) e R 2 . Hence, (0, 0) is the only
critical point. Since f(x, y) > 0 for all (x, y) and has value 0 only at (0, 0), we
see that / has a unique (global) minimum at (0, 0). â™¦
The Nature of a Critical Point: The Hessian Criterion
We illustrate our current understanding regarding extrema with the following
example:
EXAMPLE 3 We find the extrema of
f(x, y) = x 2 + xy + y 2 + 2x- 2y + 5.
266 Chapter 4 | Maxima and Minima in Several Variables
/(Â«)
Figure 4.17 An
upward-opening parabola.
Figure 4.18 A
downward-opening parabola.
Since / is a polynomial, it is differentiable everywhere, and Theorem 2.2 implies
that any extremum must occur where df/dx and df/dy vanish simultaneously.
Thus, we solve
Bf
dx
Bf
dy
= 2x + y + 2 = 0
= x+2y-2 = 0
and find that the only solution is x = â€”2, y = 2. Consequently, (â€”2, 2) is the
only critical point of this function.
To determine whether (â€”2, 2) is a maximum or minimum (or neither), we
could try graphing the function and drawing what we hope would be an obvious
conclusion. Of course, such a technique does not extend to functions of more than
two variables, so a graphical method is of limited value at best. Instead we'll see
how / changes as we move away from the critical point:
A/ = /(-2 + A,2 + *)-/(-2,2)
= [(-2 + hf + (-2 + h)(2 + k) + {2 + kf
+ 2(-2 + /!)-2(2 + Â£) + 5]- 1
= h 2 + hk + k 2 .
If the quantity Af = h 2 + hk + k 2 is nonnegative for all small values of h and k,
then (â€”2, 2) yields a local minimum. Similarly, if Af is always nonpositive, then
(â€”2, 2) must yield a local maximum. Finally, if Af is positive for some values
of h and k and negative for others, then (â€”2, 2) is a saddle point. To determine
which possibility holds, we complete the square:
Af
h 2 + hk + k 2
h 2 + hk +
,k + . k
(h + \kf + Ik 2 .
Thus, Af > 0 for all values of h and k, so (â€”2, 2) necessarily yields a local
minimum. â™¦
Example 3 with its attendant algebra clearly demonstrates the need for a better
way of determining when a critical point yields a local maximum or minimum (or
neither). In the case of a twice differentiable function /:XcR^R, you already
know a quick method namely, consideration of the sign of the second derivative.
This method derives from looking at the second-order Taylor polynomial of /
near the critical point a, namely,
f"(a) ,
f(x) Â« p 2 (x) = f(a) + / (a)(x -a)-\ â€” (x - a)
= /(a)H ^â€”(x ~ a) ,
since /' is zero at the critical point a of /. If f"{a) > 0, the graph of y = pi(x)
is an upward-opening parabola, as in Figure 4.17, whereas if f"{a) < 0, then
the graph of y = pi(x) looks like the one shown in Figure 4.18. If f"(a) = 0,
then the graph of y = pi(x) is just a horizontal line, and we would need to use
a higher-order Taylor polynomial to determine if / has an extremum at a. (You
may recall that when f"(a) = 0, the second derivative test from single-variable
calculus gives no information about the nature of the critical point a.)
The concept is similar in the context of n variables. Suppose that
/(*)= f(xi,x 2 , . . . ,x n )
4.2 | Extrema of Functions 267
is of class C 2 and that a = (a\, a 2 , . . . , aâ€ž) is a critical point of /. Then the
second-order Taylor approximation to / gives
a/ = /(x) - m Â« P2 (x) - m
= Z)/(a)(x - a) + i(x - a) r ///(a)(x - a)
when x a. (See Theorem 1.5 and formula (10) in Â§4.1.) Since / is of class C 2
and a is a critical point, all the partial derivatives vanish at a, so that we have
D/(a) = 0 and hence,
A/Â«i(x-a) r ///(a)(x-a).
(1)
The approximation in (1) suggests that we may be able to see whether the in-
crement A / remains positive (respectively, remains negative) for x near a and
hence, whether / has a local minimum (respectively, a local maximum) at a by
seeing what happens to the right side.
Note that the right side of (1), when expanded, is quadratic in the terms
(xj â€” cii). More generally, a quadratic form in h\, h 2 , â– . . , hâ€ž is a function Q
that can be written as
n
Q(h u h 2 , h n ) = ^ bijhihj,
where the bij 's are constants. The quadratic form Q can also be written in terms
of matrices as
Q(h)=[h l h 2
~ b u
b\ 2 â–
â– blâ€ž
~ h x ~
hâ€ž]
b 2 \
b 22 â–
â– b 2n
h 2
_ K\
bnl â–
â– h
u nn -
_ K _
h r fih,
(2)
where B = (pij). Note that the function Q is unchanged if we replace all bjj with
^{bij + bjj). Hence, we may always assume that the matrix B associated to Q is
symmetric, that is, that bjj = bj, (or, equivalently, that B T = B). Ignoring the
factor of 1 /2, we see that the right side of (1) is the quadratic form in h = x â€” a,
corresponding to the matrix B = Hf(a).
A quadratic form Q (respectively, its associated symmetric matrix B) is said
to be positive definite if Q(h) > 0 for all h ^ 0 and negative definite if Q(h) < 0
for all h ^ 0. Note that if Q is positive definite, then Q has a global minimum (of
0) at h = 0. Similarly, if Q is negative definite, then Q has a global maximum at
h = 0.
The importance of quadratic forms to us is that we can judge whether / has
a local extremum at a critical point a by seeing if the quadratic form in the right
side of ( 1) has a maximum or minimum at x = a. The precise result, whose proof
is given in the addendum to this section, is the following:
THEOREM 2.3 Let U c R" be open and /: U R a function of class C 2 .
Suppose that a Â£ U is a critical point of /.
1. If the Hessian Hf(a) is positive definite, then / has a local minimum at a.
2. If the Hessian Hf(a) is negative definite, then / has a local maximum at a.
3. If det Hf(a) 0 but Hf(a) is neither positive nor negative definite, then /
has a saddle point at a.
268 Chapter 4 I Maxima and Minima in Several Variables
In view of Theorem 2.3, the issue thus becomes to determine when the Hessian
Hf(a) is positive or negative definite. Fortunately, linear algebra provides an
effective means for making such a determination, which we state without proof.
Given a symmetric matrix B (which, as we have seen, corresponds to a quadratic
form Q), let B^, for k = 1, . . . , n, denote the upper leftmost k x k submatrix of
B. Calculate the following sequence of determinants:
bn b 22
det 5,
det B 7
det B 3 =
bn b i2 bn
bi\ bi2 bn
b-i\ b i2 b 3i
detÂ£â€ž = det 5.
If this sequence consists entirely of positive numbers, then B and Q are positive
definite. If this sequence is such that det B^ < 0 for k odd and det B^ > 0 for k
even, then B and Q are negative definite. Finally, if det B ^ 0, but the sequence
of determinants det B\ , det B 2 , . . . , det Bâ€ž is neither of the first two types, then B
and Q are neither positive nor negative definite. Combining these remarks with
Theorem 2.3, we can establish the following test for local extrema:
Second derivative test for local extrema. Given a critical point a of a func-
tion / of class C 2 , look at the Hessian matrix evaluated at a:
tf/(a)
/*,*,(Â») fx lX2 (*) â– â– â– /*,*â€ž(Â»)
f X2 xM) fx 2 x 2 (a) â€¢â€¢â€¢ /x 2 xâ€ž(a)
From the Hessian, calculate the sequence of principal minors of Hf(a).
This is the sequence of the determinants of the upper leftmost square sub-
matrices of Hf(a). More explicitly, this is the sequence d\, d 2 , . . . , d n ,
where dt = det Ht, and Hk is the upper leftmost k x k submatrix of Hf(a).
That is,
d\
di
/x,x,(a),
/Â«Â«(Â»)
/v!* 2 (a)
fx 2 xM)
fx 2 x 2 (&)
1
fxixM)
f xl xM
f xl xM
fx 2 xM)
fx 2 xM
fx 2 xM
/v 3 *,(a)
/v 3 * 2 (a)
inu(a)
di= fx 2 xM) fx 2 x 2 (a) fx 2 x 3 (a) , ...,dâ€ž = \Hf(a)\.
The numerical test is as follows:
Assume that dâ€ž = det Hf(a) ^ 0.
1. If dk > 0 for k = 1 , 2, . . . , n, then / has a local minimum at a.
2. If dk < 0 for k odd and dk > 0 for k even, then / has a local maximum
at a.
3. If neither case 1 nor case 2 holds, then / has a saddle point at a.
In the event that det ///(a) = 0, we say that the critical point a is degenerate
and must use another method to determine whether or not it is the site of an
extremum of /.
4.2 | Extrema of Functions 269
EXAMPLE 4 Consider the function
fxx
fxy
2 1
. fyx
fyy _
1 2
minors
is d
i = /Â«(-2, 2)
/(*, y) = x 2 + xy + y 2 + 2x - 2y + 5
in Example 3. We have already seen that (â€”2, 2) is the only critical point. The
Hessian is
Hf(x, y) =
\Hf(â€”2, 2)| = 3 (> 0). Hence, / has a minimum at (â€”2, 2), as we saw before,
but this method uses less algebra. â™¦
EXAMPLE 5 (Second derivative test for functions of two variables) Let us
generalize Example 4. Suppose that f(x, y) is a function of two variables of class
C 2 and further suppose that / has a critical point at a = (a, b). The Hessian
matrix of / evaluated at {a, b) is
Hf(a,b)
f xx (a,b) f xy (a,b)
f xy (a,b) f yy (a,b)
Note that we have used the fact that f xy = f yx (since / is of class C 2 ) in con-
structing the Hessian. The sequence of principal minors thus consists of two
numbers:
d\ = f xx (a,b) and d 2 = f xx (a, b)f yy (a, b) - f xy (a, bf .
Hence, in this case, the second derivative test tells us that
1. / has a local minimum at (a, b) if
f xx (a, b) > 0 and f xx (a, b)f yy (a, b) - f xy (a, bf > 0.
2. / has a local maximum at {a, b) if
f xx (a, b) < 0 and f xx (a, b)f yy (a, b) - f xy (a, bf > 0.
3. / has a saddle point at (a, b) if
fxxia, b)f yy (a, b) - f xy (a, bf < 0.
Note that if f xx (a, b)f yy (a, b) â€” f xy (a, bf = 0, then / has a degenerate critical
point at (a, b) and we cannot immediately determine if {a, b) is the site of a local
extremum of /. â™¦
EXAMPLE 6 Let f(x, y, z) = x 3 + xy 2 + x 2 + y 2 + 3z 2 . To find any local
extrema of /, we must first identify the critical points. Thus, we solve
Df(x,y,z)= [3x 2 + y 2 + 2x 2xy + 2y 6z] = [0 0 0] .
From this, it is not hard to see that there are two critical points: (0, 0, 0) and
(-Â§, 0,0). The Hessian of/ is
Hf(x,y,z) =
6x + 2
2y
o
2y
2x + 2
0
270 Chapter 4 I Maxima and Minima in Several Variables
At the critical point (0, 0, 0), we have
Hf(0, 0, 0) =
2 0 0
0 2 0
0 0 6
and its sequence of principal minors is d\ =2, d 2 = 4, dj = 24. Since these
determinants are all positive, we conclude that / has a local minimum at (0,0,0).
At (-Â§,0,0), we calculate that
ff/(--,0,0
-2 0 0
0 | 0
0 0 6
The sequence of minors is â€”2, â€” |, â€” 8. Hence, / has a saddle point at (â€” |, 0, 0).
+
h>0
+
+
+
1 â€”
-1
0
+
1
1
+
+
+
Figure 4.1 9 Away from the
origin, the function h of Example 7
is negative along the x-axis and
positive along the _v-axis.
EXAMPLE 7 To get a feeling for what happens in the case of a degenerate
critical point (i.e., a critical point a such that det Hf(a) = 0), consider the three
functions
/(x,y) = x 4 + x 2 + y 4 ,
g(x, y) = -x 4 - x 2
and
h(x, y) = x
x 2 + y 4 .
We leave it to you to check that the origin (0, 0) is a degenerate critical point
of each of these functions. (In fact, the Hessians themselves look very similar.)
Since / is 0 at (0, 0) and strictly positive at all (x, y) ^ (0, 0), we see that /
has a strict minimum at the origin. Similar reasoning shows that g has a strict
maximum at the origin. For h, the situation is slightly more complicated. Along
the y-axis, we have h(0, y) = y 4 , which is zero at y = 0 (the origin) and strictly
positive everywhere else. Along the x-axis,
h(x,0) = x 4
x 2 = x 2 (x
1)(JC + 1).
For â€” 1 < x < 1 and x ^ 0, h(x, 0) < 0. We have the situation depicted in Fig-
ure 4. 19. Thus, every neighborhood of (0, 0) contains some points (x, y) where h
is positive and also some points where h is negative. Therefore, h has a saddle point
at the origin. The "moral of the story" is that a degenerate critical point can exhibit
any type of behavior, and more detailed consideration of the function itself, rather
than its Hessian, is necessary to understand its nature as a site of an extremum. â™¦
Global Extrema on Compact Regions
Thus far our discussion has been limited to consideration of only local extrema.
We have said nothing about how to identify global extrema, because there really is
no general, effective method for looking at an arbitrary function and determining
whether and where it reaches an absolute maximum or minimum value. For the
purpose of applications, where finding an absolute maximum or minimum is
essential, such a state of affairs is indeed unfortunate. Nonetheless, we can say
something about global extrema for functions defined on a certain type of domain.
4.2 | Extrema of Functions 271
Figure 4.20 Compact regions.
DEFINITION 2.4 A subset X C R" is said to be compact if it is both closed
and bounded.
Recall that X is closed if it contains all the points that make up its boundary.
(See Definition 2.3 of Â§2.2.) To say that X is bounded means that there is some
(open or closed) ball B that contains it. (That is, X is bounded if there is
some positive number M such that ||x|| < M for all x e X.) Thus, compact sets
contain their boundaries (a consequence of being closed) and have only finite
extent (a consequence of being bounded). Some typical compact sets in R 2 and
R 3 are shown in Figure 4.20.
For our purposes the notion of compactness is of value because of the next
result, which we state without proof.
THEOREM 2.5 (Extreme value theorem) If X c R" is compact and /:
X -> R is continuous, then / must have both a global maximum and a global
minimum somewhere on X. That is, there must exist points a max and a m ; n in X
such that, for all x e X,
/(a m in) < /(x) < /(a max ).
We need the compactness hypothesis since a function defined over a noncom-
pact domain may increase or decrease without bound and hence, fail to have any
global extremum, as suggested by Figure 4.21. This is analogous to the situation
in one variable where a continuous function defined on an open interval may fail
to have any extrema, but one defined on a closed interval (which is a compact
subset of R) must attain both maximum and minimum values. (See Figure 4.22.)
In the one-variable case, extrema can occur either in the interior of the interval
or else at the endpoints. Therefore, you must compare the values of / at any
interior critical points with those at the endpoints to determine which is largest
and smallest. In the case of functions of n variables, we do something similar,
namely, compare the values of / at any critical points with values at any restricted
critical points that may occur along the boundary of the domain.
272 Chapter 4 I Maxima and Minima in Several Variables
x
Figure 4.21 A graph that lacks a
global minimum. Figure 4.22 The function depicted by the graph on the left has no global extrema â€” the
function is defined on the open interval (a, b). By contrast, the function defined on the
closed interval [a, b], and with the graph on the right, has both a global maximum and
minimum.
EXAMPLE 8 LetTiXcR 2
y = 2
x = -l
x = 2
y = -l
R be given by
Figure 4.23 The domain of the
function T of Example 8.
T(x, y) = x 2 -xy + y 2 + 1,
where X is the closed square in Figure 4.23. (Note that X is compact.) Think of
the square as representing a fiat metal plate and the function T as the temperature
of the plate at each point. Finding the global extrema amounts to finding the
warmest and coldest points on the plate. According to Theorem 2.5, such points
must exist.
We need to find all possible critical points of T. Momentarily considering T
as a function on all of R 2 , we find the usual critical points by setting DT(x, y)
equal to 0. The result is the system of two equations
| 2x - y = 0
|-x + 2y = 0'
which has (0, 0) as its only solution. Whether it is a local maximum or minimum
is not important for now, because we seek global extrema. Because there is only
one critical point, at least one global extremum must occur along the boundary of
X (which consists of the four edges of the square). We now find all critical points
of the restriction of T to this boundary:
1. The bottom edge of X is the set
E l ={(x,y)\y = -l,-l*

0.
12. Find the maximum value of
f(x u x 2 , ...,x n ) = (aiXi + a 2 x 2 H h a n x n ) 2 ,
subject to x\ + x\ + â€¢ â€¢ â€¢ + x 2 = I. Assume that not all
of the a, 's are zero.
1 3. Find the dimensions of the largest rectangular box that
can be inscribed in the ellipsoid x 2 + 2y 2 + 4z 2 = 12.
Assume that the faces of the box are parallel to the
coordinate planes.
14. Your company must design a storage tank for Super
Suds liquid laundry detergent. The customer's specifi-
cations call for a cylindrical tank with hemispherical
ends (see Figure 4.43), and the tank is to hold 8000 gal
of detergent. Suppose that it costs twice as much (per
square foot of sheet metal used) to machine the hemi-
spherical ends of the tank as it does to make the cylin-
drical part. What radius and height do you recommend
for the cylindrical portion so as to minimize the total
cost of manufacturing the tank?
Figure 4.43 The storage tank
of Exercise 14.
15. Find the minimum distance from the origin to the
surface x 2 â€” (y â€” z) 2 = 1 .
1 6. Determine the dimensions of the largest cone that can
be inscribed in a sphere of radius a.
17. Find the dimensions of the largest rectangular box
(whose faces are parallel to the coordinate planes) that
Chapter 4 ! Maxima and Minima in Several Variables
can be inscribed in the tetrahedron having three faces in
the coordinate planes and fourth face in the plane with
equation bcx + acy + abz = abc, where a, b, and c
are positive constants. (See Figure 4.44.)
Figure 4.44 Figure for Exercise 17.
1 8. You seek to mail a poster to your friend as a gift. You
roll up the poster and put it in a cylindrical tube of di-
ameter a' and length y. The postal regulations demand
that the sum of the length of the tube plus its girth (i.e.,
the circumference of the tube) be at most 108 in.
(a) Use the method of Lagrange multipliers to find the
dimensions of the largest- volume tube that you can
mail.
(b) Use techniques from single-variable calculus to
solve this problem in another way.
1 9. Find the distance between the line y = 2x + 2 and the
parabola x = y 2 by minimizing the distance between
a point (x\ , yi) on the line and a point (X2 , yi) on the
parabola. Draw a sketch indicating that you have found
the minimum value.
20. A ray of light travels at a constant speed in a uniform
medium, but in different media (such as air and water)
light travels at different speeds. For example, if a ray of
light passes from air to water, it is bent (or refracted)
as shown in Figure 4.45. Suppose the speed of light
A
\^ Medium
1
a i V
i V 1 !
h
i
i
Medium 2 10 Â£
i
B
in medium 1 is v\ and in medium 2 is 1)2 â– Then, by
Fermat's principle of least time, the light will strike the
boundary between medium 1 and medium 2 at a point
P so that the total time the light travels is minimized.
(a) Determine the total time the light travels in going
from point A to point B via point P as shown in
Figure 4.45.
(b) Use the method of Lagrange multipliers to estab-
lish Snell's law of refraction: that the total travel
time is minimized when
sin 6*i
sin 6*2
Vl
!'2 '
(Hint: The horizontal and vertical separations of
A and B are constant.)
21. Use Lagrange multipliers to establish the formula
\ax 0 + by 0 â€” d\
D
VÂ« 2 + b 2
for the distance D from the point (xq, yo) to the line
ax + by = d.
22. Use Lagrange multipliers to establish the formula
\ax 0 + by 0 + czo - d\
D
si a 2 + b 2 + c 2
for the distance D from the point (xq, yo, zo) to the
plane ax + by + cz = d.
23. (a) Show that the maximum value of f(x,y,z)
2 2'
x A y A z
â– 2 subject to the constraint that x 2 + y 2 +
2 2
Z = a is
27 V 3
(b) Use part (a) to show that, for all x, y, and z,
(*Vz 2 ) 1/3
x 2 + y 2 + z 2
(c) Show that, for any positive numbers x\ , x% , . . . , x n ,
i ^ x l +x 2 + ---+x n
n
The quantity on the right of the inequality is the
arithmetic mean of the numbers x\, X2, â– â– . , x n ,
and the quantity on the left is called the geomet-
ric mean. The inequality itself is, appropriately,
called the arithmetic-geometric inequality.
(d) Under what conditions will equality hold in the
arithmetic-geometric inequality?
In Exercises 24-27 you will explore how some ideas from ma-
trix algebra and the technique of Lagrange multipliers come
together to treat the problem of finding the points on the unit
hypersphere
Figure 4.45 Snell's law of refraction.
g(xu â– â– â– ,xâ€ž)
+ x 2 + -
l
4.6 | Miscellaneous Exercises for Chapter 4
that give extreme values of the quadratic form
f(*L
where the Ojj s are constants.
24. (a) Use a Lagrange multiplier X to set up a system of
n + 1 equations in w + 1 unknowns xi , . . . , x n , X
whose solutions provide the appropriate con-
strained critical points.
(b) Recall that formula (2) in Â§4.2 shows that the
quadratic form / may be written in terms of ma-
trices as
f(xi
,xâ€ž)
(1)
where the vector x is written as the Â« x 1 matrix
X\
and A is the n x n matrix whose ij th entry
is atj . Moreover, as noted in the discussion in Â§4.2,
the matrix A may be taken to be symmetric (i.e., so
that A T = A), and we will therefore assume that
A is symmetric. Show that the gradient equation
V/ = XV g is equivalent to the matrix equation
Ax = Xx.
(2)
Since the point (xi, . . . , x n ) satisfies the constraint
*]+â€¢â€¢â€¢+ x% = 1, the vector x is nonzero. If you
have studied some linear algebra, you will recog-
nize that you have shown that a constrained criti-
cal point (xi, . . . , xâ€ž) for this problem corresponds
precisely to an eigenvector of the matrix A asso-
ciated with the eigenvalue X.
(c) Now suppose that x :
x\
is one of the eigen-
vectors of the symmetric matrix A, with associated
eigenvalue X. Use equations (1) and (2) to show, if
x is a unit vector, that
/(*!
â– , X n ) = X.
Hence, the (absolute) minimum value that /
attains on the unit hypersphere must be the small-
est eigenvalue of A and the (absolute) maximum
value must be the largest eigenvalue.
25. Let n = 2 in the situation of Exercise 24, so that we are
considering the problem of finding points on the circle
x 2 + y 2 = 1 that give extreme values of the function
f(x, y) = ax 2 + 2bxy + cy 2
[* y]
(a) Find the eigenvalues of A
a b
b c
by identify-
ing the constrained critical points of the optimiza-
tion problem described above.
(b) Now use some algebra to show that the eigenval-
ues you found in part (a) must be real. It is a fact
(that you need not demonstrate here) that any n x n
symmetric matrix always has real eigenvalues.
26. In Exercise 25 you noted that the eigenvalues X\, X2
that you obtained are both real.
(a) Under what conditions does X\ = X2I
(b) Suppose that X\ and X2 are both positive. Explain
why / must be positive on all points of the unit
circle.
(c) Suppose that Xi and X2 are both negative. Explain
why / must be negative on all points of the unit
circle.
27. Let / be a general quadratic form in n variables de-
termined by an n x n symmetric matrix A, that is,
f{x\, xâ€ž) = a u x i x j = x r ^ x -
(a) Show, for any real number k, that f(kx\,...,
kxâ€ž) = k 2 f(x\, . . . , xâ€ž). (This means that a
quadratic form is a homogeneous polynomial of
degree 2 â€” see Exercises 37-44 of the Miscella-
neous Exercises for Chapter 2 for more about ho-
mogeneous functions.)
(b) Use part (a) to show that if / has a positive
minimum on the unit hypersphere, then / must
be positive for all nonzero x e R" and that if /
has a negative maximum on the unit hypersphere,
then / must be negative for all nonzero x â‚¬ R" .
(Hint: For x / 0, let u = x/||x||, so that x = ku,
where k = ||x||.)
(c) Recall from Â§4.2 that a quadratic form / is said
to be positive definite if /(x) > 0 for all nonzero
x e R" and negative definite if /(x) < 0 for all
nonzero x e R" . Use part (b) and Exercise 24 to
show that the quadratic form / is positive definite
if and only if all eigenvalues of A are positive, and
negative definite if and only if all eigenvalues of
A are negative. (Note: As remarked in part (b) of
Exercise 25, all the eigenvalues of A will be real.)
Multiple Integration
5.1 Introduction: Areas and
Volumes
5.2 Double Integrals
5.3 Changing the Order of
Integration
5.4 Triple Integrals
5.5 Change of Variables
5.6 Applications of Integration
5.7 Numerical Approximations
of Multiple Integrals
(optional)
True/False Exercises for
Chapter 5
Miscellaneous Exercises for
Chapter 5
y
Figure 5.1 The graph of
y = f(x).
y
Figure 5.2 The shaded region has
area /* f(x)dx.
5.1 Introduction: Areas and Volumes
Our purpose in this chapter is to find ways to generalize the notion of the definite
integral of a function of a single variable to the cases of functions of two or
three variables. We also explore how these multiple integrals may be used to
meaningfully represent various physical quantities.
Let / be a continuous function of one variable defined on the closed interval
[a, b] and suppose that / has only nonnegative values. Then the graph of / looks
like Figure 5.1. That / is continuous is reflected in the fact that the graph consists
of an unbroken curve. That / is nonnegative-valued means that this curve does not
dip below the x -axis. We know from one- variable calculus that the definite integral
f b f(x) dx exists and gives the area under the curve, as shown in Figure 5.2.
Now suppose that / is a continuous, nonnegative-valued function of two
variables defined on the closed rectangle
R = [( x , y) e R 2 | a < x < b, c < y < d]
in R 2 . Then the graph of / over R looks like an unbroken surface that never dips
below the xy-plane, as shown in Figure 5.3. In analogy with the single-variable
case, there should be some sort of integral that represents the volume under the
part of the graph that lies over R. (See Figure 5.4.) We can find such an integral
by using Cavalieri's principle, which is nothing more than a fancy term for the
method of slicing. Suppose we slice by the vertical plane x = xo, where xo is a
constant between a and b. Let A(xo) denote the cross-sectional area of such a
slice. Then, roughly, one can think of the quantity A(xo) dx as giving the volume
of an "infinitely thin" slab of thickness dx and cross-sectional area A(xo). (See
Figure 5.5.) Hence, the definite integral
V
-f
J Li
A(x) dx
gives a "sum" of the volumes of such slabs and can be considered to provide a
reasonable definition of the total volume of the solid.
But what about the value of A(xo)? Note that A(xo) is nothing more than the
area under the curve z = f(xo, y), obtained by slicing the surface z = f(x, y)
with the plane x = xq. Therefore,
A(x 0 )
f(x 0 , y)dy
5.1 | Introduction: Areas and Volumes 311
Figure 5.3 The graph of Figure 5.4 The region under the Figure 5.5 A slab of "volume"
z = f(x, v). portion of the graph of / lying dV = A(x 0 ) dx.
over R has volume that is given
by an integral.
Plane z = c
(a, 0, 0)
(0,6,0)
y
(a, b, 0)
Figure 5.6 Calculating the
volume of the box of Example 1 .
Figure 5.7 The graph of
z = 4 - x 2
Example 2.
4 - x 2 - y 2 of
(remember xq is a constant), and so we find that
o pupa
= / A(x)(ix = / / /(x, y)dy
J a J a lJ c
dx.
(i)
The right-hand side of formula (1) is called an iterated integral. To calculate
it, first find an "antiderivative" of f(x, y) with respect to y (by treating i as a
constant), evaluate at the integration limits y = c and y = d, and then repeat the
process with respect to x.
EXAMPLE 1 Let's make sure that the iterated integral defined in formula (1)
gives the correct answer in a case we know well, namely, the case of a box. We'll
picture the box as in Figure 5.6. That is, the box is bounded on top and bottom by
the planes z = c (where c > 0) and z = 0, on left and right by the planes y = 0
and y = b (where b > 0), and on back and front by the planes x = 0 and x = a
(a > 0). Hence, the volume of the box may be found by computing the volume
under the graph of z = c over the rectangle
R = {(x,y) | 0 < x < a,0 < y < b}.
Using formula (1), we obtain
V = y j cdydx = j (^cy\- y ~^ dx = j cbdx = cbx\]z C Q = cba.
This result checks with what we already know the volume to be, as it should. â™¦
EXAMPLE 2 We calculate the volume under the graph of z = 4 â€” x 2 â€” y 2
(Figure 5.7) over the square
R = {(x, y) | -1 < x < 1, -1 < y < 1}.
Using formula (1) once again, we calculate the volume by first integrating with
respect to y (i.e., by treating x as a constant in the inside integral) and then by
Chapter 5 | Multiple Integration
y = y 0 plane
gure 5.8
first.
Y/y
Slicing by y = yo
integrating w
â– /.
th respect to x. The details are as follows:
j (4-x 2 - y 2 )dydx = j
dx
y=-l
A-x z
8 - 2x 2
22 2 :
X X
3 3
-4 + x 2 +
dx
dx
-l
22
T
22 2
T + 3
40
y
In our development of formula (1 ), we could just as well have begun by slicing
the solid with the plane y = yo (instead of with the plane x = xq), as shown in
Figure 5.8. Then, in place of formula (1), the formula that results is
V
rd rb
= I I f( x , y )dxdy.
J c J a
(2)
Since the iterated integrals in formulas ( 1 ) and (2) both represent the volume of the
same geometric object, we can summarize the preceding discussion as follows.
PROPOSITION 1 .1 Let R be the rectangle {(x, y) \ a < x < b, c < y < d) and
let / be continuous and nonnegative on R. Then the volume V under the graph
of /' over R is
pb pel pel pb
/ / f(x,y)dydx= / f(x,y)dxdy.
J a J c J c J a
EXAMPLE 3 We find the volume under the graph of z = cosx siny over the
rectangle
R
f 7T TV 1
= \(x,y) \ 0** b, and c' < y(x), d' > S(x) for all x in [a, b]. That is, we have the situation
depicted in Figure 5.29. Since / ext is zero outside of the subrectangle R2 =
[a,b] x [c',d'],
f f f ext dA= f f f ext dA= f f f ex \x,y)dydx
by Fubini's theorem. For a fixed value of x between a and b, consider the y-
integral f ext (x, y)dy. Since f ext (x, y) = 0 unless y(x) < y < S(x) (in which
case f ext (x, y) = f(x, y)),
pel' pS(x)
/ r x \x,y)dy= / f(x,y)dy,
Jd Jy(x)
and so
f f f(x,y)dA= f f f xt dA= f f f ex \x,y)dydx
J JD J J R Ja Jc'
/ / f(x,y)dydx,
as desired.
The proof of part 2 is very similar.
Figure 5.29 The region R is the union of Ri, R2,
and^?3 .
(0,1)'
>v x + y = 1
D
(0,0)
\(1.0)
Figure
5.30 The region D of
Example 6.
;
>
\ J = 1 - X |
x = 0
D
y = 0
Figure
5.31 The region D of
Example 6 as a type 1 region.
;
>
y = l
x = 0
\x = 1 â€” y
D \.
y = 0
Figure
5.32 The region D of
Example 6 as a type 2 region.
5.2 | Double Integrals 325
We continue analyzing examples of double integral calculations.
EXAMPLE 6 Let D be the region shown in Figure 5.30 having a triangular
border. Consider f f D (\ â€” x â€” y)dA. Note that D is a type 3 elementary region,
so there should be two ways to evaluate the double integral.
Considering D as a type 1 elementary region (see Figure 5.3 1), we may apply
part 1 of Theorem 2.10 so that
/ / (1 â€” x â€” y)dA= I I (1 â€” x â€” y)dydx
J 3d Jo Jo
-L
-L
-I
y-xy
y=\-x
dx
y=0
' 7 (1 -x) 2 \
(1 â€” x) â€” x(l â€” x) I dx
(1 - xf
dx= -\{\-x)\ = \.
We can also consider D as a type 2 elementary region, as shown in Figure 5.32.
Then, using part 2 of Theorem 2.10, we obtain
/ / (1 â€” x â€” y)dA = I I (l-x-y)dxdy.
J J D Jo Jo
We leave it to you to check explicitly that this iterated integral also has a value of
t . Instead we note that
Jo Jo
y ) dy dx
can be transformed into
/ / (1 â€” x â€” y) dx dy
Jo Jo
by exchanging the roles of x and y. Hence, the two integrals must have the same
value. In any case, the double integral
(1 â€” x â€” y)dA
represents the volume under the graph of z = 1 â€” x â€” y over the triangular region
D . If we picture the situation in R 3 , as in Figure 5 . 3 3 , we see that the double integral
represents the volume of a tetrahedron. â™¦
Of course, not all regions in the plane are elementary, including even some
relatively simple ones. To integrate continuous functions over such regions, the
best advice is to attempt to subdivide the region into finitely many of elementary
type.
326 Chapter 5 | Multiple Integration
EXAMPLE 7 Let D be the annular region between the two concentric circles
of radii 1 and 2 shown in Figure 5.34. Then D is not an elementary region, but we
can break D up into four subregions that are of elementary type. (See Figure 5.35.)
If f(x, y) is any function of two variables that is continuous (hence integrable)
on D, then we may compute the double integral as the sum of the integrals over
the subregions. That is,
[[ fdA=[[ fdA+ff fdA+ff fdA+[[ fdA.
J J D J J Di J JD 2 J J D] J J D 4
For the type 1 subregions, we have the set-up shown in Figure 5.36:
f(x, y)dy dx
and
\L SiA
V3 ,-1
f(x, y)dy dx.
-VI Jâ€”jA^?
For the type 2 subregions, we use the set-up shown in Figure 5.37:
f fo 2 f-i f*/i
y = â€” V4 - x-
Figure 5.36 The subregions D\ and
Z?3 of Example 7 are of type 1 .
fix, y)dx dy
J
1
.2 _
\ âœ“
\
\
s
X
âœ“
x= VT^y2
x = V4-y 2
Figure 5.37 The subregions D2 and D4 of
Example 7 are of type 2.
5.2 | Double Integrals
and
JL<Â«-U.
(jtf)A*,, (1)
where Ax; = x ; - â€” andx* e [x,-_i, x,].Nowletc = y 0 < < â€¢ â€¢ â€¢ < = d
be a partition of [c, d]. (The partitions of [a, b] and [c, J] together give a partition
of R = [a, b] x [c, rf].) Therefore, we may write
x) = J d f(x,
y)dy
fix, y)dy +
y)dy +
t f
7=1 J y>- 1
f(x,y)dy.
+ f fix,y)dy
Jyâ€ž-i
5.2 | Double Integrals 329
By the mean value theorem for integrals, 1 on each subinterval [yj-i, yj] there
exists a number y* such that
f
Jy,
f(x, y)dy = ( yj - y } -i)f(x, y*) = f(x, y*)A yj .
The choice of y* in general depends on x, so henceforth, we will write y*(x) for
y*. Consequently,
n
/â€¢(.v)- ^./(.v.y;(.v)).\y,.
and the Riemann sum (1) may be written as
i=l I 7=1
Ax i = /(C ; ,).\.V,.\.V ( .
where cy = (xf , y*(xf)). Note that Cy e x ; ] x [y ; _i, y,-]. (See Fig-
ure 5.40.)
y;(*,*)
c â€¢
H 1 â€” h
H h
1 Â« I
Figure 5.40 The point Cy = (.Â«*, y *(*,*)) used in
the proof of Theorem 2.6.
We have thus shown that given any partition of [a,b], we can associate a
suitable partition of R = [a, b] x [c, d] such that the Riemann sum (1) that ap-
proximates f a F(x)dx is equal to a Riemann sum (namely, J2i j f( c ij)Axj Ayj)
that approximates ff R f dA. Since / is continuous, we know that
/(cy)Ajc,- Ay 7 - approaches / / f dA
U J Js
as Ax, and Ay,- tend to zero. Hence,
I F(x)dx = 11 fdA.
Ja J JR
By exchanging the roles of x and y in the foregoing argument, we can show
that
II fdA = f I f(x,y)dxdy.
J J R Jc Ja
1 The mean value theorem for integrals says that, if g is continuous on [a, b], then there is some number
c with a < c < b such that f g(x)dx = (b â€” a)g(c).
Chapter 5 | Multiple Integration
Step 2. Now we prove the general case of Theorem 2.6 (i.e., the case that
/ has discontinuities in R = [a, b] x [c, d]). By hypothesis, the set S of discon-
tinuities of / in R are such that every vertical line meets S in at most finitely
many points. Thus, for each x in [a, b], the partial function in y of f(x, y) is
continuous throughout [c, d], except possibly at finitely many points. (In other
words, the partial function is piecewise continuous.) Then, because / is bounded,
(x) = ^ f(x
F(x)= I f(x,y)dy
exists.
Now we proceed as in Step 1 . That is, we begin with a partition of [a, b] into
n subintervals and a corresponding Riemann sum
i=i
Next, we partition [c, d] into n subintervals. Hence,
Fix*) = f f(x*, y)dy = J2 f' fix*, y)dy. (2)
As in Step 1, the partitions of [a, b] and [c, d] combine to give a partition of R.
Write R as U R 2 , where R\ is the union of all subrectangles
Rij = [Xi-uXf] x [y ; _!, yj]
that intersect S and R2 is the union of the remaining subrectangles. Then we may
apply the mean value theorem for integrals to those intervals [yj-i, yj] on which
fix*, y) is continuous in y, thus replacing the integral
/ f(xf,y)dy
by
f(x*,y*(x*))Ay j = f(c i j)Ay j .
Since / is bounded, we know that
\f(x,y)\**

R 2 be a map of class C 1 that transforms the wu-plane into the xy-
plane. We are interested particularly in how certain subsets D* of the wu-plane
are distorted under T into subsets D of the xy-plane. (See Figure 5.73.)
D*
CO
D = T(D*)
Figure 5.73 The transformation T(m, v) = (x(u, v), y(u, v))
takes the subset D* in the uv-p\am to the subset D = {(x, y) |
(x, y) = T(m, v) for some (u, Â») e D*) of the xy-plane.
EXAMPLE 1 Let T(u, v) = (u + 1 , v + 2); that is, let x = u + 1, y = v + 2.
This transformation translates the origin in the Mu-plane to the point (1 , 2) in the
xy-plane and shifts all other points accordingly. The unit square D* = [0, 1] x
[0, 1 ], for example, is shifted one unit to the right and two units up but is otherwise
unchanged as shown in Figure 5.74. Thus, the image of D* is D = [1,2] x [2,3].
â™¦
EXAMPLE 2 Let S(w, v) = (2m, 3d). The origin is left fixed, but S stretches all
other points by a factor of two in the horizontal direction and by a factor of three
in the vertical direction. (See Figure 5.75.) â™¦
EXAMPLE 3 Composing the transformations in Examples 1 and 2, we obtain
(T o S)(Â«, Â«) = T(2Â«, 3u) = (2m + 1, 3v + 2).
Such a transformation must both stretch and translate as shown in Figure 5.76. â™¦
350 Chapter 5 | Multiple Integration
D
D
â–¡
= T(D')
Figure 5.74 The image of D* = [0, 1] x [0, 1] is
D = [1, 2] x [2, 3] under the translation
T(m, v) = (u + 1, v + 2) of Example 1.
Â£>* = [0, 1] x [0, 1]
D = S(D*) = [0, 2] x [0, 3]
Figure 5.75 The transformation S of Example 2 is a scaling
by a factor of 2 in the horizontal direction and 3 in the vertical
direction.
y
ToS
D
D = [1, 3] x [2, 5]
1
Figure 5.76 Composition of the transformations of
Examples 1 and 2.
EXAMPLE 4 Let T(w, v) = (u + v, u â€” v). Because each of the component
functions of T involves both variables u and v, it is less obvious how the unit
square D* = [0, 1] x [0, 1] transforms. We can begin to get some idea of the
geometry by seeing how T maps the edges of D* :
Bottom edge: (u, 0), 0 < u < 1
Top edge: (u, 1), 0 < u < 1
Left edge: (0, v), 0 < v < 1
Right edge: (1, v), 0 < v < 1
T(m, 0) = (u, u);
T(m, 1) = (u + 1, m â€” 1);
T(0, v) = (v, -v);
T(l, w) = O + l, 1 - v).
By sketching the images of the edges, it is now plausible that the image of D*
under T is as shown in Figure 5.77. â™¦
Figure 5.77 The transformation T of Example 4.
by
5.5 | Change of Variables 351
More generally, we consider linear transformations T: R 2 â€” > R 2 defined
T(w, v) = (au + bv, cu + dv)
a b
u
c d
V
where a, b, c, and d are constants. (Note: The vector (u, v) is identified with the
u
v
2 x 1 matrix
.) One general result is stated in the following proposition:
PROPOSITION 5.1 Let A =
fined by
a b
c d
, where det A ^ 0. If T: R 2 -> R 2 is de-
T(w, w) = A
then T is one-one, onto, takes parallelograms to parallelograms and the vertices
of parallelograms to vertices. (See Â§2. 1 to review the notions of one-one and onto
functions.) Moreover, if D* is a parallelogram in the i/u-plane that is mapped
onto the parallelogram D = T(D*) in the xy-plane, then
Area of D = \ det A| â– (Area of D*).
EXAMPLE 5 We may write the transformation T(u, v) = (u + v, u
Example 4 as
v) in
T(m, v)
Note that
det
u
v
= -2^0.
Hence, Proposition 5.1 tells us that the square D* = [0, 1] x [0, 1] must be
mapped to a parallelogram D = T(Z)*) whose vertices are
T(0, 0) = (0, 0), T(0, 1) = (1, -1), T(1,0) = (1,1), T(l, 1) = (2,0).
Therefore, Figure 5 .77 is indeed correct and, in view of Proposition 5.1, could have
been arrived at quite quickly. Also note that the area of D is | â€” 2| â€¢ 1 = 2. â™¦
Proof of Proposition 5.1 First we show that T is one-one. So suppose
T(w, v) = T(u', v'). We show that then u = u',v = v' . We have
T(w, v) = T(u' , v')
if and only if
(au + bu , cu + dv) = (au' + bv', cu' + du').
By equating components and manipulating, we see this is equivalent to the system
a(u â€” u') + b(v â€” v') = 0
c(u - u') + d(v - v') = 0
(1)
352 Chapter 5 | Multiple Integration
If a 7^ 0, then we may use the first equation to solve for u â€” u':
u! = â€” (v- v')
(2)
Figure 5.78 The vertices of
D* = {p + sn + tb | 0 < s,t <
1 } are at p, p + a, p + b,
p + a + b (i.e., where s and t take
on the values 0 or 1).
Figure 5.79 The image D of the
parallelogram D* under the linear
transformation T(u) = An.
Hence, the second equation in (1) becomes
be
(v - v') + d(v -v') = 0
or, equivalently,
-be + ad
(v - v') = 0.
By hypothesis, det A = ad â€” be ^ 0. Thus, we must have v â€” v' = 0 and there-
fore, u â€” u' = 0 by equation (2). If a = 0, then we must have both b 0 and
c 0, since det A ^ 0. Consequently, the system (1) becomes
J b(v -v') =0
I c(u - u') + d(v - v') = 0 "
The first equation implies v â€” v' = 0 and hence, the second becomes c(u â€” u') =
0, which in turn implies u â€” u! = 0, as desired.
To see that T is onto, we must show that, given any point (x, y) e R 2 , we can
find (u, v) e R 2 such that T(m, v) = (x, y). This is equivalent to solving the pair
of equations
Iau + bv = x
cu + dv = y
for u and v. We leave it to you to check that
dx â€” by ay â€” cx
14 = ~~a â€” TT and v = ~1 â€” TT
ad â€” be ad â€” be
will work.
Now, let D* be a parallelogram in the ww-plane. (See Figure 5.78.) Then D*
may be described as
D* = {u | u = p + + tb, 0 < s < 1, 0 < t < 1}.
Hence,
D = T(D*) = {Au ueD)
= {A(p + 5a + rb) | 0 < s < 1, 0 is also a parallelogram and moreover, the vertices of D correspond to
those of D*. (See Figure 5.79.)
Finally, note that the area of the parallelogram D* whose sides are parallel to
a 2
and b
by
b 2
5.5 | Change of Variables
may be computed as follows:
j
k
Areaof/?* = ||axb|| =
det
0
bi
b 2
0
= \a.\b 2 â€” a 2 b\
Similarly, the area of D = T(D*) whose sides are parallel to
a =
and b' =
b\
b'
is
a' 2 b\ j
a
b
a\
aa\ -
- ba2
. a 2 .
c
d
CL 2
ca\ -
\-da 2
' b\ '
a
b ~
" b x '
ab\ -
Ybb 2
, b 2_
c
d
. b2 .
cb\ -
Ydb 2
Area of D = ||a' x b'|| = \a[b' 2
Now, a' = Aa and b' = Ab. Therefore,
and
Hence, by appropriate substitution and algebra,
Area of D = \(aai + ba2)(cb\ + db 2 ) â€” (ca\ + da-Â£)(ab\ + bb 2 )\
= \{ad â€” bc)(a\b 2 â€” a 2 b\)\
= I det A] â€¢ area of D*.
Note that we have not precluded the possibility of D*'s being a "degenerate"
parallelogram, that is, such that the adjacent sides are represented by vectors a
and b, where b is a scalar multiple of a. When this happens, D will also be
a degenerate parallelogram. The assumption that det A 7^ 0 guarantees that a
nondegenerate parallelogram D* will be transformed into another nondegenerate
parallelogram, although we have not proved this fact. â–
Essentially all of the preceding comments can be adapted to the three-
dimensional case. We omit the formalism and, instead, briefly discuss an example.
EXAMPLE 6 Let T: R 3 -> R 3 be given by
T(m, v, w) = (2u, 2u + 3v + w, 3w).
Then we rewrite T by using matrix multiplication:
Note that if
- 2
0
0 "
u
w)
2
3
1
V
_ 0
0
3 _
w
" 2
0
0 "
\ =
2
3
1
Â»
0
0
3
then det A = 18^0.
354 Chapter 5 | Multiple Integration
A result analogous to Proposition 5.1 allows us to conclude that T is one-one
and onto, and T maps parallelepipeds to parallelepipeds. In particular, the unit
cube
D* = [0, 1] x [0, 1] x [0, 1]
is mapped onto some parallelepiped D = T(Â£>*) and, moreover, the volume of D
must be
| det A| - volume of D* = 18- 1 = 18.
To determine D, we need only determine the images of the vertices of the cube:
r(o, o, 0) = (o, o, o)
T(0, 0, 1) = (0, 1,3)
T(0, 1, 1) = (0,4, 3)
Both D* and its image D are shown in Figure 5.80.
7X1, 0, 0) = (2, 2, 0); T(0, 1,0) = (0, 3, 0);
T(l, 1, 0) = (2, 5, 0); 7X1, 0, 1) = (2, 3, 3);
J(l, 1, 1) = (2, 6, 3).
D*
Figure 5.80 The cube D* and its image D under the
linear transformation of Example 6.
EXAMPLE 7 Of course, not all transformations are linear ones. Consider
(x, y) = T(r, 9) = (r cos#, r sin#).
Note that T is not one-one since T(0, 0) = (0, 0) = T(0, n). (Indeed T(0, 9) =
(0, 0) for all real numbers 9.) Note that vertical lines in the r0-plane given by
r = a, where a is constant, are mapped to the points (x, y) = (a cos#, a sin#)
on a circle of radius a. Horizontal rays {(r, 9) \ 9 = a, r > 0} are mapped to
rays emanating from the origin. (See Figure 5.81.) It follows that the rectangle
D* = [5, 1] x [0, it] in the r#-plane is mapped not to a parallelogram, but bent
6
G = a
r = a
Image of
6 = a
Image of
r = a
Figure 5.81 The images of lines in the r#-plane under the
transformation T(r, 6) = (r cos 0, r sin 0).
5.5 | Change of Variables 355
D
Figure 5.82 The image of the rectangle D* = [Â±, 1] x [0, n]
under T(r, 9) = (r cos 6, r sin#).
Figure 5.83 The image of B* = [\, 1] x [0, it] x [0, 1]
under T(r, 6, z) = (r cos 6, r sin#, z).
into a region D that is part of the annular region between circles of radii \ and 1 ,
as shown in Figure 5.82.
Analogously, the transformation T: R 3 â€” > R 3 given by
(x, y, z) = T(r, 6, z) = (r cos#, r sin^, z)
bends the solid box B* = [j, 1] x [0, 7r] x [0, 1] into a horseshoe-shaped solid.
(See Figure 5.83.) â™¦
Change of Variables in Definite Integrals
Now we see what effect a coordinate transformation can have on integrals and how
to take advantage of such an effect. To begin, consider a case with which you are
already familiar, namely, the method of substitution in single-variable integrals.
EXAMPLE 8 Consider the definite integral / Q 2 2x cos(x 2 ) dx. To evaluate, one
typically makes the substitution u = x 2 (so du = 2x dx). Doing so, we have
sin 4.
/ 2x cos(x 2 ) dx = I cosudu = sinu
Jo Jo
u=4
u=0
Let's dissect this example more carefully. First of all, the substitution u = x 2
may be rewritten (restricting x to nonnegative values only) as x = *Ju. Then
dx = du /(ly/u) and
f 2 f 4 du f 4
/ 2xcos(x 2 )dx= I 2yfu cos(a/m) 2 â€” â€” = / cosMJw = sin4.
Jo Jo 2y/u Jo
In other words, the substitution is such that the 2x = 2^fu factor in the integrand
is canceled by the functional part of the differential dx = du/(2*Ju). Hence, a
simple integral results. â™¦
In general, the method of substitution works as follows: Given a (perhaps
complicated) definite integral f A f(x)dx, make the substitutions = x(u), where
x is of class C 1 . Thus, dx = x'(u)du. If A = x(a), B = x(b), and x'(u) 7^ 0 for
u between a and b, then
/â€¢ B nb
I f(x)dx = J f(x(u))x'(u)du. (3)
J A Ja
Chapter 5 | Multiple Integration
x
x = x(u)
It
Figure 5.84 As Am = du â€” >â€¢ 0,
Ax â€” > dx = x'(u) Am. Thus, the factor
x'(u) measures how length in the w-direction
relates to length in the x-direction.
Note that it is possible to have a > b in (3) above. (This happens if x(u) is
decreasing.) Although the m -integral in equation (3) may at first appear to be more
complicated than the x -integral, Example 8 shows that in fact just the opposite
can be true.
Beyond the algebraic formalism of one-variable substitution in equation (3),
it is worth noting that the term x'(u) represents the "infinitesimal length distortion
factor" involved in the changing from measurement in u to measurement in x.
(See Figure 5.84.) We next attempt to understand how these ideas may be adapted
to the case of multiple integrals.
The Change of Variables Theorem for Double Integrals
Suppose we have a differentiable coordinate transformation from the wu-plane to
the xy-plane. That is,
T: R 2 -> R 2 , T(w, v) = (x(u, v), y(u, v)).
DEFINITION 5.2 The Jacobian of the transformation T, denoted
d(x, y)
d(u, v) '
is the determinant of the derivative matrix DT(u,
v). That is,
dx dx
d(x ' y) =detDT(u, u) = det
d(u, v)
du dv
dy dy
du dv
dx dy dx dy
du dv dv du
The notation d(x, y)/d(u, v) for the Jacobian is a historical convenience. The
Jacobian is not a partial derivative, but rather the determinant of the matrix of
partial derivatives. It plays the role of an "infinitesimal area distortion factor"
when changing variables in double integrals, as in the following key result:
5.5 | Change of Variables
y
(f.f)
(8,0)
Figure 5.85 The triangular region
D of Example 9.
THEOREM 5.3 (Change of variables in double integrals) Let D and D*
be elementary regions in (respectively) the xy-plane and the Mu-plane. Suppose
T:R 2 â€” > R 2 is a coordinate transformation of class C 1 that maps D* onto D
in a one-one fashion. If /: D â€” > R is any integrable function and we use the
transformation T to make the substitution x = x(u, v), y = y(u, v), then
jj f(x,y)dxdy = j j f(x(u,v),y(u,v))
dp, y)
d(u, v)
du dv.
EXAMPLE 9 We use Theorem 5.3 to calculate the integral
cos(x + 2y) sin(x â€” y) dx dy
over the triangular region D bounded by the lines y = 0, y = x,andx + 2y = 8 as
shown in Figure 5.85. It is possible to evaluate this integral by using the relatively
straightforward methods of Â§5.2. However, this would prove to be cumbersome,
so, instead we find a suitable transformation of variables, motivated in this case by
the nature of the integrand. In particular, we let u = x + 2y, v = x â€” y. Solving
for x and y, we obtain
u + 2v
x = and
y
Therefore,
d(x, y)
d(u, v)
det
X u Xy
y u y v
det
I 2
3 3
1 _ 1
3 3
Considering the coordinate transformation as a mapping T(w , v) = (x, y) of
the plane, we need to identify a region D* that T maps in a one-one fashion
onto D. To do this, essentially all we need do is to consider the boundaries of D :
y = x
x + 2y = 8
y = 0
x - y = 0
u = 8;
u â€” v
= 0
v = 0;
v = u.
Hence, one can see that T transforms the region D* shown in Figure 5.86 onto
D. Therefore, applying Theorem 5.3,
Figure 5.86 The effect of the transformation T of Example 9.
Chapter 5 | Multiple Integration
j j cos(x + 2 v) sin(x â€” y)dx dy = j j cos u sin v
B(x, y)
du dv
d(u, v)
= jj cosw sinv | â€” 1| dudv
= J J j cos u sin v dv du
Jo Jo
,8
= / | cos u ( â€” cos v)| JJZq du
Jo
= | / cosw(â€” cosm + 1) du
Jo
, f*
= 3/ (cos u â€” cos u) du
Jo
sinwlg â€” j j(1+cos2m)Jm
sin 8 â€” [ju + \ sin2M)|pJ
= \ [sin8 - 4
sin
16]
There is another, faster way to calculate the Jacobian, namely, to calculate
d(u, v)/d(x, y) directly from the variable transformation, and then to take reci-
procals. That is, from the equations u = x + 2y, v = x â€” y, we have
d(u, v)
d(x, y)
det
I'v
Uy
Vy
det
-3.
Consequently, 3(x, y)/d(u, v) = â€” |, which checks with our previous result.
This method works because if T(h, v) = (x, v), then, under the assumptions of
Theorem 5.3, (u,v) = T _1 (jc, y). It follows from the chain rule that
DT-\x,y)= [DT(u,v)]-\
(That is, DT 1 is the inverse matrix of DT. See Exercises 30-38 in Â§ 1 .6 for more
about inverse matrices.) Hence,
^=det[DT- 1 ]=det[(DTr 1 ]
d(u, v)
detOT
EXAMPLE 10 We use Theorem 5.3 to evaluate / f D (x 2 - y 2 ) e xy dx dy, where
D is the region in the first quadrant bounded by the hyperbolas xy = 1, xy = 4
and the lines y = x, y = x + 2. (See Figure 5.87.)
5.5 | Change of Variables 359
y y=x+2
Figure 5.87 The region D of Example 10.
D*
Figure 5.88 The region D*
corresponding to the region D of
Example 10.
Both the integrand and the region present complications for evaluation. There
would seem to be two natural choices for ways to transform the variables. One
would be
u = x 2 â€” y 2 and v = xy,
motivated by the nature of the integrand. However, the region D of integration
will not be easy to describe in terms of this particular choice of wu-coordinates.
Another possible transformation of variables, motivated instead by the shape of
D, is
u = xy and v = y â€” x .
Now this change of variables would not seem to help much with the integrand,
but, as we shall see, it turns out to be just what we need.
First note that the boundary hyperbolas xy = 1 and xy = 4 correspond re-
spectively, to the lines u = 1 and u = 4; the lines y = x and y = x + 2 correspond
to v = 0 and v = 2. Thus, the region D* in the Mu-plane that corresponds to D
(see Figure 5.88) is
D* = {(u, v) | 1 < u < 4, 0 < v < 2}.
Next, we calculate that the Jacobian of the variable transformation is
9(w, u)
= det
y x
-1 1
= x + y.
d(x, y)
Hence, the Jacobian we require in order to use Theorem 5.3 is
d(x, y) = 1
9(w, v) x + y
Moreover, since we will be working in the first quadrant (where x and y are both
positive), |3(jc, y)/9(w, v)\ = l/(x + y).
Chapter 5 | Multiple Integration
At last we are ready to compute:
jji* 2 - y 2 ) e * y dx d y = jj (x 2 - y 2 ) e * y (v^) du dv
= f 2 C (*-?) (* + y)
Jo J i x
+ y
du dv
e xy dudv
-u(e 4 - e l )dv = -y 0 - e )
2
k'
2(e - A
Note that the insertion of the Jacobian in the integrand caused precisely the can-
celation needed to make the evaluation straightforward. We cannot always expect
this to happen, but the lesson here is to be willing to carry through calculations
that may not at first appear to be so easy. â™¦
EXAMPLE 11 (Double integrals in polar coordinates) In Example 9, a coor-
dinate transformation was chosen primarily to simplify the integrand of the double
integral. In this example we change variables by using a coordinate system better
suited to the geometry of the region of integration.
For example, suppose that the region D is a disk of radius a :
D={(x,y)\x 2 + y 2 < a}
= |( x > y) I â€” V a 2 â€” x 2 < y < \l a 1 â€” x 2 , â€”a < x < a J .
Then, to integrate any (integrable) function / over D in Cartesian coordinates,
one would write
/ / f(x,y)dxdy= I I f(x,y)dydx.
J Jd J -a J-Ja^l?
Even if it is easy initially to find a partial antiderivative of the integrand, the limits
in the preceding double integral may complicate matters considerably. This is be-
cause the disk is described rather awkwardly by Cartesian coordinates. We know,
however, that it has a much more convenient description in polar coordinates as
{(r, 0) | 0 y> z)dxdydz
J J Jw
-III
f(x(p, ip, 0), y(p, dp d(p dO. (See Figure 5. 103.) â™¦
EXAMPLE 17 The volume of a ball is easy to calculate in spherical coordi-
nates. A solid ball of radius a may be described as
B = {(p, (p,9)\0 so du = sec 2 (p dep. Then the last integral becomes
h 3 f 2n [ a,h h 3 f 2n 1 /a\ 2 h 3 a 2 f 2lt
â€” / ududO = â€” - - ) dO = â€” - / dO
3 Jo Jo 3 Jo 2 \hj 6h* Jo
3
a 2 h
-{lit)
Tt
-a h,
as expected.
The use of spherical coordinates in Example 18 is not the most appropri-
ate. We merely include the example so that you can develop some facility with
"thinking spherically." Further practice can be obtained by considering some of
the applications in the next section as well as, of course, some of the exercises.
Summary: Change of Variables Formulas
Change of variables in double integrals:
jj f(x,y)dxdy = j j f(x(u,v),y(u,v
))
d(x, y)
d(u, v)
du dv
Area elements:
dA = dx dy
= rdrd0
djpc, y)
d(u, v)
(Cartesian)
(polar)
dudv (general)
5.5 I Exercises 371
Change of variables in triple integrals:
/ / / /(*> J> z)dxdydz
J J Jw
=fll
f(x(u, v, w), y(u, v, w), z(u, v, w))
d(x, y, z)
d(u, v, w)
du dvdw
Volume elements:
dV = dxdydz (Cartesian)
= r dr d6 dz (cylindrical)
= p 2 sin (p dp d

*â– R 3 be the transformation given by
T(p, *

w jx 2 + y 2
W is the solid region bounded by the plane z = 12 and
the paraboloid z = 2x 2 + 2y 2 â€” 6.
33. Find the volume of the region W bounded on top by
z = y/a 2 â€” x 2 â€” y 2 , on the bottom by the xy-plane,
and on the sides by the cylinder x 2 + y 2 = b 2 , where
0 < b < a.
In Exercises 34 and 35, determine the values of the given in-
tegrals, where W is the region bounded by the two spheres
x 2 + y 2 + z 2 = a 2 and x 2 + y 2 + z 2 = b 2 , for 0 < a < b.
Â«â€¢ SSL
dV
35
w jx 2 + y 2 + z 2
â– III*
x 1 + y L + z z e x
36. Let W denote the solid region in the first octant be-
tween the spheres x 2 + y 2 + z 2 = a 2 and x 2 + y 2 +
z 2 = b 2 , where 0 < a < b. Determine the value of
fff w (x + y + z)dv.
37. Determine the value of fff w z 2 dV, where W is the
solid region lying above the cone z = ^3x 2 + 3y 2 and
inside the sphere x 2 + y 2 + z 2 = 6z.
38. Determine
where W
(l + Jx 2 + y 2 ) dV,
(x,y,z)\ y/x 2 + y 2

*)
and Eâ€ž denotes the error involved in using T n to approximate the value of the
definite integral. In addition, if / is of class C 2 on [a, b], then there is some
number f in (a , b) such that
F â€”
(Ax) 2 /"(0 =
12
12n 2
In particular, Theorem 7.1 shows that if /" is bounded that is, if \f"(x)\ < M
for all x in [a, b], then \E n \ < (b â€” a) 3 M/(l2n 2 ). This inequality is very useful
in estimating the accuracy of the approximation.
EXAMPLE 1 We approximate f 0 sin(x 2 ) dx using the trapezoidal rule with
n = 4. Thus, we have
1 - 0
Ax = = 0.25,
4
so that, using (1), we have
I
1 0 25
sin(x 2 )Jx RÂ« T 4 = â€” [sin0 + 2sin(0.25 2 ) + 2sin(0.5 2 ) + 2sin(0.75 2 )
+ sinl]
0.25
2
0.25
[0 + 0.12492 + 0.49481 + 1.06661 + 0.84147]
[2.52780] = 0.31598.
Chapter 5 | Multiple Integration
Note that the second derivative of sin(x 2 ) is 2 cos (x 2 ) â€” Ax 2 sin(x 2 ) so that, for
0 < x < 1,
|2cos(x 2 )-4x 2 sin(x 2 )| < 2| cos(x 2 )| + 4x 2 | sin(x 2 )| < 2 + 4 = 6.
Hence, using Theorem 7.1,
(1 - 0) 3 6 1
|Â£ 4 | < - V- = â€” = 0.03125.
1 1 " 12 -4 2 32
Thus, the true value ofthe integral must be between 0.3 1598 - 0.03125 = 0.2847
(rounding to four decimal places) and 0.31598 + 0.03125 = 0.3473. Of course,
a more accurate approximation may be obtained by taking a finer partition (i.e.,
a larger value for n). â™¦
Another numerical technique for approximating the value of f(x)dx with
which you may be familiar is Simpson's rule. As with the trapezoidal rule, we
partition the interval [a, b] into equal subintervals, only now we require that the
number of subintervals be even, which we will write as 2n. Thus, we take
b â€” a
Ax = , a = xo < X\ < . . . < X2 n = b, where x,- = a + / Ax.
2n
Then the Simpson's rule approximation Sin is
Ax
Sm = â€” If (a) + 4/(xi) + 2/(x 2 ) + 4/(x 3 ) + â€¢ â– â– + 2/(x 2 â€ž_ 2 )
+ 4/(x 2n _0 + /(&)]
n â€” 1 n
f(a) + 2 Â£ /(x 2( ) + 4 Â£ /(x 2l _.) + f(b)
Ax
(2)
EXAMPLE 2 We approximate the value of f 0 sin(x 2 ) dx of Example 1 using
Simpson's rule with n = 2 (i.e., four subintervals). As before, we have
1 - 0
Ax = = 0.25,
4
so (2) gives
â– ! o 25
sin (x 2 ) dx RÂ» 5 4 = â€” [sin 0 + 4 sin (0.25 2 ) + 2 sin (0.5 2 ) + 4 sin (0.75 2 )
/
Jo
+ sin 1]
0.25
3
0.25
[0 + 0.24984 + 0.49481 + 2.13321 + 0.84147]
[3.71933] = 0.30994.
3
Note that this value is in the range predicted by the trapezoidal rule. In fact, it is
a more accurate approximation to the value of the definite integral. â™¦
Simpson's rule is obtained by approximating the function / with n quadratic
functions that pass through triples of points (x 2 ,_ 2 , /(x 2( _ 2 )), (x 2 ;_i, /fe-i)),
(x 2i , f(xii)) for i = 1 , . . . , n. As with the trapezoidal rule, we have the following
summary result.
5.7 | Numerical Approximations of Multiple Integrals (optional)
THEOREM 7.2 Given a function / that is integrable on [a, b], we have
b
f(x)dx = S 2n + E 2n ,
where
nâ€” 1 n
f(a) + 2 Â£ /(xa) + 4 J] /(x 2i _0 + /(*) ,
i=i i=i
and Ein denotes the error involved in using S^i to approximate the value of the
definite integral. In addition, if / is of class C 4 on [a, b], then there is some
number Â£ in (a , b) such that
E, - - b ~ a (Ax) 4 f^>(f) - - (fc - a) f (4) m
2 " " 180 { ' 1 (;) ~ 2880/2 4 1 {;) -
L
Sin =
Ax
EXAMPLE 3 Consider (x 3 + 3x 2 ) dx. We compare the trapezoidal rule
and Simpson's rule approximations with 4 subintervals. We thus have Ax =
(3 - l)/4 = 0.5 and
0.5
T 4 = â€” [4 + 2(10.125 + 20 + 34.375) + 54] = 46.75;
0.5
S 4 = â€” [4 + 4(10.125) + 2(20) + 4(34.375) + 54] = 46.
Note that
(x 3 + 3x 2 ) dx = (I* 4 + x 3 ) \\ = (f + 27) - (i + 1) = 46,
so Simpson's rule agrees with the exact answer. This should not be a surprise,
since for f(x) = x 3 + 3x 2 , we have that f {4 \x) is identically zero, which means
that the error term Â£4 for Simpson's rule must be zero. â™¦
Approximating Double Integrals over Rectangles
Now let / be a function of two variables that is continuous on the rectangle
R = [a, b] x [c, d] in R 2 . We adapt the previous ideas to provide methods for
approximating the value of the double integral Jj R fdA.
Because we assume that / is continuous on R, Fubini's theorem applies to
give
/ / f(x,y)dA= f f f(x,y)dydx. (3)
J J R J a Jc
Next we partition the x-interval [a, b] into m equal subintervals. Thus,
b â€” a
Ax = , a = xo < x\ < . . . < x m = b, where x, â– = a + / Ax.
m
Similarly, we partition the y-interval [c, d] into n equal subintervals; hence,
d â€” c
Ay = , c = yo < v\ < . . . < yâ€ž = d, where y ; - = c + j Ay.
n
In the inner integral f d f(x, y) dy of the iterated integral in (3), the variable x is
held constant. Therefore, we may approximate this integral using the trapezoidal
Chapter 5 i Multiple Integration
rule of Theorem 7.1. We obtain
f{x,y)dy
Ay
n-l
/(ac,c) + 2^)/(x,y;) + /(x,d)
7 = 1
Next, integrate each function of x appearing on the right side, so that
-b pel
J a J c
Ay
2
y)dydx &
r-b n â€” 1 pb r-b
J f{x, c)dx + 2 I fi x ^yj)dx+ I fix,d)dx
J a ,-_ i J a J a
(4)
Now use the trapezoidal rule again on each integral appearing on the right. This
means that, for j = 0, . . . , n,
f
J (I
fix, yj)dx
Ax
f(a t yj) + 2j2f(Xi,yj) + m yj)
(5)
Putting (4) and (5) together, we obtain
p b pel
/ / /(*,
J a J c
y) dy dx
Ay Ax
mâ€” 1
n-l
Ax
fia,c) + 2 Y J fi.x i ,c) + fib, c)
mâ€” 1
f(a.y,) â€¢ 2j./|v,.v / ) â€¢ /(/>. V;)
+
7=1
Ax
i=i
fia,d) + 2Y J f(x i ,d) + fib,d)
Therefore, the trapezoidal rule approximation Tâ€ž hn to ff R f dA is
T
1 m.n
Ax Ay
-l
fia,c) + 2j2f(xi,c) + fQ>,c)
( = 1
nâ€” 1 m â€” 1
nâ€” 1 nâ€” 1 mâ€” 1 ftâ€” 1
+ 2 J] /(a, y,) + 4Â£Â£ /(x, , yj ) + 2 J] fib, yj) (6)
7 = 1 7 = 1 Â«=1 7 = 1
1-1
+ fia,d) + 2Y J f{x i ,d) + fib,d)
i=i
The expression appearing in (6) is not too memorable as it stands. However,
we may interpret it as follows:
Ax Ay -r-^ ^-^
T m ,â€ž = â€”â€” 2^ yj)>
j=0 1=0
5.7 | Numerical Approximations of Multiple Integrals (optional)
where
1 if (xi , yj) is one of the four vertices of R ;
2 if (xi , yj) is a point on an edge of R, but not a vertex;
4 if (x, , yj) is a point in the interior of R.
EXAMPLE 4 We approximate the value of L j x {xy + 3x)dydx with 7^2.
Thus, the ^-interval [3, 6] is partitioned using Ajc = (6 â€” 3)/3 = 1 and the y-
interval [1, 2] is partitioned using Ay = (2 â€” l)/2 = 0.5. See Figure 5.126 for
the rectangle [3,6] x [1,2] with partition points marked.
Hence, we have
(xy + 3x) dy dx Â«
1(0.5)
[12 + 24 + 30+ 15
1
+ 2(16 + 20 + 27 + 20 + 25 + 13.5) + 4(18 + 22.5)]
-(486) = 60.75.
y
2 --
1.5
1 +
15
13.5,
12
20
25
30
18
22.5
27
â€¢
â€¢
16
20
24
H 1 1 h
H h
Figure 5.126 The rectangle
[3, 6] x [1, 2] of Example 4 with
partition points marked with the
values of f(x, y) = xy + 3x.
For comparison, we calculate the iterated integral directly:
2
dx
y=l
f 6 9 9 2
= / -x dx = -x
J 3 2 4
Note that T3 2 gives the exact answer in this case. â™¦
In the derivation above of the trapezoidal rule we assumed that the func-
tion / was continuous. Nonetheless, we may use formula (6) to approximate
ff R f dA whenever / is integrable on R. However, in order to make estimates
of the accuracy of trapezoidal approximations, we must assume more about /, as
the following result (whose proof involves use of the intermediate value theorem
and the mean value theorem for integrals) indicates.
if;
(xy + 3x)dy dx
= f(
xy-
2
3xy
243
= 60.75.
394 Chapter 5 | Multiple Integration
THEOREM 7.3 (TRAPEZOIDAL RULE FOR DOUBLE INTEGRALS OVER RECTANGLES)
Given a function / that is integrable on the rectangle R = [a , b] x [c, d], we have
//.
fix, y)dA = T m%n + E m>n ,
R
where Tâ€ž u â€ž is given by (6) and E mj , denotes the error involved in using Tâ€ž un to
approximate the value of the double integral. Moreover, if / is of class C 2 on R,
then there exist points (fi , rj\) and (Â£2, Vi) in R such that
E m , n = _ ( fc -Â°X^- c ) [(A*) 2 /^, , x ) + (Av) 2 / yy fe, ift)] â€¢
EXAMPLE 5 For /(*, y) = xy + 3x, we have that d 2 f/dx 2 and d 2 f/dy 2 are
both identically zero. Hence, Theorem 7.3 implies that the trapezoidal rule ap-
proximation is exact, as we already observed in Example 4. â™¦
In a manner entirely analogous to our derivation of the trapezoidal rule, we
may also produce a Simpson's rule approximation to the value of the double
integral ff R f dA. As in the case of Simpson's rule for approximating single-
variable definite integrals, we partition both the x- and y-intervals into even
numbers of equal subintervals. Thus, we take
b â€” a
Ax = , a = xq < x\ < . . . < X2 m = b, where x,- = a + / Ax,
2m
and
d â€” c
Ay = , c = yo < y\ < . . . < ym = d, where y 7 - = c + j Ay.
2n
The resulting Simpson's rule approximation S2 m ,2n is given by the expression
Ax Ay
mâ€” 1 m
fia, c) + 2j2 f(*2i ,c) + 4J2 f(.X2i-u c) + fib, c)
(=1 (=1
â€” 1 nâ€” 1 mâ€” 1 nâ€” 1 m
+ 2 f(a, yij) f(*Â»> *y) + 8 E E
;=1 7=1 i=l ;=1 t=l
n-1
+ 2J2fib,y 2 j) (7)
7 = 1
+ fia, yij-x) + /(**. yy-i)
7=1 ;=1 i=l
n m n
+ 16 E E /c^-i. ^-i) + 4 E ^-0
7=1 i=l 7=1
m â€” 1 m
+ /(fl, J) + 2 J] /(x 2 â€ž rf) + 4Â£ f(x2i-i,d) + fib, d)
/=! i=i
5.7 | Numerical Approximations of Multiple Integrals (optional)
Just as in the case of the trapezoidal rule for double integrals over rectangles, the
intermediate value theorem and the mean value theorem for integrals provide the
following result.
THEOREM 7.4 (Simpson's rule for double integrals over rectangles)
Let / be a function that is integrable on the rectangle R = [a, b] x [c, d]. Then
f(x, y)dA = S2,â€ž,2n + ^2m,2n>
where S2m,2n is given by (7) and E 2m ,2â€ž denotes the error involved in using S% m ^ n
to approximate the value of the double integral. Moreover, if / is of class C 4 on
R, then there exist points (Â£i , r]\) and (Â£2, m) in R such that
E 2m ,2n = ~ (b ~ a ^ K AX ) 4 + (Ay)Vvvvv(Â£2, m)] â–
EXAMPLE 6 We compare approximations to the value of / 0 Â°' 5 /â€ž' e x+y dydx
using both the trapezoidal rule and Simpson's rule with four subintervals in
each of the x- and y-intervals. Thus, the x-interval [0, 0.5] is partitioned using
Ax = (0.5 â€” 0)/4 = 0.125 and the y-interval [0, 1] is partitioned using Ay =
(1 -0)/4 = 0.25.
The trapezoidal rule gives
T 4A = ( Q - 125 XÂ°- 25 > yO+0 + e 0+0.5 + e l+0 + e 0.5+l + 2 ^0+0.25 + e 0+0.5
_|_ ^0+0.75 _j_ e 0.125+0 _j_ ^0.25+0 _|_ ^0.375+0 _|_ ^0.125+1 _|_ ^0.25+1
_|_ ^0.375+1 _|_ ^0.5+0.25 _j_ e 0.5+0.5 _j_ ^0.5+0.75^
0.125+0.25 | â€ž0.125+0. 5 _|_ g 0. 125+0.75 _|_ g 0.25+0.25
_|_ g 0.375+0.5 + ^0.375+0.75^ j
+ 4(e"
, ^0. 25+0. 75 , g 0.37
1
= (143.608854) = 1.121944.
128
Note that d 2 /dx 2 (e x+y ) = d 2 /dy 2 (e x+y ) = e x+y . Theorem 7.3 says that there
exist points rji) and (Â£2, V2) in the rectangle [0, 0.5] x [0, 1] such that
Eaa = _ (0-5-0)(l -0) ^ (0 _ 125) 2 efl+1J1 + (0 . 2 5)V 2 +" 2 ] .
On the rectangle [0, 0.5] x [0, 1], the smallest possible value of e x+y is e 0+0 = 1
and the largest possible value is e 0 5+l = e LS . Hence, we must have
_(Ogl) j- (0 125) 2 e i. 5 + (0.25)2^-5] < e 4A < [ (0 . 125 ) 2 + (0.25) 2 ]
or
-0.0145888 < Â£44 < -0.003255.
Chapter 5 | Multiple Integration
|e 0+0.25 + ^0+0.75 + g 0.125+0 + ^0.125+1 + ^0.25+0.5
g 0.375+0 . g 0.375+l . g 0.5+0.25 , ^0.5+0.75-,
Hence, the true value of the double integral lies between 1.121944 â€” 0.0145888 =
1.10735 and 1.121944 - 0.003255 = 1.11869.
On the other hand, Simpson's rule gives
5 4 ,4 = (0 - 125 ^ (0 - 25) + + e H0 + ^0.5+1
+ 2 ( e Â°+ 0 - 5 + e Â°- 25+0 + e Â°- 25+1 + e Â°- 5+0 - 5 )
+ 4(e
" 5 )
_|_ g ^0.125+0.5 _|_ ^0.25+0.25 _|_ g 0.25+0.75 _|_ g 0.375+0.5^
_|_ ! g ^0.125+0.25 _|_ ^0.125+0.75 _|_ ^0.375+0.25 _|_ ^0.375+0.75^
= â€”(321.036910) = 1.114711.
288
In this case, we note that d 4 /dx 4 (e x+y ) = d 4 /dy 4 (e x+y ) = e x+y , so, as before,
the minimum and maximum values of these partial derivatives on [0, 0.5] x [0, 1]
are, respectively, 1 and e . Therefore, Theorem 7.4 implies that
[(Â°- 125 ) V - 5 + (0-25)V- 5 ] < Â£ 4 .4 < [(0.125) 4 + (0.25) 4 ]
or
-0.0000516688 < Â£ 4 ,4 < -0.0000115289.
Hence, the true value of the double integral lies between 1.114711 â€”
0.0000516688 = 1.11466 and 1.114711 - 0.0000115289 = 1.1147.
For comparison, we may calculate the iterated integral exactly:
/ / e x+y dydx=\ (e x+y )f dx = / (e x+l - e x ) dx
Jo Jo Jo ' Jo
= (e x+l - e x )\Â° Q 5 = e hS - e 0 - 5 - e + \ ^ 1.114686.
Thus, we see that Simpson's rule gives a highly accurate approximation with a
very coarse partition. â™¦
Approximating Double Integrals over Elementary Regions
We can modify the methods for approximating double integrals over rectangles
to approximate double integrals over more general regions. Suppose first that D
is an elementary region of type 1 ; that is,
D = {{x, y) 6 R 2 I y(x) < y < S(x), a U.J
/ {y 4 - xy 2 ) dy dx
-o.i Jo
pOA p2
JO Jl
1 +X'
dy dx
Ii Jo
Jo Jn
? x+2}, dvdx
-7i/4 rn/2
10. / / sin2x cos3y dy dx
rr/4
px/4 rn/4
11.1 / sin (x + y) dy dx
Jo Jo
1.1 rx/4
Ii Jo
e x cos v dy dx
13. In Chapter 7 we will see that the area of the portion of
the graph of f(x, y) for (x, y) in a region D in R 2 is
given by the double integral / f D J f 2 + f 2 + 1 dA.
(a) Set up an appropriate iterated integral to compute
the surface area of the portion of the paraboloid
z = 4 - x 2 - 3y 2 where (x, y) e [0, 1] x [0, 1].
(b) Use the trapezoidal rule approximation 74 4 to es-
timate the surface area.
14. Concerning the iterated integral
f\' 5 f\.4 ln ( 2x + y) dydx:
(a) Calculate the trapezoidal rale approximation T^.
(b) Use Theorem 7.3 to estimate the error in your ap-
proximation in part (a).
(c) Calculate the Simpson's rule approximation 52,4.
True/False Exercises for Chapter 5 401
(d) Use Theorem 7.4 to estimate the error in your ap-
proximation in part (a).
15. Without either evaluating or estimating the integral
fi 4 Jo s ln(xy)dy dx, which approximation is more
accurate: 74 4 or 52,2? Explain your answer.
1 6. Suppose that the trapezoidal rule is used to estimate the
value of / 0 Â°' 2 /^qj e xl+2y dydx. Determine the small-
est value of n so that the resulting approximation Tâ€ž t â€ž
is accurate to within 10~ 4 of the actual value of the
integral.
17. Consider J 0 Â°' 3 J 0 Â°' 4 e x ~ y dy dx.
(a) If the trapezoidal rule approximation T n â€ž is used
to estimate the value of this integral, what is
the smallest value of n so that the resulting
approximation is accurate to within 10~ 5 of the
actual value?
(b) If the Simpson's rule approximation S2,,, iâ€ž is used
to estimate the value of this integral, what is the
smallest value of n so that the resulting approx-
imation is accurate to within 10~ 5 of the actual
value?
18. Concerning the iterated integral JqJq(3x +
5y)dy dx:
(a) Calculate the trapezoidal rule approximation 7/2,2.
(b) Compare your result in part (a) with the exact
answer.
(c) Use Theorem 7.3 to explain your results in parts
(a) and (b).
19. Consider the iterated integral j\ f^ 2 x 3 y 3 dy dx:
(a) Calculate the Simpson's rule approximation Â£2,2-
(b) Compare your result in part (a) with the exact an-
swer.
(c) Use Theorem 7.4 to explain your results in parts
(a) and (b).
In Exercises 20-25, (a) use the approximation T3 3 from the
trapezoidal rule to estimate the values of the given integrals.
20. J J (x 3 + 2y 2 ) dy dx
/*jr/4 pcosx
21 . / / (2x cos y + sin 2 x) dy dx
Jo J sinx
,.0.3 fix
22. J J (xy - x 2 ) dy dx
23. / / dydx
Jo Jo v 1 â€” y 2
24. J j sin x dx dy (Note the order of integration.)
/1.6 />2y
J In (xy)dxdy
26. In this problem, you will develop another way to think
about the trapezoidal rule approximation given in equa-
tion (6).
(a) Let L denote a general linear function of two vari-
ables, that is, L(x, y) = Ax + By + C, where A,
B, and C are constants. Set R = [a, b] x [c, d].
Show that
/ / LdA = (area of W)(average of the values of
â€¢> J R L taken at the four vertices of/?).
(Note that this gives an exact expression for the
double integral.)
(b) Suppose that / is any function of two variables that
is integrable on R. Show that the trapezoidal rule
approximation Ti \ to ff R f dA is
7^1 = (area of /?)(average of the values off
taken at the four vertices of R).
(c) Now let Ax = (b â€” a)/m, Ay = (d â€” c)/n,
and, for i = 1 , . . . , m, j = 1, . . . , n, let /?,â€¢_,â€¢ =
[Xi-i, x{\ x [y,_i, y,], where x t = a + iAx and
y/ = c + j Ay. Then we have
///<" = Â£Â£ // f dA -
Use 7i t i to approximate each integral ff R fdA
and sum the results to obtain the formula for Tâ€ž un
given by equation (6).
True/False Exercises for Chapter 5
1. Every rectangle in R 2 may be denoted [a, b] x [c, d].
f 1/2 f 2 3 2
. I y sin {jix ) dy dx
2. If / is a continuous function and f(x, y) > 0 on a
region D in R 2 , then the volume of the solid in R 3 r 2 Z" 1 / 2
under the graph of the surface z = f(x, y) and above = J i Jo ^ s ^ n ^ nx ^ dx
the region D in the xy-plane is / / f(x,y)dA.
402 Chapter 5 | Multiple Integration
/ / 3 dy dx = J J 3 dx dy
Jo Jo Jo Jo
10 Jo
-2 r y
5- / / f(x,y)dxdy= I I f(x, y)dy dx for
Jo Jo Jo Jo
all continuous functions /.
6. 0 < j J sin(x 4 + y 4 )dx dy < jt, where D denotes
the unit disk {(x, y) \ x 1 + y 2 < 1}.
7. f [ sfx 3 + ydydx = [ [ ^x 3 + y dx dy.
Jo Jo Jo Jo
8. J x 2 e x+y dydx=(^J x 2 e x dx\(j^ e y dy^j.
9. The region in R 2 bounded by the graphs of y = x 3
and y = ~Jx is a type 3 elementary region in the
plane.
10. The region in R 2 bounded by the graphs of y = sinx,
y = cosx,x = u j A, andx = 5 jr/4 is a type 2 elemen-
tary region in the plane.
1 1 . The region in R 3 bounded by the graphs of y 2 â€” x 2 â€”
z 2 = 1 and 9x 2 + 4z 2 = 36 is a type 3 elementary
region in space.
12. j j (y 3 + l)dxdy gives the area of the region D,
where D = {(x, y) | (x - 2) 2 + 3y 2 < 5}.
' // <
angle with vertices (-1,0), (1, 0), (0, 1).
13. / / (y 2 sin (x 3 ) + 3) dx dy = 3 /2, where D is the tri-
(x + y 3 +z 5 )dV = 0.
[-2,2]x[-l,l]x[-3,3]
. [ [ I (x + z)dV = 0.
/[-2,2]x[0,l]x[-l,l]
SSL
â– SSL
l l l[-2,2]
Jo Jv/4 J-
x[0,l]x[-l,l]
4 n^y/2 f 2
(y + z)dV = 0.
sin yz dz dx dy = 0.
18. / / / (y -x 2 )dzdydx = 0.
J-l J-JT^x 2 J-^/l-x 2 -/
19. If T(n, v) = (2m â€” v, u + 3d), then the area of the im-
age/) = T(D*)ofthe unit square/)* = [0, 1] x [0, 1]
is 7 square units.
20. If T(m, v) = (v â€” u,3u + 2v), then the area of the im-
age D = T(D*) of the rectangle D* = [0, 3] x [0, 2]
is 5 square units.
21
77
JO Jy
2 r5-2y
12
e 2x y cos(x â€” 3y) dx dy
10 nu/2
I f
JO Ju
-5e" cos vdvdu.
22. If D is the disk {(x, y) | x 2 + y 2 < 9}, then
jj v / 9-x 2 -y 2 dA = 36jr.
/>2jt f>2 p4
23. The iterated integral / / / dzdrdO represents
Jo Jo J2r
the volume enclosed by the cone of height 4 and
radius 2.
24. The iterated integral
ffZS
J-2 J-V^x 2 JO
{2 + Jx 2 + y 2 )dzdy dx
is given by an equivalent integral in cylindrical coor-
dinates as
2m p2 /.79-r 2
pill pi r>
Jo Jo Jo
(r 2 + 2r)dz dr d6.
10 Jo Jo
25. The iterated integral
fV2=^ n^4-x 2 -y 2
/ / / 7x 2 + y 2 +z 2 +5dzdydx
J-s/2.Jo J^/x 2 +y 2
is given by an equivalent integral in spherical coordi-
nates as
/ / / Pv P 2 + 5 sin*