Section 13.3 Arc Length and Curvature 869
; 34–35 Use a graphing calculator or computer to graph both the 47–48 Find the vectors T, N, and B at the given point.
curve and its curvature function sxd on the same screen. Is the
graph of what you would expect? 47. rstd − k t 2, 2 t 3, t l , s1, 2 , 1d
3 3
34. y − x 4 2 2x 2 35. y − x22
48. rstd − k cos t, sin t, ln cos t l, s1, 0, 0d
CAS 36–37 Plot the space curve and its curvature function std. 49–50 Find equations of the normal plane and osculating plane
Comment on how the curvature reflects the shape of the curve. of the curve at the given point.
36. rstd − k t 2 sin t, 1 2 cos t, 4 cossty2d l, 0 < t < 8 49. x − sin 2t, y − 2cos 2t, z − 4t; s0, 1, 2d
37. rstd − k tet, e2t, s2 t l , 25 < t < 5 50. x − ln t, y − 2t, z − t 2; s0, 2, 1d
38–39 Two graphs, a and b, are shown. One is a curve y − f sxd ; 51. F ind equations of the osculating circles of the ellipse
and the other is the graph of its curvature function y − sxd. 9x 2 1 4y 2 − 36 at the points s2, 0d and s0, 3d. Use a graph-
Identify each curve and explain your choices. ing calculator or computer to graph the ellipse and both
osculating circles on the same screen.
38. 39. ; 52. F ind equations of the so0s,c0udlaatnindgsc1i,r12clde. sGorfapthhebpoatrhaobsoclau
y y 1
y − 2 x2 at the points
lating circles and the parabola on the same screen.
a a 53. A t what point on the curve x − t 3, y − 3t, z − t 4 is the
b b normal plane parallel to the plane 6x 1 6y 2 8z − 1?
x x CAS 54. Is there a point on the curve in Exercise 53 where the
osculating plane is parallel to the plane x 1 y 1 z − 1?
[Note: You will need a CAS for differentiating, for simplify-
ing, and for computing a cross product.]
CAS 40. (a) Graph the curve rstd − ksin 3t, sin 2t, sin 3t l. At how 55. Find equations of the normal and osculating planes of the
many points on the curve does it appear that the curva- curve of intersection of the parabolic cylinders x − y 2 and
z − x 2 at the point s1, 1, 1d.
ture has a local or absolute maximum?
56. Show that the osculating plane at every point on the curve
(b) Use a CAS to find and graph the curvature function. kt l1
r std 2, 1 t, t 2 is the same plane. What can you
Does this graph confirm your conclusion from part (a)? 2
41. The graph of rstd kt 3 sin t, 1 3 cos t, tl is shown − 1 2
2 2
CAS − 2 2 conclude about the curve?
in Figure 13.1.12(b). Where do you think the curvature is 57. Show that at every point on the curve
largest? Use a CAS to find and graph the curvature function.
For which values of t is the curvature largest? rstd − ke t cos t, e t sin t, e t l
42. U se Theorem 10 to show that the curvature of a plane para- the angle between the unit tangent vector and the z-axis is
metric curve x − f std, y − tstd is the same. Then show that the same result holds true for the
unit normal and binormal vectors.
| |
− ?x?y? 2 y? ?x? 58. The rectifying plane of a curve at a point is the plane that
fx? 2 1 y? 2 g3y2
contains the vectors T and B at that point. Find the recti-
where the dots indicate derivatives with respect to t. fying plane of the curve rstd − sin t i 1 cos t j 1 tan t k at
43–45 Use the formula in Exercise 42 to find the curvature. the point (s2 y2, s2 y2, 1).
43. x − t 2, y − t 3
44. x − a cos t, y − b sin t 59. Show that the curvature is related to the tangent and
45. x − e t cos t, y − e t sin t normal vectors by the equation
dT − N
ds
| | 60. Show that the curvature of a plane curve is − dyds ,
where is the angle between T and i; that is, is the angle
46. Consider the curvature at x − 0 for each member of the of inclination of the tangent line. (This shows that the
family of functions f sxd − e cx. For which members is s0d
largest? definition of curvature is consistent with the definition for
plane curves given in Exercise 10.2.69.)
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870 chapter 13 Vector Functions
61. (a) Show that d Byds is perpendicular to B. 64. Show that the circular helix rstd − k a cos t, a sin t, bt l,
(b) Show that d Byds is perpendicular to T. where a and b are positive constants, has constant curvature
(c) Deduce from parts (a) and (b) that d Byds − 2ssdN
and constant torsion. [Use the result of Exercise 63(d).]
for some number ssd called the torsion of the curve.
(The torsion measures the degree of twisting of a curve.) 65. U cusrevethrestfdor−muklta, 1lin Exercise1 63(d) to find the torsion of the
(d) Show that for a plane curve the torsion is ssd − 0. 2 3
t 2, t 3 .
62. The following formulas, called the Frenet-Serret formulas, 66. Find the curvature and torsion of the curve x − sinh t,
are of fundamental importance in differential geometry: y − cosh t, z − t at the point s0, 1, 0d.
1. dTyds − N 67. T he DNA molecule has the shape of a double helix (see
2. dNyds − 2T 1 B Figure 3 on page 850). The radius of each helix is about
3. dByds − 2 N 10 angstroms (1 Å − 1028 cm). Each helix rises about 34 Å
(Formula 1 comes from Exercise 59 and Formula 3 comes during each complete turn, and there are about 2.9 3 108
from Exercise 61.) Use the fact that N − B 3 T to
deduce Form ula 2 from Formulas 1 and 3. complete turns. Estimate the length of each helix.
68. Let’s consider the problem of designing a railroad track to
make a smooth transition between sections of straight track.
Existing track along the negative x-axis is to be joined
63. Use the Frenet-Serret formulas to prove each of the follow- smoothly to a track along the line y − 1 for x > 1.
ing. (Primes denote derivatives with respect to t. Start as in (a) Find a polynomial P − Psxd of degree 5 such that the
the proof of Theorem 10.)
Hfunction F defined by if x < 0
(a) r0 − s0T 1 ss9d2 N 0 if 0 , x , 1
Fsxd − Psxd
(b) r9 3 r0 − ss9d3 B 1 if x > 1
(c) r- − f s- 2 2ss9d3g T 1 f 3s9s0 1 9ss9d2 g N 1 ss9d3B is continuous and has continuous slope and continuous
curvature.
s r9 3 r0 d ? r-
r9 3 r0 2 ; (b) Graph F.
| | (d) −
z r ( t+h )-r ( t ) In this section we show how the ideas of tangent and normal vectors and curvature can
C rª(t) h be used in physics to study the motion of an object, including its velocity and accelera-
O Q
tion, along a space curve. In particular, we follow in the footsteps of Newton by using
x
these methods to derive Kepler’s First Law of planetary motion.
FIGURE 1 Suppose a particle moves through space so that its position vector at time t is rstd.
Notice from Figure 1 that, for small values of h, the vector
P 1 rst 1 hd 2 rstd
h
r(t) r(t+h)
approximates the direction of the particle moving along the curve rstd. Its magnitude mea-
sures the size of the displacement vector per unit time. The vector (1) gives the average
velocity over a time interval of length h and its limit is the velocity vector vstd at time t:
y 2 vstd − lim rst 1 hd 2 rstd − r9std
h
hl0
Thus the velocity vector is also the tangent vector and points in the direction of the tan-
gent line.
| |The speed of the particle at time t is the magnitude of the velocity vector, that is,
vstd . This is appropriate because, from (2) and from Equation 13.3.7, we have
ds
−| | | |vstdr9std − dt − rate of change of distance with respect to time
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Section 13.4 Motion in Space: Velocity and Acceleration 871
y As in the case of one-dimensional motion, the acceleration of the particle is defined as
v(1) the derivative of the velocity:
a(1)
astd − v9std − r0std
(1, 1)
0 Example 1 The position vector of an object moving in a plane is given by
FIGURE 2 rstd − t3 i 1 t2 j. Find its velocity, speed, and acceleration when t − 1 and illustrate
geometrically.
SOLUTION The velocity and acceleration at time t are
vstd − r9std − 3t2 i 1 2t j
x and the speed is astd − r0std − 6t i 1 2 j
| |vstd − ss3t2 d2 1 s2td2 − s9t 4 1 4t 2
TEC Visual 13.4 shows animated When t − 1, we have ■
velocity and acceleration vectors for ■
objects moving along various curves. | |vs1d − 3 i 1 2 j as1d − 6 i 1 2 j vs1d − s13
Figure 3 shows the path of the particle These velocity and acceleration vectors are shown in Figure 2.
in Example 2 with the velocity and
acceleration vectors when t − 1. Example 2 Find the velocity, acceleration, and speed of a particle with position
z a(1) vector rstd − k t 2, e t, te t l.
v(1) SOLUTION
vstd − r9std − k 2t, et, s1 1 tde tl
astd − v9std − k 2, et, s2 1 tde tl
| | vstd − s4t 2 1 e 2t 1 s1 1 td2e 2t
1 The vector integrals that were introduced in Section 13.2 can be used to find position
x y vectors when velocity or acceleration vectors are known, as in the next example.
FIGURE 3 Example 3 A moving particle starts at an initial position rs0d − k1, 0, 0 l with initial
velocity vs0d − i 2 j 1 k. Its acceleration is astd − 4t i 1 6t j 1 k. Find its velocity
and position at time t.
SOLUTION Since astd − v9std, we have
vstd − y astd dt − y s4t i 1 6t j 1 kd dt
− 2t2 i 1 3t2 j 1 t k 1 C
To determine the value of the constant vector C, we use the fact that vs0d − i 2 j 1 k.
The preceding equation gives vs0d − C, so C − i 2 j 1 k and
vstd − 2t 2 i 1 3t 2 j 1 t k 1 i 2 j 1 k
− s2t 2 1 1d i 1 s3t 2 2 1d j 1 st 1 1d k
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872 Chapter 13 Vector Functions
The expression for rstd that we obtained Since vstd − r9std, we have
in Example 3 was used to plot the path
of the particle in Figure 4 for 0 < t < 3. rstd − y vstd dt
y− fs2t2 1 1d i 1 s3t2 2 1d j 1 st 1 1d kg dt
6 − ( 2 t3 1 t) i 1 st 3 2 td j 1 ( 1 t2 1 t) k 1 D
3 2
z4
2 (1, 0, 0) 0 Putting t − 0, we find that D − rs0d − i, so the position at time t is given by
20 x
00 5 10 15 20 rstd − ( 2 t3 1 t 1 1) i 1 st 3 2 td j 1 ( 1 t2 1 t) k ■
y 3 2
FIGURE 4 In general, vector integrals allow us to recover velocity when acceleration is known
and position when velocity is known:
y yvstd − vst0d 1 t asud du rstd − rst0d 1 t vsud du
t0 t0
If the force that acts on a particle is known, then the acceleration can be found from
Newton’s Second Law of Motion. The vector version of this law states that if, at any
time t, a force Fstd acts on an object of mass m producing an acceleration astd, then
Fstd − mastd
The object moving with position P has Example 4 An object with mass m that moves in a circular path with constant angular
angular speed − dydt, where is
the angle shown in Figure 5. speed has position vector rstd − a cos t i 1 a sin t j. Find the force acting on the
object and show that it is directed toward the origin.
y SOLUTION To find the force, we first need to know the acceleration:
P
vstd − r9std − 2a sin t i 1 a cos t j
astd − v9std − 2a2 cos t i 2 a2 sin t j
¨ x Therefore Newton’s Second Law gives the force as
0 Fstd − mastd − 2m2sa cos t i 1 a sin t jd
Notice that Fstd − 2m2 rstd. This shows that the force acts in the direction opposite
to the radius vector rstd and therefore points toward the origin (see Figure 5). Such a
FIGURE 5
force is called a centripetal (center-seeking) force. ■
y d Projectile Motion
v¸
Example 5 A projectile is fired with angle of elevation ␣ and initial velocity v0. (See
a Figure 6.) Assuming that air resistance is negligible and the only external force is due to
0 gravity, find the position function rstd of the projectile. What value of ␣ maximizes the
range (the horizontal distance traveled)?
FIGURE 6
x SOLUTION We set up the axes so that the projectile starts at the origin. Since the force
due to gravity acts downward, we have
F − ma − 2mt j
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Section 13.4 Motion in Space: Velocity and Acceleration 873
| |where t − a < 9.8 mys2. Thus
a − 2t j
Since v9std − a, we have
vstd − 2tt j 1 C
where C − vs0d − v0. Therefore
r9std − vstd − 2tt j 1 v0
Integrating again, we obtain
rstd − 221 tt 2 j 1 t v0 1 D
But D − rs0d − 0, so the position vector of the projectile is given by
3 rstd − 221 tt 2 j 1 t v0
| |If we write v0 − v0 (the initial speed of the projectile), then
v0 − v0 cos ␣ i 1 v0 sin ␣ j
and Equation 3 becomes
f grstd − sv0 cos ␣dt i 1 sv0 sin ␣dt 2 1 tt2 j
2
The parametric equations of the trajectory are therefore
If you eliminate t from Equations 4, you 4 x − sv0 cos ␣dt y − sv0 sin ␣dt 2 1 tt 2
will see that y is a quadratic function 2
of x. So the path of the projectile is part
of a parabola.
The horizontal distance d is the value of x when y − 0. Setting y − 0, we obtain t − 0
or t − s2v0 sin ␣dyt. This second value of t then gives
d − x − sv0 cos ␣d 2v0 sin ␣ − v02s2 sin ␣ cos ␣d − v02 sin 2␣
t t t
Clearly, d has its maximum value when sin 2␣ − 1, that is, ␣ − 45°. ■
Example 6 A projectile is fired with muzzle speed 150 mys and angle of elevation 45°
from a position 10 m above ground level. Where does the projectile hit the ground, and
with what speed?
SOLUTION If we place the origin at ground level, then the initial position of the projec-
tile is (0, 10) and so we need to adjust Equations 4 by adding 10 to the expression for y.
With v0 − 150 mys, ␣ − 45°, and t − 9.8 mys2, we have
x − 150 coss45°dt − 75s2 t
y − 10 1 150 sins45°dt 2 1 s9.8dt 2 − 10 1 75s2 t 2 4.9t 2
2
Impact occurs when y − 0, that is, 4.9t2 2 75s2 t 2 10 − 0. Using the quadratic
formula to solve this equation (and taking only the positive value of t), we get
t − 75s2 1 s11,250 1 196 < 21.74
9.8
Then x < 75s2 s21.74d < 2306, so the projectile hits the ground about 2306 m away.
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874 Chapter 13 Vector Functions
The velocity of the projectile is ■
vstd − r9std − 75s2 i 1 s75s2 2 9.8td j
So its speed at impact is
| | vs21.74d − s(75s2 )2 1 (75s2 2 9.8 ? 21.74)2 < 151 mys
Tangential and Normal Components of Acceleration
When we study the motion of a particle, it is often useful to resolve the acceleration into
| |two components, one in the direction of the tangent and the other in the direction of the
normal. If we write v − v for the speed of the particle, then
Tstd − r9std − vstd − v
v
| r9std | | vstd |
and so v − vT
If we differentiate both sides of this equation with respect to t, we get
5 a − v9 − v9T 1 vT9
If we use the expression for the curvature given by Equation 13.3.9, then we have
6 − | T9| − | T9| | so T9| − v
| r9|
v
| |The unit normal vector was defined in the preceding section as N − T9y T9 , so (6) gives
T9 − | T9|N − vN
and Equation 5 becomes
7 a − v9T 1 v2 N
aT Writing aT and aN for the tangential and normal components of acceleration, we have
T
a a − aT T 1 aN N
N
where
aN 8 aT − v9 and aN − v2
FIGURE 7 This resolution is illustrated in Figure 7.
Let’s look at what Formula 7 says. The first thing to notice is that the binormal vector
B is absent. No matter how an object moves through space, its acceleration always lies in
the plane of T and N (the osculating plane). (Recall that T gives the direction of motion
and N points in the direction the curve is turning.) Next we notice that the tangential
component of acceleration is v9, the rate of change of speed, and the normal component
of acceleration is v2, the curvature times the square of the speed. This makes sense if we
think of a passenger in a car—a sharp turn in a road means a large value of the curvature
, so the component of the acceleration perpendicular to the motion is large and the pas-
senger is thrown against a car door. High speed around the turn has the same effect; in
fact, if you double your speed, aN is increased by a factor of 4.
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Section 13.4 Motion in Space: Velocity and Acceleration 875
Although we have expressions for the tangential and normal components of accelera-
tion in Equations 8, it’s desirable to have expressions that depend only on r, r9, and r0.
To this end we take the dot product of v − vT with a as given by Equation 7:
v ? a − vT ? sv9T 1 v2 Nd
− vv9T ? T 1 v3 T ? N
− vv9 (since T ? T − 1 and T ? N − 0)
Therefore
v ?a r9std ? r0std
v r9std
| |9 aT − v9 − −
Using the formula for curvature given by Theorem 13.3.10, we have
10 aN − v2 − | r9std 3 r99std | | r9std | 2 − | r9std 3 r99std |
| r9std |3 | r9std |
Example 7 A particle moves with position function rstd − kt2, t2, t3 l . Find the
tangential and normal components of acceleration.
SOLUTION rstd − t 2 i 1 t 2 j 1 t 3 k
r9std − 2t i 1 2t j 1 3t2 k
r0std − 2 i 1 2 j 1 6t k
| |r9std − s8t2 1 9t4
Therefore Equation 9 gives the tangential component as
r9std ? r0std 8t 1 18t 3
r9std s8t 2 1 9t 4
| |aT− −
Since Z Zi j k
r9std 3 r0std − 2t 2t 3t 2 − 6t 2 i 2 6t 2 j
2 2 6t
Equation 10 gives the normal component as
| | | |
aN − r9std 3 r0std − 6 s2 t 2 ■
r9std s8t 2 1 9t 4
Kepler’s Laws of Planetary Motion
We now describe one of the great accomplishments of calculus by showing how the
material of this chapter can be used to prove Kepler’s laws of planetary motion. After 20
years of studying the astronomical observations of the Danish astronomer Tycho Brahe,
the German mathematician and astronomer Johannes Kepler (1571–1630) formulated
the following three laws.
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876 Chapter 13 Vector Functions
Kepler’s Laws
1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
2. The line joining the sun to a planet sweeps out equal areas in equal times.
3. T he square of the period of revolution of a planet is proportional to the cube of
the length of the major axis of its orbit.
In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show that
these three laws are consequences of two of his own laws, the Second Law of Motion and
the Law of Universal Gravitation. In what follows we prove Kepler’s First Law. The
remaining laws are left as exercises (with hints).
Since the gravitational force of the sun on a planet is so much larger than the forces
exerted by other celestial bodies, we can safely ignore all bodies in the universe except the
sun and one planet revolving about it. We use a coordinate system with the sun at the
origin and we let r − rstd be the position vector of the planet. (Equally well, r could be
the position vector of the moon or a satellite moving around the earth or a comet moving
around a star.) The velocity vector is v − r9 and the acceleration vector is a − r0. We use
the following laws of Newton:
Second Law of Motion: F − ma
Law of Gravitation: F − 2 GMm r − 2 GMm u
r3 r2
| |where F is the gravitational force on the planet, m and M are the masses of the planet and
the sun, G is the gravitational constant, r − r , and u − s1yrdr is the unit vector in the
direction of r.
We first show that the planet moves in one plane. By equating the expressions for F in
Newton’s two laws, we find that
a − 2 GM r
r3
and so a is parallel to r. It follows that r 3 a − 0. We use Formula 5 in Theorem 13.2.3 to
write
d sr 3 vd − r9 3 v 1 r 3 v9
dt
−v3v1r3a−010−0
Therefore r3v−h
where h is a constant vector. (We may assume that h ± 0; that is, r and v are not paral-
lel.) This means that the vector r − rstd is perpendicular to h for all values of t, so the
planet always lies in the plane through the origin perpendicular to h. Thus the orbit of the
planet is a plane curve.
To prove Kepler’s First Law we rewrite the vector h as follows:
h − r 3 v − r 3 r9 − r u 3 sr ud9
− r u 3 sr u9 1 r9ud − r 2su 3 u9d 1 rr9su 3 ud
− r 2su 3 u9d
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Section 13.4 Motion in Space: Velocity and Acceleration 877
Then
a 3 h − 2GM u 3 sr 2 u 3 u9d − 2GM u 3 su 3 u9d
r2
− 2GM fsu ? u9du 2 su ? udu9g (by Theorem 12.4.11, Property 6)
| | | |But u ? u − u 2 − 1 and, since ustd − 1, it follows from Example 13.2.4 that
u ? u9 − 0
Therefore a 3 h − GM u9
and so sv 3 hd9 − v9 3 h − a 3 h − GM u9
Integrating both sides of this equation, we get
11 v 3 h − GM u 1 c
z y where c is a constant vector.
v
h u At this point it is convenient to choose the coordinate axes so that the standard
c basis vector k points in the direction of the vector h. Then the planet moves in the
xy-plane. Since both v 3 h and u are perpendicular to h, Equation 11 shows that c lies
¨ in the xy-plane. This means that we can choose the x- and y-axes so that the vector i lies
r in the direction of c, as shown in Figure 8.
x If is the angle between c and r, then sr, d are polar coordinates of the planet. From
Equation 11 we have
FIGURE 8
r ? sv 3 hd − r ? sGM u 1 cd − GM r ? u 1 r ? c
where c − | c |. Then − GMr u ? u 1 | r | | c | cos − GMr 1 rc cos
r− r ? sv 3 hd − 1 r ? sv 3 hd
GM 1 c cos GM 1 1 e cos
where e − cysGMd. But
| |r ? sv 3 hd − sr 3 vd ? h − h ? h − h 2 − h2
where h − | h |. So
r − h 2ysGM d − 1 eh 2yc
1 1 e cos 1 e cos
Writing d − h 2yc, we obtain the equation
12 r − 1 1 ed
e cos
Comparing with Theorem 10.6.6, we see that Equation 12 is the polar equation of a conic
section with focus at the origin and eccentricity e. We know that the orbit of a planet is a
closed curve and so the conic must be an ellipse.
This completes the derivation of Kepler’s First Law. We will guide you through
the derivation of the Second and Third Laws in the Applied Project on page 880. The
proofs of these three laws show that the methods of this chapter provide a powerful tool
for describing some of the laws of nature.
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878 Chapter 13 Vector Functions
1. T he table gives coordinates of a particle moving through 11. rstd − s2 t i 1 e t j 1 e2t k
space along a smooth curve.
12. rstd − t 2 i 1 2t j 1 ln t k
(a) Find the average velocities over the time intervals
[0, 1], [0.5, 1], [1, 2], and [1, 1.5]. 13. rstd − e tscos t i 1 sin t j 1 t kd
(b) Estimate the velocity and speed of the particle at t − 1. 14. rstd − k t 2, sin t 2 t cos t, cos t 1 t sin t l , t > 0
t xyz 15–16 Find the velocity and position vectors of a particle that has
the given acceleration and the given initial velocity and position.
0 2.7 9.8 3.7
0.5 3.5 7.2 3.3 15. astd − 2 i 1 2t k, vs0d − 3 i 2 j, rs0d − j 1 k
1.0 4.5 6.0 3.0
1.5 5.9 6.4 2.8 16. astd − sin t i 1 2 cos t j 1 6t k,
2.0 7.3 7.8 2.7 vs0d − 2k, rs0d − j 2 4 k
2. T he figure shows the path of a particle that moves with 17–18
position vector rstd at time t. (a) Find the position vector of a particle that has the given
(a) Draw a vector that represents the average velocity of acceleration and the specified initial velocity and position.
the particle over the time interval 2 < t < 2.4. ; (b) Use a computer to graph the path of the particle.
(b) Draw a vector that represents the average velocity over 17. astd − 2t i 1 sin t j 1 cos 2t k, vs0d − i, rs0d − j
the time interval 1.5 < t < 2.
18. astd − t i 1 e t j 1 e2t k, vs0d − k, rs0d − j 1 k
(c) Write an expression for the velocity vector vs2d.
(d) Draw an approximation to the vector vs2d and estimate 19. T he position function of a particle is given by
rstd − k t 2, 5t, t 2 2 16t l . When is the speed a minimum?
the speed of the particle at t − 2.
20. What force is required so that a particle of mass m has the
y position function rstd − t 3 i 1 t 2 j 1 t 3 k?
r(2.4) 21. A force with magnitude 20 N acts directly upward from the
2 r(2) xy-plane on an object with mass 4 kg. The object starts at the
origin with initial velocity vs0d − i 2 j. Find its position
1 r(1.5) function and its speed at time t.
0 12 x 22. S how that if a particle moves with constant speed, then the
velocity and acceleration vectors are orthogonal.
3–8 Find the velocity, acceleration, and speed of a particle with
the given position function. Sketch the path of the particle and 23. A projectile is fired with an initial speed of 200 mys and
draw the velocity and acceleration vectors for the specified value angle of elevation 60°. Find (a) the range of the projectile,
of t. (b) the maximum height reached, and (c) the speed at impact.
3. rstd − k221 t 2, tl , t − 2 24. R ework Exercise 23 if the projectile is fired from a position
100 m above the ground.
4. rstd − kt 2, 1yt 2l, t − 1
25. A ball is thrown at an angle of 45° to the ground. If the ball
5. rstd − 3 cos t i 1 2 sin t j, t − y3 lands 90 m away, what was the initial speed of the ball?
6. rstd − e t i 1 e 2t j, t − 0 26. A projectile is fired from a tank with initial speed 400 mys.
Find two angles of elevation that can be used to hit a target
7. rstd − t i 1 t 2 j 1 2 k, t − 1 3000 m away.
8. rstd − t i 1 2 cos t j 1 sin t k, t − 0 27. A rifle is fired with angle of elevation 36°. What is the muzzle
speed if the maximum height of the bullet is 1600 ft?
9 –14 Find the velocity, acceleration, and speed of a particle with
the given position function. 28. A batter hits a baseball 3 ft above the ground toward the
center field fence, which is 10 ft high and 400 ft from home
9. rstd − k t 2 1 t, t 2 2 t, t 3l plate. The ball leaves the bat with speed 115 ftys at an
angle 508 above the horizontal. Is it a home run? (In other
10. rstd − k 2 cos t, 3t, 2 sin t l words, does the ball clear the fence?)
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Section 13.4 Motion in Space: Velocity and Acceleration 879
29. A medieval city has the shape of a square and is protected 3 7–40 Find the tangential and normal components of the
by walls with length 500 m and height 15 m. You are the acceleration vector.
commander of an attacking army and the closest you can 37. rstd − st 2 1 1d i 1 t 3 j, t > 0
get to the wall is 100 m. Your plan is to set fire to the city by 38. rstd − 2t2 i 1 (2 t 3 2 2t) j
3
catapulting heated rocks over the wall (with an initial speed
of 80 mys). At what range of angles should you tell your 39. rstd − cos t i 1 sin t j 1 t k
men to set the catapult? (Assume the path of the rocks is
40. rstd − t i 1 2e t j 1 e 2t k
perpendicular to the wall.)
30. S how that a projectile reaches three-quarters of its maxi- 4 1–42 Find the tangential and normal components of the
mum height in half the time needed to reach its maximum acceleration vector at the given point.
height.
31. A ball is thrown eastward into the air from the origin (in 41. rstd − ln t i 1 st 2 1 3td j 1 4 st k, s0, 4, 4d
the direction of the positive x-axis). The initial velocity is 42. rstd − 1 i 1 1 j 1 1 k, s1, 1, 1d
50 i 1 80 k, with speed measured in feet per second. The t t2 t3
spin of the ball results in a southward acceleration of 43. The magnitude of the acceleration vector a is 10 cmys2. Use
4 ftys2, so the acceleration vector is a − 24 j 2 32 k. the figure to estimate the tangential and normal components
Where does the ball land and with what speed? of a.
32. A ball with mass 0.8 kg is thrown southward into the air y
with a speed of 30 mys at an angle of 30° to the ground.
A west wind applies a steady force of 4 N to the ball in an a
easterly direction. Where does the ball land and with what
speed?
; 33. W ater traveling along a straight portion of a river normally
flows fastest in the middle, and the speed slows to almost
zero at the banks. Consider a long straight stretch of river 0x
flowing north, with parallel banks 40 m apart. If the maxi-
mum water speed is 3 mys, we can use a quadratic function
as a basic model for the rate of water flow x units from the 44. I f a particle with mass m moves with position vector
3 rstd, then its angular momentum is defined as
west bank: f sxd − 400 xs40 2 xd. Lstd − mrstd 3 vstd and its torque as t std − mrstd 3 astd.
Show that L9std − tstd. Deduce that if t std − 0 for all t,
(a) A boat proceeds at a constant speed of 5 mys from a then Lstd is constant. (This is the law of conservation of
angular momentum.)
point A on the west bank while maintaining a heading
perpendicular to the bank. How far down the river on
the opposite bank will the boat touch shore? Graph the
path of the boat. S D 45. The position function of a spaceship is
(b) Suppose we would like to pilot the boat to land at the
point B on the east bank directly opposite A. If we rstd − s3 1 td i 1 s2 1 ln td j 1 7 2 4 k
1
maintain a constant speed of 5 mys and a constant head- t2 1
ing, find the angle at which the boat should head. Then and the coordinates of a space station are s6, 4, 9d. The
captain wants the spaceship to coast into the space station.
graph the actual path the boat follows. Does the path
When should the engines be turned off?
seem realistic?
34. Another reasonable model for the water speed of the river 46. A rocket burning its onboard fuel while moving through
in Exercise 33 is a sine function: f sxd − 3 sins xy40d. If space has velocity vstd and mass mstd at time t. If the
a boater would like to cross the river from A to B with con- exhaust gases escape with velocity ve relative to the rocket,
stant heading and a constant speed of 5 mys, determine the
angle at which the boat should head. it can be deduced from Newton’s Second Law of Motion
that dv dm
dt dt
35. A particle has position function rstd. If r9std − c 3 rstd, m − ve
where c is a constant vector, describe the path of the
particle. ms0d
mstd
36. (a) If a particle moves along a straight line, what can you (a) Show that vstd − vs0d 2 ln ve.
say about its acceleration vector?
(b) For the rocket to accelerate in a straight line from rest to
(b) If a particle moves with constant speed along a curve,
what can you say about its acceleration vector? twice the speed of its own exhaust gases, what fraction
of its initial mass would the rocket have to burn as fuel?
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880 Chapter 13 Vector Functions
applied Project Kepler’s laws
Johannes Kepler stated the following three laws of planetary motion on the basis of massive
amounts of data on the positions of the planets at various times.
Kepler’s Laws
1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
2. The line joining the sun to a planet sweeps out equal areas in equal times.
3. T he square of the period of revolution of a planet is proportional to the cube of the
length of the major axis of its orbit.
K epler formulated these laws because they fitted the astronomical data. He wasn’t able to see
why they were true or how they related to each other. But Sir Isaac Newton, in his Principia
Mathemati ca of 1687, showed how to deduce Kepler’s three laws from two of Newton’s own
laws, the Seco nd Law of Motion and the Law of Universal Gravitation. In Section 13.4 we
proved Kepler’s First Law using the calculus of vector functions. In this project we guide you
through the proofs of Kepler’s Second and Third Laws and explore some of their consequences.
1. U se the following steps to prove Kepler’s Second Law. The notation is the same as in
the proof of the First Law in Section 13.4. In particular, use polar coordinates so that
r − sr cos d i 1 sr sin d j.
(a) Show that h − r2 d k.
dt
(b) Deduce that r 2 d − h.
dt
y (c) If A − Astd is the area swept out by the radius vector r − rstd in the time interval ft0, tg as
r(t) in the figure, show that
A(t) r(t¸) dA − 1 r 2 d
x dt 2 dt
0
(d) Deduce that
dA − 1 h − constant
dt 2
T his says that the rate at which A is swept out is constant and proves Kepler’s Second
Law.
2. L et T be the period of a planet about the sun; that is, T is the time required for it to travel once
around its elliptical orbit. Suppose that the lengths of the major and minor axes of the ellipse
are 2a and 2b.
(a) Use part (d) of Problem 1 to show that T − 2abyh.
(b) Show that h2 − ed − b2 .
GM a
(c) Use parts (a) and (b) to show that T 2 − 4 2 a 3.
GM
T his proves Kepler’s Third Law. [Notice that the proportionality constant 4 2ysGMd is inde-
pendent of the planet.]
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chapter 13 Review 881
3. T he period of the earth’s orbit is approximately 365.25 days. Use this fact and Kepler’s Third
Law to find the length of the major axis of the earth’s orbit. You will need the mass of the sun,
M − 1.99 3 1030 kg, and the gravitational constant, G − 6.67 3 10211 N ∙ m2ykg 2.
4. I t’s possible to place a satellite into orbit about the earth so that it remains fixed above a given
location on the equator. Compute the altitude that is needed for such a satellite. The earth’s
mass is 5.98 3 1024 kg; its radius is 6.37 3 106 m. (This orbit is called the Clarke Geosyn-
chronous Orbit after Arthur C. Clarke, who first proposed the idea in 1945. The first such
satellite, Syncom II, was launched in July 1963.)
13 Review Answers to the Concept Check can be found on the back endpapers.
CONCEPT CHECK 6. (a) What is the definition of curvature?
(b) Write a formula for curvature in terms of r9std and T9std.
1. W hat is a vector function? How do you find its derivative and (c) Write a formula for curvature in terms of r9std and r0std.
its integral? (d) Write a formula for the curvature of a plane curve with
2. What is the connection between vector functions and space equation y − f sxd.
curves?
7. (a) Write formulas for the unit normal and binormal vectors of
3. How do you find the tangent vector to a smooth curve at a a smooth space curve rstd.
point? How do you find the tangent line? The unit tangent
vector? (b) What is the normal plane of a curve at a point? What is the
osculating plane? What is the osculating circle?
4. I f u and v are differentiable vector functions, c is a scalar, and
f is a real-valued function, write the rules for differentiating 8. (a) How do you find the velocity, speed, and acceleration of a
the following vector functions. particle that moves along a space curve?
(a) ustd 1 vstd (b) custd (c) f std ustd (b) Write the acceleration in terms of its tangential and normal
(d) ustd ? vstd (e) ustd 3 vstd (f) us f stdd components.
5. How do you find the length of a space curve given by a vector 9. State Kepler’s Laws.
function rstd?
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true, explain | | 7. I f Tstd is the unit tangent vector of a smooth curve, then the
why. If it is false, explain why or give an example that disproves the curvature is − dTydt .
statement. 8. The binormal vector is Bstd − Nstd 3 Tstd.
9. S uppose f is twice continuously differentiable. At an inflection
1. T he curve with vector equation rstd − t 3 i 1 2t 3 j 1 3t 3 k is
a line. point of the curve y − f sxd, the curvature is 0.
10. If std − 0 for all t, the curve is a straight line.
2. The curve rstd − k 0, t 2, 4t l is a parabola.
3. The curve rstd − k 2t, 3 2 t, 0 l is a line that passes through the
origin.
4. T he derivative of a vector function is obtained by differen- | | | | 11. I f rstd − 1 for all t, then r9std is a constant.
tiating each component function. | | 12. If rstd − 1 for all t, then r9std is orthogonal to rstd for all t.
5. If ustd and vstd are differentiable vector functions, then 13. The osculating circle of a curve C at a point has the same tan-
gent vector, normal vector, and curvature as C at that point.
d fustd 3 vstdg − u9std 3 v9std
dt 14. Different parametrizations of the same curve result in identical
tangent vectors at a given point on the curve.
6. If rstd is a differentiable vector function, then
d | rstd | − | r9std |
dt
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
882 chapter 13 Vector Functions
EXERCISES
1. (a) Sketch the curve with vector function 16. The figure shows the curve C traced by a particle with posi-
tion vector rstd at time t.
rstd − t i 1 cos t j 1 sin t k t > 0
(a) Draw a vector that represents the average velocity of the
(b) Find r9std and r0std. particle over the time interval 3 < t < 3.2.
2. L et rstd − ks2 2 t , set 2 1dyt, lnst 1 1dl . (b) Write an expression for the velocity vs3d.
(c) Write an expression for the unit tangent vector Ts3d and
(a) Find the domain of r.
(b) Find limtl0 rstd. draw it.
(c) Find r9std.
y
3. F ind a vector function that represents the curve of intersec- C
tion of the cylinder x 2 1 y 2 − 16 and the plane x 1 z − 5.
; 4. F ind parametric equations for the tangent line to the curve 1
sx1−, s23 s,i2nd.t,Gyra−ph2tshien 2t , z −2 sin 3t at the point r(3)
curve and the tangent line on a com-
r(3.2)
mon screen.
5. If rstd − t 2 i 1 t cos t j 1 sin t k, evaluate y01 rstd dt. 01 x
6. Let C be the curve with equations x − 2 2 t 3, y − 2t 2 1, 17. A particle moves with position function
z − ln t. Find (a) the point where C intersects the xz-plane, rstd − t ln t i 1 t j 1 e2t k. Find the velocity, speed, and
(b) parametric equations of the tangent line at s1, 1, 0d, and acceleration of the particle.
(c) an equation of the normal plane to C at s1, 1, 0d.
18. Find the velocity, speed, and acceleration of a particle mov-
7. Use Simpson’s Rule with n − 6 to estimate the length of ing with position function rstd − s2t 2 2 3d i 1 2t j. Sketch
the arc of the curve with equations x − t 2, y − t 3, z − t 4, the path of the particle and draw the position, velocity, and
0 < t < 3. acceleration vectors for t − 1.
8. Find the length of the curve rstd − k2t 3y2, cos 2t, sin 2t l, 19. A particle starts at the origin with initial velocity
0 < t < 1. i 2 j 1 3 k. Its acceleration is astd − 6t i 1 12t 2 j 2 6t k.
Find its position function.
9. T he helix r1std − cos t i 1 sin t j 1 t k intersects the curve
r2std − s1 1 td i 1 t 2 j 1 t 3 k at the point s1, 0, 0d. Find the 20. An athlete throws a shot at an angle of 458 to the horizontal
angle of intersection of these curves. at an initial speed of 43 ftys. It leaves his hand 7 ft above
the ground.
10. R eparametrize the curve rstd − e t i 1 e t sin t j 1 e t cos t k
with respect to arc length measured from the point s1, 0, 1d (a) Where is the shot 2 seconds later?
in the direction of increasing t. (b) How high does the shot go?
(c) Where does the shot land?
11. For the curve given by rstd − ksin 3t, cos 3t, sin 2tl,
0 < t < y2, find 21. A projectile is launched with an initial speed of 40 mys
from the floor of a tunnel whose height is 30 m. What
(a) the unit tangent vector, angle of elevation should be used to achieve the maximum
(b) the unit normal vector, possible horizontal range of the projectile? What is the
(c) the unit binormal vector, and maximum range?
(d) the curvature.
22. F ind the tangential and normal components of the accelera-
12. F ind the curvature of the ellipse x − 3 cos t, y − 4 sin t at tion vector of a particle with position function
the points s3, 0d and s0, 4d.
rstd − t i 1 2t j 1 t 2 k
13. Find the curvature of the curve y − x 4 at the point s1, 1d.
23. A disk of radius 1 is rotating in the counterclockwise
; 14. Find an equation of the osculating circle of the curve direction at a constant angular speed . A particle starts at
y − x 4 2 x 2 at the origin. Graph both the curve and its the center of the disk and moves toward the edge along a
osculating circle. fixed radius so that its position at time t, t > 0, is given by
15. F ind an equation of the osculating plane of the curve
x − sin 2t, y − t, z − cos 2t at the point s0, , 1d.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
chapter 13 Review 883
rstd − t Rstd, where shown in the figure. It looks reasonable at first glance.
Rstd − cos t i 1 sin t j
HShow that the function
(a) Show that the velocity v of the particle is 1 if x < 0
Fsxd − s1 2 x 2 if 0 , x , 1ys2
v − cos t i 1 sin t j 1 t vd
s2 2 x if x > 1ys2
where vd − R9std is the velocity of a point on the edge is continuous and has continuous slope, but does not
of the disk. have continuous curvature. Therefore f is not an appro-
priate transfer curve.
(b) Show that the acceleration a of the particle is
a − 2 vd 1 t ad y y
y=F(x)
where ad − R0std is the acceleration of a point on y=
the edge of the disk. The extra term 2 vd is called the 1
Coriolis acceleration; it is the result of the interaction transfer
of the rotation of the disk and the motion of the particle. 01 y=0 1
One can obtain a physical demonstration of this accel- x0
eration by walking toward the edge of a moving œ„2
merry-go-round.
; (b) Find a fifth-degree polynomial to serve as a transfer
(c) Determine the Coriolis acceleration of a particle that curve between the following straight line segments:
moves on a rotating disk according to the equation y − 0 for x < 0 and y − x for x > 1. Could this be
done with a fourth-degree polynomial? Use a graphing
rstd − e2t cos t i 1 e2t sin t j calculator or computer to sketch the graph of the “con-
nected” function and check to see that it looks like the
24. In designing transfer curves to connect sections of straight one in the figure.
railroad tracks, it’s important to realize that the accelera-
tion of the train should be continuous so that the reactive
force exerted by the train on the track is also continuous. y
Because of the formulas for the components of acceleratioyn y=x
y=F(x)
in Section 13.4, this will be the case if the curvature varies1
continuously.
(a) A logical candidate for a transfer curve to join existing
tracks given by y − 1 for x < 0 and y − s2 2 x for y=0 transfer curve
x > 1ys2 might be the function f sxd − s1 2 x 2 , 0 1 x0 1x
0 , x , 1ys2 , whose graph is the arc of the circle
œ„2
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Problems Plus 1. A particle P moves with constant angular speed around a circle whose center is at the
y origin and whose radius is R. The particle is said to be in uniform circular motion. Assume
that the motion is counterclockwise and that the particle is at the point sR, 0d when t − 0.
v vt The position vector at time t > 0 is rstd − R cos t i 1 R sin t j.
r (a) Find the velocity vector v and show that v ? r − 0. Conclude that v is tangent to the
circle and points in the direction of the motion.
| | (b) Show that the speed v of the particle is the constant R. The period T of the particle
is the time required for one complete revolution. Conclude that
x 2R 2
T − |v| −
FIGURE for problem 1 (c) Find the acceleration vector a. Show that it is proportional to r and that it points toward
the origin. An acceleration with this property is called a centripetal acceleration. Show
| |that the magnitude of the acceleration vector is a − R2.
(d) Suppose that the particle has mass m. Show that the magnitude of the force F that is
required to produce this motion, called a centripetal force, is
|F| − m| v |2
R
F 2. A circular curve of radius R on a highway is banked at an angle so that a car can safely
traverse the curve without skidding when there is no friction between the road and the tires.
¨ mg
The loss of friction could occur, for example, if the road is covered with a film of water
FIGURE for problem 2 or ice. The rated speed vR of the curve is the maximum speed that a car can attain without
skidding. Suppose a car of mass m is traversing the curve at the rated speed vR. Two forces
are acting on the car: the vertical force, mt, due to the weight of the car, and a force F
exerted by, and normal to, the road (see the figure).
| | The vertical component of F balances the weight of the car, so that F cos − mt. The
horizontal component of F produces a centripetal force on the car so that, by Newton’s
Second Law and part (d) of Problem 1,
| |F sin − m vR2
R
(a) Show that vR2 − Rt tan .
(b) Find the rated speed of a circular curve with radius 400 ft that is banked at an angle
of 128.
(c) Suppose the design engineers want to keep the banking at 128, but wish to increase the
rated speed by 50%. What should the radius of the curve be?
3. A projectile is fired from the origin with angle of elevation ␣ and initial speed v0. Assum-
ing that air resistance is negligible and that the only force acting on the projectile is gravity,
t, we showed in Example 13.4.5 that the position vector of the projectile is
f grstd − sv0 cos ␣dt i 1 sv0 sin ␣dt 2 1 tt2 j
y 2
y We also showed that the maximum horizontal distance of the projectile is achieved when
v02yt.
_R 0 (␣a)− A4t5w° haantdainngtlheisshcoausledtthhheerparnogjeecistilRe − fired to achieve maximum height and what is the
be
FIGURE for problem 3
884 maximum height?
Rx 0(b) FshixowthneiinnitthiealfisgpuereedDavt0tahnedlecfot.nSsxihdoewr tthheatpathraebporloajexc2ti1le 2Ry 2 R2 − 0, whose graph is
can hit any target inside or on
the boundary of the region bounded by the parabola and the x-axis, and that it can’t hit
any target outside this region.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
(c) Suppose that the gun is elevated to an angle of inclination ␣ in order to aim at a target
that is suspended at a height h directly over a point D units downrange (see the figure
below). The target is released at the instant the gun is fired. Show that the projectile
always hits the target, regardless of the value v0, provided the projectile does not hit the
ground “before” D.
yy
h
_R 0 Rx 0 Dx
y 4. ( a) A projectile is fired from the origin down an inclined plane that makes an angle with
v¸ the horizontal. The angle of elevation of the gun and the initial speed of the projectile
a are ␣ and v0, respectively. Find the position vector of the projectile and the parametric
¨ equations of the path of the projectile as functions of the time t. (Ignore air resistance.)
x (b) Show that the angle of elevation ␣ that will maximize the downhill range is the angle
halfway between the plane and the vertical.
(c) Suppose the projectile is fired up an inclined plane whose angle of inclination is .
Show that, in order to maximize the (uphill) range, the projectile should be fired in the
direction halfway between the plane and the vertical.
(d) In a paper presented in 1686, Edmond Halley summarized the laws of gravity and
FIGURE for problem 4 projectile motion and applied them to gunnery. One problem he posed involved firing
a projectile to hit a target a distance R up an inclined plane. Show that the angle at
which the projectile should be fired to hit the target but use the least amount of energy
is the same as the angle in part (c). (Use the fact that the energy needed to fire the
projectile is proportional to the square of the initial speed, so minimizing the energy is
equivalent to minimizing the initial speed.)
3.5 ft ¨¨ 5. A ball rolls off a table with a speed of 2 ftys. The table is 3.5 ft high.
(a) Determine the point at which the ball hits the floor and find its speed at the instant of
FIGURE for problem 5 impact.
(b) Find the angle between the path of the ball and the vertical line drawn through the
point of impact (see the figure).
(c) Suppose the ball rebounds from the floor at the same angle with which it hits the floor,
but loses 20% of its speed due to energy absorbed by the ball on impact. Where does
the ball strike the floor on the second bounce?
6. F ind the curvature of the curve with parametric equations
t s1 2d d y s dt cos 1
0y yx −sin2 − 2 2 d
0
; 7. If a projectile is fired with angle of elevation ␣ and initial speed v, then parametric equa-
tions for its trajectory are
x − sv cos ␣dt y − sv sin ␣dt 2 1 tt 2
2
( See Example 13.4.5.) We know that the range (horizontal distance traveled) is maximized
when ␣ − 45°. What value of ␣ maximizes the total distance traveled by the projectile?
(State your answer correct to the nearest degree.)
8. A cable has radius r and length L and is wound around a spool with radius R without over-
lapping. What is the shortest length along the spool that is covered by the cable?
9. S how that the curve with vector equation
rstd − ka1t 2 1 b1t 1 c1, a2t 2 1 b2t 1 c2, a3t 2 1 b3t 1 c3l
l ies in a plane and find an equation of the plane.
885
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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14 Partial Derivatives
In 2008 Speedo introduced
the LZR Racer and, because
it reduced drag in the water,
many swimming records
were broken. In the project
on page 936 you are asked
to use partial derivatives to
explain why a small decrease
in drag can have a big effect
on performance.
Courtesy Speedo and ANSYS, Inc.
So far we have dealt with the calculus of functions of a single variable. But, in the real world,
physical quantities often depend on two or more variables, so in this chapter we turn our atten-
tion to functions of several variables and extend the basic ideas of differential calculus to such
functions.
887
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
888 Chapter 14 Partial Derivatives
In this section we study functions of two or more variables from four points of view:
● verbally (by a description in words)
● numerically (by a table of values)
● algebraically (by an explicit formula)
● visually (by a graph or level curves)
Functions of Two Variables
The temperature T at a point on the surface of the earth at any given time depends on the
longitude x and latitude y of the point. We can think of T as being a function of the
two variables x and y, or as a function of the pair sx, yd. We indicate this functional
dependence by writing T − f sx, yd.
The volume V of a circular cylinder depends on its radius r and its height h. In fact,
we know that V − r 2h. We say that V is a function of r and h, and we write
Vsr, hd − r 2h.
Definition A function f of two variables is a rule that assigns to each ordered
pair of real numbers sx, yd in a set D a unique real number denoted by f sx, yd.
|The set D is the domain of f and its range is the set of values that f takes on,
that is, h f sx, yd sx, yd [ Dj.
y z We often write z − f sx, yd to make explicit the value taken on by f at the general
f (x, y) point sx, yd. The variables x and y are independent variables and z is the dependent
(x, y) x variable. [Compare this with the notation y − f sxd for functions of a single variable.]
0
0 f (a, b) A function of two variables is just a function whose domain is a subset of R2 and
D (a, b)
whose range is a subset of R. One way of visualizing such a function is by means of an
FFIIGGUURREE11
arrow diagram (see Figure 1), where the domain D is represented as a subset of the
xy-plane and the range is a set of numbers on a real line, shown as a z-axis. For instance,
if f sx, yd represents the temperature at a point sx, yd in a flat metal plate with the shape
of D, we can think of the z-axis as a thermometer displaying the recorded temperatures.
If a function f is given by a formula and no domain is specified, then the domain of f
is understood to be the set of all pairs sx, yd for which the given expression is a well-
defined real number.
7et140101 Example 1 For each of the following functions, evaluate f s3, 2d and find and sketch
04/23/10
MasterID: 03044 the domain.
(a) f sx, yd − sx 1 y 1 1 (b) f sx, yd − x lns y2 2 xd
x21
SOLUTION
(a) f s3, 2d − s3 1 2 1 1 − s6
321 2
The expression for f makes sense if the denominator is not 0 and the quantity under the
square root sign is nonnegative. So the domain of f is
|D − hsx, yd x 1 y 1 1 > 0, x ± 1j
The inequality x 1 y 1 1 > 0, or y > 2x 2 1, describes the points that lie on or
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 889
x+y+1=0 above the line y − 2x 2 1, while x ± 1 means that the points on the line x − 1 must
y be excluded from the domain. (See Figure 2.)
x=1 (b) f s3, 2d − 3 lns22 2 3d − 3 ln 1 − 0
_1 0 Since lns y2 2 xd is defined only when y2 2 x . 0, that is, x , y2, the domain of f is
_1 D − hsx, yd x , y2 j. This is the set of points to the left of the parabola x − y2.
|x
(See Figure 3.) ■
FIGGUURREE22 Not all functions can be represented by explicit formulas. The function in the next
Domain of ff(sxx,, yy)d=−œs„xx+„x-„x1„y2„1+y„„111 1 example is described verbally and by numerical estimates of its values.
7et140102 y x=¥ Example 2 In regions with severe winter weather, the wind-chill index is often used
04/23/10 x
MasterID: 01526 to describe the apparent severity of the cold. This index W is a subjective temperature
0 that depends on the actual temperature T and the wind speed v. So W is a function of
T and v, and we can write W − f sT, vd. T able 1 records values of W compiled by the
US National Weather Service and the Meteorological Service of Canada.
Table 1 Wind-chill index as a function of air temperature and wind speed
Wind speed (km/h)
v 5 10 15 20 25 30 40 50 60 70 80
T
FFIIGGUURREE33 Ϫ3
Domain of ff(sxx,, yy)d=−x xlnl(n¥s y-2 2x) xd 5 4 3 2 1 1 0 Ϫ1 Ϫ1 Ϫ2 Ϫ2 Ϫ10
Ϫ17
7et140103 Actual temperature (°C) 0 Ϫ2 Ϫ3 Ϫ4 Ϫ5 Ϫ6 Ϫ6 Ϫ7 Ϫ8 Ϫ9 Ϫ9 Ϫ24
04/23/10 Ϫ31
MTheaWstinedr-ICDhi:ll 0In1d5ex27 Ϫ5 Ϫ7 Ϫ9 Ϫ11 Ϫ12 Ϫ12 Ϫ13 Ϫ14 Ϫ15 Ϫ16 Ϫ16 Ϫ38
Ϫ45
The wind-chill index measures how Ϫ10 Ϫ13 Ϫ15 Ϫ17 Ϫ18 Ϫ19 Ϫ20 Ϫ21 Ϫ22 Ϫ23 Ϫ23 Ϫ52
cold it feels when it’s windy. It is based Ϫ60
on a model of how fast a human face Ϫ15 Ϫ19 Ϫ21 Ϫ23 Ϫ24 Ϫ25 Ϫ26 Ϫ27 Ϫ29 Ϫ30 Ϫ30 Ϫ67
loses heat. It was developed through
clinical trials in which volunteers were Ϫ20 Ϫ24 Ϫ27 Ϫ29 Ϫ30 Ϫ32 Ϫ33 Ϫ34 Ϫ35 Ϫ36 Ϫ37
exposed to a variety of temperatures
and wind speeds in a refrigerated wind Ϫ25 Ϫ30 Ϫ33 Ϫ35 Ϫ37 Ϫ38 Ϫ39 Ϫ41 Ϫ42 Ϫ43 Ϫ44
tunnel.
Ϫ30 Ϫ36 Ϫ39 Ϫ41 Ϫ43 Ϫ44 Ϫ46 Ϫ48 Ϫ49 Ϫ50 Ϫ51
Ϫ35 Ϫ41 Ϫ45 Ϫ48 Ϫ49 Ϫ51 Ϫ52 Ϫ54 Ϫ56 Ϫ57 Ϫ58
Ϫ40 Ϫ47 Ϫ51 Ϫ54 Ϫ56 Ϫ57 Ϫ59 Ϫ61 Ϫ63 Ϫ64 Ϫ65
For instance, the table shows that if the temperature is 258C and the wind speed is
50 kmy7he, tth1e4n0su1btj0ec1tively it would feel as cold as a temperature of about 2158C with
no wind0.4S/o23/10
MasterID: 01528 f s25, 50d − 215 ■
Example 3 In 1928 Charles Cobb and Paul Douglas published a study in which they
modeled the growth of the American economy during the period 1899–1922. They
considered a simplified view of the economy in which production output is determined
by the amount of labor involved and the amount of capital invested. While there are
many other factors affecting economic performance, their model proved to be remark-
ably accurate. The function they used to model production was of the form
1 PsL, K d − bL␣K 12␣
where P is the total production (the monetary value of all goods produced in a year),
L is the amount of labor (the total number of person-hours worked in a year), and K is
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
890 Chapter 14 Partial Derivatives
Year Table 2 K the amount of capital invested (the monetary worth of all machinery, equipment, and
buildings). In Section 14.3 we will show how the form of Equation 1 follows from
1899 PL 100 certain economic assumptions.
1900 107
1901 100 100 114 Cobb and Douglas used economic data published by the government to obtain
1902 101 105 122 Table 2. They took the year 1899 as a baseline and P, L, and K for 1899 were each
1903 112 110 131 assigned the value 100. The values for other years were expressed as percentages of the
1904 122 117 138 1899 figures.
1905 124 122 149
1906 122 121 163 Cobb and Douglas used the method of least squares to fit the data of Table 2 to the
1907 143 125 176 function
1908 152 134 185
1909 151 140 198 2 PsL, K d − 1.01L0.75K 0.25
1910 126 123 208
1911 155 143 216 (See Exercise 81 for the details.)
1912 159 147 226 If we use the model given by the function in Equation 2 to compute the production
1913 153 148 236
1914 177 155 244 in the years 1910 and 1920, we get the values
1915 184 156 266
1916 169 152 298 Ps147, 208d − 1.01s147d0.75s208d0.25 < 161.9
1917 189 156 335 Ps194, 407d − 1.01s194d0.75s407d0.25 < 235.8
1918 225 183 366
1919 227 198 387 which are quite close to the actual values, 159 and 231.
1920 223 201 407
1921 218 196 417 The production function (1) has subsequently been used in many settings, ranging
1922 231 194 431
179 146 from individual firms to global economics. It has become known as the Cobb-Douglas
240 161
|production function. Its domain is hsL, Kd L > 0, K > 0j because L and K represent
labor and capital and are therefore never negative. ■
y Example 4 Find the domain and range of tsx, yd − s9 2 x2 2 y2 .
≈+¥=9
SOLUTION The domain of t is
_3 3 x
| |D − hsx, yd 9 2 x2 2 y2 > 0j − hsx, yd x2 1 y2 < 9j
which is the disk with center s0, 0d and radius 3. (See Figure 4.) The range of t is
|5z z − s9 2 x2 2 y2, sx, yd [ D6
Since z is a positive square root, z > 0. Also, because 9 2 x2 2 y2 < 9, we have
FFIIGGUURREE44 So the range is s9 2 x 2 2 y 2 < 3 ■
DDoommaaiinn ooff gts(xx,, yy)d=−œs„9„9-„2„≈„x„-2„2„¥„y 2
|hz 0 < z < 3j − f0, 3g
7et14z0104 Graphs
04/23/10 {x, y, f(x, y)} Another way of visualizing the behavior of a function of two variables is to consider its
MasterID: 0S1529 graph.
f(x, y) Definition If f is a function of two variables with domain D, then the graph of
f is the set of all points sx, y, zd in R3 such that z − f sx, yd and sx, yd is in D.
0 y Just as the graph of a function f of one variable is a curve C with equation y − f sxd,
D so the graph of a function f of two variables is a surface S with equation z − f sx, yd.
We can visualize the graph S of f as lying directly above or below its domain D in the
x (x, y, 0)
xy-plane (see Figure 5).
FFIIGGUURREE55
7 e t 1 4 010 5Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 893
ferent vantage points. In parts (a) and (b) the graph of f is very flat and close to the
xy-plane except near the origin; this is because e2x 22y 2 is very small when x or y is large.
Level Curves
So far we have two methods for visualizing functions: arrow diagrams and graphs. A
third method, borrowed from mapmakers, is a contour map on which points of constant
elevation are joined to form contour curves, or level curves.
Definition The level curves of a function f of two variables are the curves
with equations f sx, yd − k, where k is a constant (in the range of f ).
A level curve f sx, yd − k is the set of all points in the domain of f at which f takes
on a given value k. In other words, it shows where the graph of f has height k.
You can see from Figure 11 the relation between level curves and horizontal traces. The
level curves f sx, yd − k are just the traces of the graph of f in the horizontal plane
z − k projected down to the xy-plane. So if you draw the level curves of a function and
visualize them being lifted up to the surface at the indicated height, then you can men-
tally piece together a picture of the graph. The surface is steep where the level curves are
close together. It is somewhat flatter where they are farther apart.
z
45
4000 4500
0 5000
/21(620( 071
x y 5500
5000 4500 A
k=45 FFIGGUURREE1122
f(x, y)=20 k=40 B
k=35 Lonesome Creek
k=30
k=25
k=20
FIGURE 11
TEC Visual 14.1A animates Figure 11 One common example of level curves occurs in topographic maps of mountainous
by showing level curves being lifted regions, such as the map in Figure 12. The level curves are curves of constant elevation
up to graphs of functions. above sea level. If you walk along one of these contour lines, you neither ascend nor
descend. Another common example is the temperature function introduced in the open-
ing paragraph of this section. Here the level curves are called isothermals and join loca-
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 891
z Example 5 Sketch the graph of the function f sx, yd − 6 2 3x 2 2y.
(0, 0, 6)
SOLUTION The graph of f has the equation z − 6 2 3x 2 2y, or 3x 1 2y 1 z − 6,
which represents a plane. To graph the plane we first find the intercepts. Putting
y − z − 0 in the equation, we get x − 2 as the x-intercept. Similarly, the y-intercept
is 3 and the z-intercept is 6. This helps us sketch the portion of the graph that lies in the
(0, 3, 0) first octant in Figure 6. ■
(2, 0, 0) y The function in Example 5 is a special case of the function
f sx, yd − ax 1 by 1 c
x
which is called a linear function. The graph of such a function has the equation
FFIIGGUURREE66
z − ax 1 by 1 c or ax 1 by 2 z 1 c − 0
7et140106
04/23/10 so it is a plane. In much the same way that linear functions of one variable are important
MasterID: 01531 in single-variable calculus, we will see that linear functions of two variables play a cen-
tral role in multivariable calculus.
Example 6 Sketch the graph of tsx, yd − s9 2 x2 2 y2 .
z SOLUTION The graph has equation z − s9 2 x2 2 y 2 . We square both sides of this
(0, 0, 3) equation to obtain z2 − 9 2 x 2 2 y 2, or x 2 1 y 2 1 z2 − 9, which we recognize as an
equation of the sphere with center the origin and radius 3. But, since z > 0, the graph
0 (0, 3, 0) of t is just the top half of this sphere (see Figure 7). ■
(3, 0, 0) y Note An entire sphere can’t be represented by a single function of x and y. As we
saw in Example 6, the upper hemisphere of the sphere x2 1 y2 1 z2 − 9 is represented
x
b y the function tsx, yd − s9 2 x2 2 y2 . The lower hemisphere is represented by the
FIGGUURREE77
Graph of tgs(xx,, yyd)=−œs„99„-„2„≈„x„-2„2„¥„y 2 function hsx, yd − 2s9 2 x 2 2 y 2 .
7et140107 Example 7 Use a computer to draw the graph of the Cobb-Douglas production
04/23/10
MasterID: 01532 function PsL, K d − 1.01L0.75K 0.25.
SOLUTION Figure 8 shows the graph of P for values of the labor L and capital K that
lie between 0 and 300. The computer has drawn the surface by plotting vertical traces.
We see from these traces that the value of the production P increases as either L or K
increases, as is to be expected.
300
200
P
100
0
300 200
FFIIGGUURREE8 8 K 100 00 100 200 300 ■
L
Example 8 Find the domain and range and sketch the graph of hsx, yd − 4x2 1 y2.
snSxOo,nLynU07de,T4egsIat/Oot12Nitvh43 eeN0/rd1oe1oat00imlc8neauitnmhaibstehRrss2x.,,[tNhydeoitesincdteiertfehinaxetydx-pf2ol>arnae0l.l possible ordered pairs of real numbers
MasterID: 01533 The range of h is the set f0, `d of all
and y2 > 0, so hsx, yd > 0 for all x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
892 Chapter 14 Partial Derivatives
and y.] The graph of h has the equation z − 4x2 1 y2, which is the elliptic paraboloid
that we sketched in Example 12.6.4. Horizontal traces are ellipses and vertical traces
are parabolas (see Figure 9).
z
FIGURE 9 x y ■
Graph of hsx, yd − 4x 2 1 y 2
Computer programs are readFiIlGyUaRvEa9ilable for graphing functions of two variables.
In most such programs, traces Ginrapthheofvhe(rxti,c ya)=l p4la≈n+es¥x − k and y − k are drawn for
equally spaced values of k and parts of the graph are eliminated using hidden line
removal.
an Figu re 10 shows pciocmtupreutoefr-ag07efn4euent/rc12atti46eod0/n11gwr00ahp9ehnsrooftasteiovneraisl functions. Notice that we get
especially good used to give views from dif-
z MasterID: 01534 z
x
x y
(a) f(x, y)=(≈+3¥)e_≈_¥
z (b) f(x, y)=(≈+3¥)e_≈_¥
z
xy
xy
FIGURE 10 (c) f(x, y)=sin x+sin y (d) f(x, y)= sin x sin y
xy
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
7et140110Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
894 Chapter 14 Partial Derivatives
© 2016 Cengage Learning®
FIGURE 13 Average air temperature near sea level in July (°F) KEY °
°
° Precipitation (cm/yr)
° °
Under to
° °
to to °
° °
° °
° °
° °
° °
° °
° °
° °
° °
° °
°
° to
Over
© 2016 Cengage Learning®
FIGURE 14
Precipitation
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 895
tions with the same temperature. Figure 13 shows a weather map of the world indicating
the average July temperatures. The isothermals are the curves that separate the colored
bands.
In weather maps of atmospheric pressure at a given time as a function of longitude
and latitude, the level curves are called isobars and join locations with the same pres-
sure. (See Exercise 34.) Surface winds tend to flow from areas of high pressure across the
isobars toward areas of low pressure, and are strongest where the isobars are tightly
packed.
A contour map of world-wide precipitation is shown in Figure 14. Here the level
curves are not labeled but they separate the colored regions and the amount of precipita-
tion in each region is indicated in the color key.
y Example 9 A contour map for a function f is shown in Figure 15. Use it to estimate
50
the values of f s1, 3d and f s4, 5d.
5
SOLUTION The point (1, 3) lies partway between the level curves with z-values 70
4 and 80. We estimate that
3 f s1, 3d < 73
2 80 50 80 Similarly, we estimate that f s4, 5d < 56
70 70
1 60 60 ■
0 12345 x Example 10 Sketch the level curves of the function f sx, yd − 6 2 3x 2 2y for the
FFIIGGUURREE114 5 values k − 26, 0, 6, 12.
y SOLUTION The level curves are
0x 6 2 3x 2 2y − k or 3x 1 2y 1 sk 2 6d − 0
This is a family of lines with slope 2 32. The four particular level curves with
k − 26, 0, 6, and 12 are 3x 1 2y 2 12 − 0, 3x 1 2y 2 6 − 0, 3x 1 2y − 0, and
3x 1 2y 1 6 − 0. They are sketched in Figure 16. The level curves are equally spaced
parallel lines because the graph of f is a plane (see Figure 6).
k=_6
k=0
k=6
k=12
k=_6
k=0
k=6
k=12
FIGURE 15 y
Contour map of ■
f(x, y)=6-3x-2y 0x
FFIIGGURREE 1156
70e4t/12460/11014-f1sfx5(,x,y ydC)−Co=not6n6ot-u2oru3mr3xxma-p2apo2f2oyyf Example 11 Sketch the level curves of the function
MasterID: 01539-40 7et140114-15
04/26/10tsx, yd − s9 2 x2 2 y2 for k − 0, 1, 2, 3
SOLUTIOMN aTshteelrevIDel:c0ur1ve5s3a9re-40
s9 2 x 2 2 y 2 − k or x 2 1 y 2 − 9 2 k 2
This is a family of concentric circles with center s0, 0d and radius s9 2 k2 . The cases
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
896 Chapter 14 Partial Derivatives
k − 0, 1, 2, 3 are shown in Figure 17. Try to visualize these level curves lifted up to
form a surface and compare with the graph of t (a hemisphere) in Figure 7. (See TEC
Visual 14.1A.)
y k=3
k=2
k=1
k=0
FIGURE 17 0 (3, 0) x ■
ConFtoIGurUmREap16of
Contour map oftsgx(,x,y yd )−=sœ9„9„-2„„≈x„2-„2„„¥„y 2
7et140116 Example 12 Sketch some level curves of the function hsx, yd − 4x2 1 y2 1 1.
04/26/10
MasterID: 015S4O1LUTION The level curves are
4x2 1 y2 1 1 − k or 1 sk x2 1d 1 k y2 − 1
4 2 21
which, for k . 1, describes a family of ellipses with semiaxes 1 sk 2 1 and sk 2 1.
2
Figure 18(a) shows a contour map of h drawn by a computer. Figure 18(b) shows these
level curves lifted up to the graph of h (an elliptic paraboloid) where they become
horizontal traces. We see from Figure 18 how the graph of h is put together from the
level curves.
y
z
TEC Visual 14.1B demonstrates the x
connection between surfaces and their
contour maps.
FFIIGGUURREE 1178 (a) Contour map x ■
ThTehgergaprahpohfohfshx(,xy,d y−)=44x≈2 1+y¥2+111 y
isisfoformrmeeddbbyylilfitfitninggththeelelveveel lccuurvrvees.s.
(b) Horizontal traces are raised level curves
7et140117 Example 13 Plot level curves for the Cobb-Douglas production function of Example 3.
06/10/10
MasterID: 01660 SOLUTION In Figure 19 we use a computer to draw a contour plot for the Cobb-
Douglas production function
PsL, K d − 1.01L0.75K 0.25
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 897
K Level curves are labeled with the value of the production P. For instance, the level
300
curve labeled 140 shows all values of the labor L and capital investment K that result
in a production of P − 140. We see that, for a fixed value of P, as L increases K
decreases, and vice versa. ■
200
180 220 For some purposes, a contour map is more useful than a graph. That is certainly
140 true in Example 13. (Compare Figure 19 with Figure 8.) It is also true in estimating func-
100 tion values, as in Example 9.
100 Figure 20 shows some computer-generated level curves together with the corre
sponding computer-generated graphs. Notice that the level curves in part (c) crowd
100 200 300 L together near the origin. That corresponds to the fact that the graph in part (d) is very
steep near the origin.
FFIIGGUURREE1189
y z z
y
7et140118
04/26/10
MasterID: 01543
x
x
(a) Level curves of f(x, y)=_xye_≈_¥ (b) Two views of f(x, y)=_xye_≈_¥
y z
x
y
x
(c) Level curves of f(x, y)= _3y (d) f(x, y)= _3y
≈+¥+1 ≈+¥+1
FIGURE 20
Functions of Three or More Variables
A function of three variables, f , is a rule that assigns to each ordered triple sx, y, zd in a
domain D ʚ R 3 a unique real number denoted by f sx, y, zd. For instance, the tempera-
ture T at a point on the surface of the earth depends on the longitude x and latitude y of
the point and on the time t, so we could write T − f sx, y, td.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
898 Chapter 14 Partial Derivatives
Example 14 Find the domain of f if
f sx, y, zd − lnsz 2 yd 1 xy sin z
SOLUTION The expression for f sx, y, zd is defined as long as z 2 y . 0, so the domain
of f is
|D − hsx, y, zd [ R 3 z . yj
This is a half-space consisting of all points that lie above the plane z − y. ■
It’s very difficult to visualize a function f of three variables by its graph, since that
would lie in a four-dimensional space. However, we do gain some insight into f by
examining its level surfaces, which are the surfaces with equations f sx, y, zd − k, where
k is a constant. If the point sx, y, zd moves along a level surface, the value of f sx, y, zd
remains fixed.
z ≈+¥+z@=9 Example 15 Find the level surfaces of the function
≈+¥+z@=4 f sx, y, zd − x 2 1 y 2 1 z2
SOLUTION The level surfaces are x 2 1 y 2 1 z2 − k, where k > 0. These form a family
of concentric spheres with radius sk . (See Figure 21.) Thus, as sx, y, zd varies over any
sphere with center O, the value of f sx, y, zd remains fixed.
■
x y Functions of any number of variables can be considered. A function of n vari
≈+¥+z@=1 ables is a rule that assigns a number z − f sx1, x2, . . . , xn d to an n-tuple sx1, x2, . . . , xn d
FFIIGGUURREE 2201 of real numbers. We denote by Rn the set of all such n-tuples. For example, if a company
uses n different ingredients in making a food product, ci is the cost per unit of the ith
ingredient, and xi units of the ith ingredient are used, then the total cost C of the ingredi-
ents is a function of the n variables x1, x2, . . . , xn:
7et140120 3 C − f sx1, x2, . . . , xn d − c1 x1 1 c2 x2 1 ∙ ∙ ∙ 1 cn xn
04/26/10
MasterID: 01545 The function f is a real-valued function whose domain is a subset of R n. Sometimes we
will use vector notation to write such functions more compactly: If x − kx1, x2, . . . , xnl,
we often write f sxd in place of f sx1, x2, . . . , xn d. With this notation we can rewrite the
function defined in Equation 3 as
f sxd − c ? x
where c − k c1, c2, . . . , cnl and c ? x denotes the dot product of the vectors c and x in Vn.
In view of the one-to-one correspondence between points sx1, x2, . . . , xnd in R n and
their position vectors x − k x1, x2, . . . , xnl in Vn, we have three ways of looking at a func-
tion f defined on a subset of Rn:
1. As a function of n real variables x1, x2, . . . , xn
2. As a function of a single point variable sx1, x2, . . . , xn d
3. As a function of a single vector variable x − k x1, x2, . . . , xnl
We will see that all three points of view are useful.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 899
1. In Example 2 we considered the function W − f sT, vd, where discussed in Example 3 that the production will be doubled
W is the wind-chill index, T is the actual temperature, and v is if both the amount of labor and the amount of capital are
the wind speed. A numerical representation is given in Table 1 doubled. Determine whether this is also true for the general
on page 889. production function
(a) What is the value of f s215, 40d? What is its meaning? PsL, K d − bL␣K 12␣
(b) Describe in words the meaning of the question “For what
5. A model for the surface area of a human body is given by the
value of v is f s220, vd − 230?” Then answer the question. function
(c) Describe in words the meaning of the question “For what
S − f sw, hd − 0.1091w 0.425h 0.725
value of T is f sT, 20d − 249?” Then answer the question.
(d) What is the meaning of the function W − f s25, vd? where w is the weight (in pounds), h is the height (in inches),
and S is measured in square feet.
Describe the behavior of this function.
(e) What is the meaning of the function W − f sT, 50d? (a) Find f s160, 70d and interpret it.
(b) What is your own surface area?
Describe the behavior of this function.
6. T he wind-chill index W discussed in Example 2 has been
2. The temperature-humidity index I (or humidex, for short) is the modeled by the following function:
perceived air temperature when the actual temperature is T and
the relative humidity is h, so we can write I − f sT, hd. The fol-
lowing table of values of I is an excerpt from a table compiled
by the National Oceanic & Atmospheric Administration.
Table 3 Apparent temperature as a function WsT, vd − 13.12 1 0.6215T 2 11.37v 0.16 1 0.3965Tv 0.16
of temperature and humidity
Check to see how closely this model agrees with the values in
Relative humidity (%) Table 1 for a few values of T and v.
h 20 30 40 50 60 70 7. The wave heights h in the open sea depend on the speed v
T 83 of the wind and the length of time t that the wind has been
Actual temperature (°F) blowing at that speed. Values of the function h − f sv, td are
80 77 78 79 81 82 Wi nd speed (knots) recorded in feet in Table 4.
85 82 84 86 88 90 93 (a) What is the value of f s40, 15d? What is its meaning?
(b) What is the meaning of the function h − f s30, td?
90 87 90 93 96 100 106
Describe the behavior of this function.
95 93 96 101 107 114 124 (c) What is the meaning of the function h − f sv, 30d?
100 99 104 110 120 132 144 Describe the behavior of this function.
(a) What is the value of f s95, 70d? What is its meaning? Table 4
(b) For what value of h is f s90, hd − 100? Duration (hours)
(c) For what value of T is f sT, 50d − 88?
(d) What are the meanings of the functions I − f s80, hd √t 5 10 15 20 30 40 50
and I − f s100, hd? Compare the behavior of these two 10 2 2 2 2 2 2 2
functions of h.
15 4 4 5 5 5 5 5
3. A manufacturer has modeled its yearly production function P
(the monetary value of its entire production in millions of 20 5 7 8 8 9 9 9
dollars) as a Cobb-Douglas function
30 9 13 16 17 18 19 19
PsL, Kd − 1.47L 0.65K 0.35
40 14 21 25 28 31 33 33
where L is the number of labor hours (in thousands) and K is 50 19 29 36 40 45 48 50
the invested capital (in millions of dollars). Find Ps120, 20d 60 24 37 47 54 62 67 69
and interpret it.
4. Verify for the Cobb-Douglas production function 8. Am7ecdoeimutmp1a,4nayn0dm1latakrxge0es.t7hItreceosstisze$s2.o5f0ctaordmbaokaerdabsomxaelsl:bsomxa,ll,
PsL, K d − 1.01L 0.75K 0.25 04/26/10
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
MasterID: 01547Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
900 chapter 14 Partial Derivatives
$4.00 for a medium box, and $4.50 for a large box. Fixed costs Iz II z
are $8000.
(a) Express the cost of making x small boxes, y medium x y
IV z
boxes, and z large boxes as a function of three variables: x y
C − f sx, y, zd. III z
(b) Find f s3000, 5000, 4000d and interpret it.
(c) What is the domain of f ? x y x y
Vz VI z
9. Let tsx, yd − cossx 1 2yd.
(a) Evaluate ts2, 21d. y
(b) Find the domain of t.
(c) Find the range of t. x x y
10. Let Fsx, yd − 1 1 s4 2 y2. 33. A contour map for a function f is shown. Use it to estim ate the
(a) Evaluate F s3, 1d. values of f s23, 3d and f s3, 22d. What can you say about the
(b) Find and sketch the domain of F. shape of the graph?
(c) Find the range of F.
11. Let f sx, y, zd − sx 1 sy 1 sz 1 lns4 2 x 2 2 y 2 2 z 2d.
(a) Evaluate f s1, 1, 1d.
(b) Find and describe the domain of f.
12. Let ts x, y, zd − x 3y 2zs10 2 x 2 y 2 z .
(a) Evaluate ts1, 2, 3d.
(b) Find and describe the domain of t.
13–22 Find and sketch the domain of the function.
13. f sx, yd − sx 2 2 1 sy 2 1 y
14. f sx, yd − s4 x 2 3y
15. f sx, yd − lns9 2 x 2 2 9y2 d 16. f sx, yd − sx 2 1 y2 2 4
17. tsx, yd − x 2 y 18. tsx, yd − 1l2ns2x 2 xd 2
x 1 y 22y
19. f sx, yd − sy 2 x 2 1 70 60 50 40
1 2 x2
01 30 x
20. f sx, yd − sin21sx 1 yd
20
21. f sx, y, zd − s4 2 x 2 1 s9 2 y2 1 s1 2 z 2
22. f sx, y, zd − lns16 2 4x 2 2 4y2 2 z2 d 10
23–31 Sketch the graph of the function. 34. Shown is a contour map of atmospheric pressure in North
23. f sx, yd − y 24. f sx, yd − x 2
25. f sx, yd − 10 2 4x 2 5y 26. f sx, yd − cos y America o7n eAutg1u4st0112,x230038. On the level curves (called
27. f sx, yd − sin x 28. f sx, yd − 2 2 x 2 2 y 2 isobars) th0e 4pr/es2su6r/e1is0indicated in millibars (mb).
29. f sx, yd − x 2 1 4y 2 1 1 30. f sx, yd − s4x 2 1 y 2 (a) ES s(tSimanaMtFeraathnsecitpserceors)sI,uDarne:da0Vt 1C(5V(Ca5nh6cicoaugvoe)r,).N (Nashville),
31. f sx, yd − s4 2 4x 2 2 y 2
(b) At which of these locations were the winds strongest?
1016
32. M atch the function with its graph (labeled I–VI). Give reasons V
1016
for your choices.
1012
(a) f sx, yd − 1 1 1 1 y 2 (b) f sx, yd − 1 1 1008 C
x2 1 x2y2
S 1012
(c) f sx, yd − lnsx 2 1 y2d (d) f sx, yd − cos sx 2 1 y2 1008 N
1004
(e) f sx, yd − | xy | (f ) f sx, yd − coss xyd
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
7et1401x34Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.1 Functions of Several Variables 901
35. Level curves (isothermals) are shown for the typical water 18.5; optimal if the BMI lies between 18.5 and 25; overweight
temperature sin 8Cd in Long Lake (Minnesota) as a function of if the BMI lies between 25 and 30; and obese if the BMI
depth and time of year. Estimate the temperature in the lake on exceeds 30. Shade the region corresponding to optimal BMI.
Does someone who weighs 62 kg and is 152 cm tall fall into
June 9 (day 160) at a depth of 10 m and on June 29 (day 180) this category?
at a depth of 5 m.
0 20 40. The body mass index is defined in Exercise 39. Draw the level
12 16 curve of this function corresponding to someone who is 200 cm
Depth (m) tall and weighs 80 kg. Find the weights and heights of two
8 other people with that same level curve.
5
20 16 12 41–44 A contour map of a function is shown. Use it to make a
10 rough sketch of the graph of f .
15 8 41. y 42. y
120
160 200 240 14
Day of the year
280 13
12 _8
11
_6
36. Tpgarwaroapbhcooilsnotaiodcu.orW07nmehe4.aicTp/ths1h2eai46sroe0w/ths11hhei0oxrcwhi3s,n5af.onOrdnawefhuisyn?cfotiroan function f whose x _4
t whose graph is a
I My asterID: 016I5I 9 y x
8
43. y 44. y
7et1401x40
5 M04a/s2t6e/r1ID02:1_001_2_13552
xx
7et144 01x39
04/236/10 3 x
Mas2terID: 01551
1
00 1 2 3 45
37. L(NFoeicagaru07trBe4ee?t1/th21e2)p4.6oH0/ino11tws0xAw3oa6unlddByooun the map of Lonesome Mountain
describe the terrain near A?
0
x
38. MakMe aarosutgeh rskIDetc:h0o1f a5c4o9ntour map for the function whose
graph is shown. c44u55r–.v 5e2fs s .xDM70, yre4adaw/t−s12atx46ce2o0/2nr11tIoD0xyu2r4: m10a1p5o45f t3he fu6n. ct07Mfisoe4xna,/tsys12hdto46−we0/irxn11IygD0xs4:ev20er1al5le5v4el
z 47. f sx, yd − sx 1 y 48. f sx, yd − lnsx 2 1 4y 2d
49. f sx, yd − ye x 50. f sx, yd − y 2 arctan x
51. f sx, yd − s3 x 2 1 y2 52. f sx, yd − yysx 2 1 y2d
y
x 53–54 Sketch both a contour map and a graph of the function and
compare them.
53. f sx, yd − x 2 1 9y 2 54. f sx, yd − s36 2 9x 2 2 4y 2
39. The body mass index (BMI) of a person is defined by
Bsm, hd − m 55. A thin metal plate, located in the xy-plane, has temperature
h2 Tsx, yd at the point sx, yd. Sketch some level curves (isother-
where m7isetth1e 4pe0rs1oxn’3s m8ass (in kilograms) and h is the mals) if the temperature function is given by
height (i0n4m/e2te6rs/).1D0raw the level curves Bsm, hd − 18.5,
Bsm, hdM− a25s,tBesrmI,Dhd: −01306,6an1d Bsm, hd − 40. A rough 100
Tsx, yd − 1 1 x2 1 2y 2
guideline is that a person is underweight if the BMI is less than
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
902 chapter 14 Partial Derivatives
56. If Vsx, yd is the electric potential at a point sx, yd in the 59. f sx, yd − e2sx 21y 2dy3ssinsx 2d 1 coss y 2dd
xy-plane, then the level curves of V are called equipotential 60. f sx, yd − cos x cos y
curves because at all points on such a curve the electric
potential is the same. Sketch some equipotential curves if
Vsx, yd − cysr 2 2 x 2 2 y 2 , where c is a positive constant. 6 1–66 Match the function (a) with its graph (labeled A–F below)
; 57–60 Use a computer to graph the function using various and (b) with its contour map (labeled I–VI). Give reasons for your
domains and viewpoints. Get a printout of one that, in your choices.
opinion, gives a good view. If your software also produces level 61. z − sinsxyd 62. z − e x cos y
curves, then plot some contour lines of the same function and
compare with the graph. 63. z − sinsx 2 yd 64. z − sin x 2 sin y
57. f sx, yd − xy2 2 x 3 (monkey saddle)Graphs and Contour Maps for Exercises 35–40 2d 66. z − 1 1x 2 y
58. f sx, yd − xy3 2 yx 3 (dog saddle) 65. z − s1 2 x 2ds1 x2 1
2 y y2
Az Bz Cz
y x z y x y
x E F z y
x
Dz x yx
II III
7et1401x59-3 y
04/26/10 y
MasxterID: 01555 x
y
I
y
7et1401x59-2
04/26/10
MasterID: 01555
x
IV y Vy VI y
x xx
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.2 Limits and Continuity 903
67–70 Describe the level surfaces of the function. ; 78. Use a computer to investigate the family of surfaces
67. f sx, y, zd − x 1 3y 1 5z
68. f sx, y, zd − x 2 1 3y 2 1 5z2 z − sax 2 1 by 2 de 2x 22y 2
69. f sx, y, zd − y 2 1 z2
70. f sx, y, zd − x 2 2 y 2 2 z2 How does the shape of the graph depend on the numbers a
and b?
7 1–72 Describe how the graph of t is obtained from the graph
of f . ; 79. Use a computer to investigate the family of surfaces
z − x 2 1 y 2 1 cxy. In particular, you should determine the
71. (a) tsx, yd − f sx, yd 1 2 transitional values of c for which the surface changes from
(b) tsx, yd − 2 f sx, yd one type of quadric surface to another.
(c) tsx, yd − 2f sx, yd
(d) tsx, yd − 2 2 f sx, yd ; 80. Graph the functions
f sx, yd − sx 2 1 y 2
72. (a) tsx, yd − f sx 2 2, yd
(b) tsx, yd − f sx, y 1 2d f sx, yd − esx 21y 2
(c) tsx, yd − f sx 1 3, y 2 4d f sx, yd − lnsx 2 1 y 2
f sx, yd − sinssx 2 1 y 2 d
and f sx, yd − 1
sx2 1 y2
; 73–74 Use a computer to graph the function using various In general, if t is a function of one variable, how is the
domains and viewpoints. Get a printout that gives a good view graph of
of the “peaks and valleys.” Would you say the function has a
maximum value? Can you identify any points on the graph that f sx, yd − tssx 2 1 y 2 d
you might consider to be “local maximum points”? What about
“local minimum points”? obtained from the graph of t?
; 81. (a) Show that, by taking logarithms, the general Cobb-
73. f sx, yd − 3x 2 x 4 2 4y 2 2 10xy
Douglas function P − bL␣K 12␣ can be expressed as
74. f sx, yd − xye2x 22y 2
ln P − ln b 1 ␣ ln L
K K
; 75–76 Graph the function using various domains and view-
points. Comment on the limiting behavior of the function. What (b) If we let x − lnsLyK d and y − lnsPyK d, the equation
happens as both x and y become large? What happens as sx, yd in part (a) becomes the linear equation y − ␣x 1 ln b.
approaches the origin? Use Table 2 (in Example 3) to make a table of values
of lnsLyKd and lnsPyKd for the years 1899–1922.
75. f sx, yd − x 1 y 76. f sx, yd − x 2 xy y 2 Then use a graphing calculator or computer to find
x2 1 y2 1
the least squares regression line through the points
; 77. Investigate the family of functions f sx, yd − e cx 21y 2 . How slnsLyKd, lnsPyKdd.
does the shape of the graph depend on c? (c) Deduce that the Cobb-Douglas production function is
P − 1.01L0.75K 0.25.
Let’s compare the behavior of the functions
f sx, yd sinsx 21 y 2 d and ts yd x2 2 y2
x2 1y x2 1 y2
− 2 x, −
as x and y both approach 0 [and therefore the point sx, yd approaches the origin].
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
904 Chapter 14 Partial Derivatives
Tables 1 and 2 show values of f sx, yd and tsx, yd, correct to three decimal places, for
points sx, yd near the origin. (Notice that neither function is defined at the origin.)
Table 1 Values of f sx, yd Table 2 Values of tsx, yd
y Ϫ1.0 Ϫ0.5 Ϫ0.2 0 0.2 0.5 1.0 y Ϫ1.0 Ϫ0.5 Ϫ0.2 0 0.2 0.5 1.0
x 0.455 0.759 0.829 0.841 0.829 0.759 0.455 x 0.000 0.600 0.923 1.000
Ϫ1.0 Ϫ1.0 0.923 0.600 0.000
Ϫ0.5 0.759 0.959 0.986 0.990 0.986 0.959 0.759 Ϫ0.5 Ϫ0.600 0.000 0.724 1.000 0.724 0.000 Ϫ0.600
Ϫ0.2 0.829 0.986 0.999 1.000 0.999 0.986 0.829 Ϫ0.2 Ϫ0.923 Ϫ0.724 0.000 1.000 0.000 Ϫ0.724 Ϫ0.923
0 0.841 0.990 1.000 1.000 0.990 0.841 0 Ϫ1.000 Ϫ1.000 Ϫ1.000 Ϫ1.000 Ϫ1.000 Ϫ1.000
0.2 0.829 0.986 0.999 1.000 0.999 0.986 0.829 0.2 Ϫ0.923 Ϫ0.724 0.000 1.000 0.000 Ϫ0.724 Ϫ0.923
0.5 0.759 0.959 0.986 0.990 0.986 0.959 0.759 0.5 Ϫ0.600 0.000 0.724 1.000 0.724 0.000 Ϫ0.600
1.0 0.455 0.759 0.829 0.841 0.829 0.759 0.455 1.0 0.000 0.600 0.923 1.000 0.923 0.600 0.000
7et1402t01 It appears that as sx, yd a7pperto1ac4h0es2t(00,20), the values of f sx, yd are approaching 1
04/28/10 whereas the values of tsx, y0d 4ar/e2n8’t/a1p0proaching any number. It turns out that these
MasterID: 01557 guesses based on numerical eMvidaesntceerarIeDc:o0rr1ec5t,5a8nd we write
lim sinsx 2 1 y 2 d − 1 and lim x 2 2 y2 does not exist
x2 1 y2 s x, yd l s0, 0d x 2 1 y2
s x, yd l s0, 0d
In general, we use the notation
lim f sx, yd − L
s x, yd l s a, bd
to indicate that the values of f sx, yd approach the number L as the point sx, yd approaches
the point sa, bd along any path that stays within the domain of f. In other words, we can
make the values of f sx, yd as close to L as we like by taking the point sx, yd sufficiently
close to the point sa, bd, but not equal to sa, bd. A more precise definition follows.
1 Definition Let f be a function of two variables whose domain D includes
points arbitrarily close to sa, bd. Then we say that the limit of f sx, yd as sx, yd
approaches sa, bd is L and we write
lim f sx, yd − L
sx, yd l sa, bd
if for every number « . 0 there is a corresponding number ␦ . 0 such that
| |if sx, yd [ D and 0 , ssx 2 ad2 1 sy 2 bd2 , ␦ then f sx, yd 2 L , «
Other notations for the limit in Definition 1 are
lim f sx, yd − L and f sx, yd l L as sx, yd l sa, bd
xla
ylb
| |Notice that f sx, yd 2 L is the distance between the numbers f sx, yd and L, and
ssx 2 ad 2 1 s y 2 bd 2 is the distance between the point sx, yd and the point sa, bd. Thus
Definition 1 says that the distance between f sx, yd and L can be made arbitrarily small by
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.2 Limits and Continuity 905
making the distance from sx, yd to sa, bd sufficiently small (but not 0). Figure 1 illustrates
Definition 1 by means of an arrow diagram. If any small interval sL 2 «, L 1 «d is given
around L, then we can find a disk D␦ with center sa, bd and radius ␦ . 0 such that f maps
all the points in D␦ [except possibly sa, bd] into the interval sL 2 «, L 1 «d.
yy zz zz
D D∂ (x, y) (x, y) L+∑ L+∑
00 ∂ L L
(a, b) (a, b) L-∑ L-∑ SS
x ff )(L+∑ L+∑
)( LL
x L-∑ L-∑
00
00 xx (a, b) (a, b) D∂ D∂ y y
FIGURFFEIIGG1UURREE11 FFIGUURRFEIEG22URE 2
7et174e0t210410-20021-02 Another illustration of Definition 1 is given in Figure 2 where the surface S is the
04/2084//1208/10 graph of f. If « . 0 is given, we can find ␦ . 0 such that if sx, yd is restricted to lie in the
MasMtearsIDte:r0ID30: 4053/004155/6001560 disk D␦ and sx, yd ± sa, bd, then the corresponding part of S lies between the horizontal
planes z − L 2 « and z − L 1 «.
y
For functions of a single variable, when we let x approach a, there are only two pos-
b x
0a sible directions of approach, from the left or from the right. We recall from Chapter 2 that
if limx la2 f sxd ± limx la1 f sxd, then limx l a f sxd does not exist.
FFIIGGUURREE33
For functions of two variables the situation is not as simple because we can let sx, yd
7et140203 approach sa, bd from an infinite number of directions in any manner whatsoever (see
04/28/10 Figure 3) as long as sx, yd stays within the domain of f.
MasterID: 01561
Definition 1 says that the distance between f sx, yd and L can be made arbitrarily small
by making the distance from sx, yd to sa, bd sufficiently small (but not 0). The definition
refers only to the distance between sx, yd and sa, bd. It does not refer to the direction of
approach. Therefore, if the limit exists, then f sx, yd must approach the same limit no
matter how sx, yd approaches sa, bd. Thus, if we can find two different paths of approach
along which the function f sx, yd has different limits, then it follows that limsx, yd l sa, bd f sx, yd
does not exist.
If f sx, yd l L1 as sx, yd l sa, bd along a path C1 and f sx, yd l L2 as
sx, yd l sa, bd along a path C2, where L1 ± L2, then limsx, yd l sa, bd f sx, yd does
not exist.
EXAMPLE 1 Show that lim x2 2 y2 does not exist.
x2 1 y2
s x, yd l s0, 0d
y SOLUTION Let f sx, yd − sx 2 2 y 2 dysx 2 1 y 2 d. First let’s approach s0, 0d along the
f=_1 x-axis. Then y − 0 gives f sx, 0d − x2yx2 − 1 for all x ± 0, so
FFIIGGUURREE44 f sx, yd l 1 as sx, yd l s0, 0d along the x-axis
f=1 x We now approach along the y-axis by putting x 0. Then f s0, yd 2y 2 21 for all
y ± 0, so y2
− − −
f sx, yd l 21 as sx, yd l s0, 0d along the y-axis
(See Figure 4.) Since f has two different limits along two different lines, the given limit
7et140204Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
906 Chapter 14 Partial Derivatives
does not exist. (This confirms the conjecture we made on the basis of numerical evi-
dence at the beginning of this section.) ■
EXAMPLE 2 If f sx, yd − xyysx2 1 y2 d, does lim f sx, yd exist?
sx, ydl s0, 0d
SOLUTION If y − 0, then f sx, 0d − 0yx2 − 0. Therefore
f sx, yd l 0 as sx, yd l s0, 0d along the x-axis
y If x − 0, then f s0, yd − 0yy2 − 0, so
f=0 f sx, yd l 0 as sx, yd l s0, 0d along the y-axis
y=x Although we have obtained identical limits along the axes, that does not show that the
given limit is 0. Let’s now approach s0, 0d along another line, say y − x. For all x ± 0,
f= 1
2
f=0 x f sx, xd − x2 x2 x2 − 1
1 2
Therefore f sx, yd l 21 as sx, yd l s0, 0d along y − x
(See Figure 5.) Since we have obtained different limits along different paths, the given
FFIIGGUURREE55 limit does not exist. ■
7et140205 Figure 6 sheds some light on Example 2. The ridge that occurs above the line y − x
04/28/10 1
MasterID: 01563 corresponds to the fact that f sx, yd − 2 for all points sx, yd on that line except the origin.
TEC In Visual 14.2 a rotating line on zy
the surface in Figure 6 shows differ-
ent limits at the origin from different
directions.
x
FFIGUURREE66
xxyy
f sfx(x, ,y yd)−= x≈2+1¥y 2
EXAMPLE 3 If f sx, yd − x 2 xy2 4 , does lim f sx, yd exist?
1y s x, yd l s0, 0d
7et140206
04/28/10 SOLUTION With the solution of Example 2 in mind, let’s try to save time by letting
sx, yd l s0, 0d along any line through the origin. If the line is not the y-axis, then
MasterID: 01564 y − mx, where m is the slope, and
f sx, yd − f sx, mxd − xsmxd2 − x2 m 2x 3 − 1 m 2x
x 2 1 smxd4 1 m 4x 4 1 m 4x 2
So f sx, yd l 0 as sx, yd l s0, 0d along y − mx
We get the same result as sx, yd l s0, 0d along the line x − 0. Thus f has the same
limiting value along every line through the origin. But that does not show that the given
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Section 14.2 Limits and Continuity 907
Figure 7 shows the graph of the func- limit is 0, for if we now let sx, yd l s0, 0d along the parabola x − y2, we have
tion in Example 3. Notice the ridge f sx, yd − f s y 2, yd − y2 ? y2 − y4 − 1
above the parabola x − y 2. s y 2 d2 1 y 4 2y 4 2
0.5 so f sx, yd l 12 as sx, yd l s0, 0d along x − y2
Since different paths lead to different limiting values, the given limit does not exist. ■
z0
2 0 _2 Now let’s look at limits that do exist. Just as for functions of one variable, the calcula
_0.5 y tion of limits for functions of two variables can be greatly simplified by the use of prop-
2 x0 _2 erties of limits. The Limit Laws listed in Section 2.3 can be extended to functions of two
variables: the limit of a sum is the sum of the limits, the limit of a product is the product
FFIGUURREE77 of the limits, and so on. In particular, the following equations are true.
7et140207 2 lim x − a lim y − b lim c − c
04/28/10 sx, yd l sa, bd sx, yd l sa, bd sx, yd l sa, bd
MasterID: 01565 The Squeeze Theorem also holds.
EXAMPLE 4 Find lim 3x 2 y if it exists.
x2 1 y2
sx, yd l s0, 0d
SOLUTION As in Example 3, we could show that the limit along any line through the
origin is 0. This doesn’t prove that the given limit is 0, but the limits along the parabo-
las y − x2 and x − y2 also turn out to be 0, so we begin to suspect that the limit does
exist and is equal to 0.
Let « . 0. We want to find ␦ . 0 such that
Z Zif 0 , sx2 1 y2
, ␦ then 3x 2 y 20 ,«
x2 1 y2
| |if 0
that is, , sx2 1 y2 , ␦ then 3x 2 y , «
x2 1 y2
But x 2 < x 2 1 y 2 since y 2 > 0, so x 2ysx 2 1 y 2 d < 1 and therefore
| | | |3
3x 2 y < 3 y − 3 sy 2 < 3 sx 2 1 y 2
x2 1 y2
Thus if we choose ␦ − «y3 and let 0 , sx2 1 y2 , ␦, then
Another way to do Example 4 is to Z Z S D3x2y «
use the Squeeze Theorem instead of x2 1 y2 3
Definition 1. From (2) it follows that 2 0 < 3 sx 2 1 y 2 , 3␦ − 3 −«
| |lim 3 y − 0 Hence, by Definition 1, lim 3x 2 y − 0 ■
x2 1 y2
sx, yd l s0, 0d sx, yd l s0, 0d
and so the first inequality in (3) shows
that the given limit is 0.
Continuity
Recall that evaluating limits of continuous functions of a single variable is easy. It can
be accomplished by direct substitution because the defining property of a continuous
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
908 Chapter 14 Partial Derivatives
function is limxla f sxd − f sad. Continuous functions of two variables are also defined
by the direct substitution property.
4 Definition A function f of two variables is called continuous at sa, bd if
lim f sx, yd − f sa, bd
sx, yd l sa, bd
We say f is continuous on D if f is continuous at every point sa, bd in D.
The intuitive meaning of continuity is that if the point sx, yd changes by a small
amount, then the value of f sx, yd changes by a small amount. This means that a surface
that is the graph of a continuous function has no hole or break.
Using the properties of limits, you can see that sums, differences, products, and quo-
tients of continuous functions are continuous on their domains. Let’s use this fact to give
examples of continuous functions.
A polynomial function of two variables (or polynomial, for short) is a sum of
terms of the form cxmyn, where c is a constant and m and n are nonnegative integers. A
rational function is a ratio of polynomials. For instance,
f sx, yd − x 4 1 5x 3y 2 1 6xy 4 2 7y 1 6
is a polynomial, whereas
tsx, yd − 2xy 1 1
x2 1 y2
is a rational function.
The limits in (2) show that the functions f sx, yd − x, tsx, yd − y, and hsx, yd − c are
continuous. Since any polynomial can be built up out of the simple functions f , t, and h
by multiplication and addition, it follows that all polynomials are continuous on R 2.
Likewise, any rational function is continuous on its domain because it is a quotient of
continuous functions.
EXAMPLE 5 Evaluate lim sx2y3 2 x3y2 1 3x 1 2yd.
sx, yd l s1, 2d
SOLUTION Since f sx, yd − x 2 y 3 2 x 3y 2 1 3x 1 2y is a polynomial, it is continuous
everywhere, so we can find the limit by direct substitution:
lim sx 2y 3 2 x 3y 2 1 3x 1 2yd − 12 ? 23 2 13 ? 22 1 3 ? 1 1 2 ? 2 − 11 ■
sx, yd l s1, 2d
EXAMPLE 6 Where is the function f sx, yd − x2 2 y2 continuous?
x2 1 y2
SOLUTION The function f is discontinuous at s0, 0d because it is not defined there. ■
|Since f is a rational function, it is continuous on its domain, which is the set
D − hsx, yd sx, yd ± s0, 0dj.
EXAMPLE 7 Let H x2 2 y2
tsx, yd − x 2 1 y 2 if sx, yd ± s0, 0d
0 if sx, yd − s0, 0d
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.2 Limits and Continuity 909
Here t is defined at s0, 0d but t is still discontinuous there because limsx, yd l s0, 0d tsx, yd
does not exist (see Example 1). ■
Figure 8 shows the graph of the EXAMPLE 8 Let H 3x2y
continuous function in Example 8.
f sx, yd − x 2 1 y 2 if sx, yd ± s0, 0d
z 0 if sx, yd − s0, 0d
We know f is continuous for sx, yd ± s0, 0d since it is equal to a rational function there.
y Also, from Example 4, we have
x lim fsx, yd − lim 3x 2 y − 0 − f s0, 0d
sx, yd l s0, 0d x2 1 y2
sx, yd l s0, 0d
Therefore f is continuous at s0, 0d, and so it is continuous on R 2. ■
FFIIGGUURREE88
7et140208 Just as for functions of one variable, composition is another way of combining two
04/28/10 continuous functions to get a third. In fact, it can be shown that if f is a continuous func-
MasterID: 01566 tion of two variables and t is a continuous function of a single variable that is defined on
the range of f , then the composite function h − t 8 f defined by hsx, yd − ts f sx, ydd is
2 also a continuous function.
z0 EXAMPLE 9 Where is the function hsx, yd − arctansyyxd continuous?
_2 _2 SOLUTION The function f sx, yd − yyx is a rational function and therefore continuous
_2 except on the line x − 0. The function tstd − arctan t is continuous everywhere. So the
_1 composite function
_1 0 x ts f sx, ydd − arctansyyxd − hsx, yd
y0 1
1
22
FFIIGGUURREE99 is continuous except where x − 0. The graph in Figure 9 shows the break in the graph
TThhee ffuunnccttiioonn h(sxx,, yy)=d −arcatracnt(ayn/sxy) yxd
iiss ddiissccoonnttiinnuuoouuss wwhheerree x=−0.0. of h above the y-axis. ■
7et140209 Functions of Three or More Variables
04/28/10 Everything that we have done in this section can be extended to functions of three or
MasterID: 01567 more variables. The notation
lim f sx, y, zd − L
sx, y, zd l sa, b, cd
means that the values of f sx, y, zd approach the number L as the point sx, y, zd approaches
the point sa, b, cd along any path in the domain of f. Because the distance between two
points sx, y, zd and sa, b, cd in R 3 is given by ssx 2 ad 2 1 s y 2 bd 2 1 sz 2 cd 2 , we can
write the precise definition as follows: for every number « . 0 there is a corresponding
number ␦ . 0 such that
if sx, y, zd is in the domain of f and 0 , ssx 2 ad 2 1 s y 2 bd 2 1 sz 2 cd 2 , ␦
| |then f sx, y, zd 2 L , «
The function f is continuous at sa, b, cd if
lim f sx, y, zd − f sa, b, cd
sx, y, zd l sa, b, cd
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
910 Chapter 14 Partial Derivatives
For instance, the function
f sx, y, zd − x2 1 1 z2 2 1
y2 1
is a rational function of three variables and so is continuous at every point in R 3 except
where x2 1 y2 1 z2 − 1. In other words, it is discontinuous on the sphere with center
the origin and radius 1.
If we use the vector notation introduced at the end of Section 14.1, then we can write
the definitions of a limit for functions of two or three variables in a single compact form
as follows.
5 If f is defined on a subset D of Rn, then lim x l a f sxd − L means that for every
number « . 0 there is a corresponding number ␦ . 0 such that
| | | |if x [ D and 0 , x 2 a , ␦ then f sxd 2 L , «
Notice that if n − 1, then x − x and a − a, and (5) is just the definition of a limit for
functions of a single variable. For the case n − 2, we have x − kx, y l, a − ka, b l,
| |and x 2 a − ssx 2 ad2 1 sy 2 bd2 , so (5) becomes Definition 1. If n − 3, then
x − kx, y, z l, a − ka, b, c l, and (5) becomes the definition of a limit of a function of
three variables. In each case the definition of continuity can be written as
lim f sxd − f sad
xla
1. Suppose that limsx, yd l s3, 1d f sx, yd − 6. What can you say 7. lim y sinsx 2 yd 8. lim e s2x2y
about the value of f s3, 1d? What if f is continuous? sx, yd l s, y2d sx, yd l s3, 2d
2. Explain why each function is continuous or discontinuous. 9. lim x4 2 4y 2 10. sx, yldilms0, 0d 5xy44 cos2x
(a) The outdoor temperature as a function of longitude, x2 1 2y 2 1 y4
sx, yd l s0, 0d
latitude, and time
(b) Elevation (height above sea level) as a function of 11. lim y2 sin2x 12. sx, yldilms1, 0d sx 2xy12d2 y y2
x4 1 y4 1
longitude, latitude, and time sx, ydls0, 0d
(c) The cost of a taxi ride as a function of distance traveled
13. lim xy y 2 14. sx, yldilms0, 0d x 2 x3 2 y3
and time sx 2 1 1 xy 1
sx, ydl s0, 0d
3–4 Use a table of numerical values of f sx, yd for sx, yd near the y2
origin to make a conjecture about the value of the limit of f sx, yd
as sx, yd l s0, 0d. Then explain why your guess is correct. 15. lim xy 2 cos y 16. sx, ydlilms0, 0d x 4 xy4 y 4
x2 1 y 1
sx, ydl s0, 0d 4
3. f sx, yd − x2y3 1 x3y2 2 5 4. f sx, yd − 2xy x2 1 y2
2 2 xy x2 1 2y2
17. lim
sx, ydl s0, 0d sx 2 1 y 2 1 1 2 1
5–22 Find the limit, if it exists, or show that the limit does 18. lim xy4
x2 1 y8
not exist. sx, yd l s0, 0d
5. lim sx2y3 2 4y 2d 6. lim x2y 1 xy2 19. lim e y 2tansx zd 20. lim xy 1 yz z2
sx, ydls3, 2d sx, yd l s2, 21d x2 2 y2 sx, y, zd l s, 0, 1y3d sx, y, zdls0, 0, 0d x2 1 y2 1
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.3 Partial Derivatives 911
21. lim x y 1 yz 2 1 xz 2 H xy if sx, yd ± s0, 0d
x2 1 y2 1 z4 if sx, yd − s0, 0d
sx, y, zd l s0, 0, 0d 38. f sx, yd − x 2 1 x y 1 y 2
0
22. lim x 2 y 2z 2
x2 1 y2 1 z2
sx, y, zd l s0, 0, 0d
; 2 3–24 Use a computer graph of the function to explain why the 39–41 Use polar coordinates to find the limit. [If sr, d are
limit does not exist. polar coordinates of the point sx, yd with r > 0, note that
r l 01 as sx, yd l s0, 0d.]
2 x 2 1 3x y 1 4 y 2 xy3 39. lim x3 1 y3
3x 2 1 5y 2 x2 1 y6 sx, yd l s0, 0d x2 1 y2
23. lim 24. lim 0d
sx, yd l s0,
sx, yd l s0, 0d
40. lim sx2 1 y2 d lnsx2 1 y2 d
sx, yd l s0, 0d
2 5–26 Find hsx, yd − ts f sx, ydd and the set of points at which h 41. lim e2x 22y 2 2 1
is continuous. x2 1 y2
sx, yd l s0, 0d
25. tstd − t 2 1 st , f sx, yd − 2 x 1 3y 2 6
26. tstd − t 1 ln t, f sx, yd − 1 2 xy ; 42. At the beginning of this section we considered the function
1 1 x2y2
sinsx 2 1 y 2 d
f sx, yd − x2 1 y2
; 2 7–28 Graph the function and observe where it is discontinu- and guessed on the basis of numerical evidence that
ous. Then use the formula to explain what you have observed. f sx, yd l 1 as sx, yd l s0, 0d. Use polar coordinates to
confirm the value of the limit. Then graph the function.
27. f sx, yd − e 1ysx2yd 28. f sx, yd − 1
1 2 x2 2 y2
29–38 Determine the set of points at which the function is H; 43. G raph and discuss the continuity of the functionsin xy
xy if xy ± 0
continuous. f sx, yd −
29. Fsx, yd − 1 x y x2y 30. Fsx, yd − cos s1 1 x 2 y 1 if xy − 0
1 e
31. Fsx, yd − 1 1 x2 1 y 2 32. Hsx, yd − ex 1 ey H 44. Let 0 if y < 0 or y > x 4
1 2 x2 2 y 2 exy 2 1 1 if 0 , y , x 4
f sx, yd −
33. Gsx, yd − sx 1 s1 2 x 2 2 y 2
34. Gsx, yd − lns1 1 x 2 yd (a) Show that f sx, yd l 0 as sx, yd l s0, 0d along any path
through s0, 0d of the form y − mx a with 0 , a , 4.
35. f sx, y, zd − arcsinsx 2 1 y 2 1 z 2d
(b) Despite part (a), show that f is discontinuous at s0, 0d.
H 36. f sx, y, zd − sy 2 x2 ln zx2y3 if sx, yd ± s0, 0d (c) Show that f is discontinuous on two entire curves.
37. f sx, yd − 2x2 1 y2
| | | | 45. S how that the function f given by f sxd − x is continuous
1 if sx, yd − s0, 0d on R n. [Hint: Consider x 2 a 2 − sx 2 ad ? sx 2 ad.]
46. If c [ Vn, show that the function f given by f sxd − c ? x is
continuous on R n.
On a hot day, extreme humidity makes us think the temperature is higher than it really
is, whereas in very dry air we perceive the temperature to be lower than the thermom-
eter indicates. The National Weather Service has devised the heat index (also called the
temperature-humidity index, or humidex, in some countries) to describe the combined
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
912 Chapter 14 Partial Derivatives
effects of temperature and humidity. The heat index I is the perceived air tempera-
ture when the actual temperature is T and the relative humidity is H. So I is a function of
T and H and we can write I − f sT, Hd. The following table of values of I is an excerpt
from a table compiled by the National Weather Service.
Table 1 Heat index I as a function of temperature and humidity
Relative humidity (%)
TH 50 55 60 65 70 75 80 85 90
90 96 98 100 103 106 109 112 115 119
128
92 100 103 105 108 112 115 119 123 137
Actual 146
temperature 94 104 107 111 114 118 122 127 132 157
168
(°F) 96 109 113 116 121 125 130 135 141
98 114 118 123 127 133 138 144 150
100 119 124 129 135 141 147 154 161
If7weet1co4n0ce3ntt0ra1te on the highlighted column of the table, which corresponds to a rela-
tive 0hu4m/i2d8ity/1o0f H − 70%, we are considering the heat index as a function of the single
tvhaerihaMebaleat isTntdfeeoxrrIIaDifni:xc0ered1av5sea6slu8aes of H. Let’s write tsT d − f sT, 70d. Then tsT d des cribes how
the actual temperature T increases when the relative humid-
ity is 70%. The derivative of t when T − 968F is the rate of change of I with respect to
T when T − 968F:
t9s96d − lim ts96 1 hd 2 ts96d − lim f s96 1 h, 70d 2 f s96, 70d
hl0 h hl0 h
We can approximate t9s96d using the values in Table 1 by taking h − 2 and 22:
t9s96d < ts98d 2 ts96d − f s98, 70d 2 f s96, 70d − 133 2 125 − 4
2 2 2
t9s96d < ts94d 2 ts96d − f s94, 70d 2 f s96, 70d − 118 2 125 − 3.5
22 22 22
Averaging these values, we can say that the derivative t9s96d is approximately 3.75. This
means that, when the actual temperature is 968F and the relative humidity is 70%, the
apparent temperature (heat index) rises by about 3.758F for every degree that the actual
temperature rises!
Now let’s look at the highlighted row in Table 1, which corresponds to a fixed temper
ature of T − 968F. The numbers in this row are values of the function GsH d − f s96, Hd,
which describes how the heat index increases as the relative humidity H increases when
the actual temperature is T − 968F. The derivative of this function when H − 70% is
the rate of change of I with respect to H when H − 70%:
G9s70d − lim Gs70 1 hd 2 Gs70d − lim f s96, 70 1 hd 2 f s96, 70d
hl0 h h
hl0
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.3 Partial Derivatives 913
By taking h − 5 and 25, we approximate G9s70d using the tabular values:
G9s70d < Gs75d 2 Gs70d − f s96, 75d 2 f s96, 70d − 130 2 125 − 1
5 5 5
G9s70d < Gs65d 2 Gs70d − f s96, 65d 2 f s96, 70d − 121 2 125 − 0.8
25 25 25
By averaging these values we get the estimate G9s70d < 0.9. This says that, when the
temperature is 968F and the relative humidity is 70%, the heat index rises about 0.98F for
every percent that the relative humidity rises.
In general, if f is a function of two variables x and y, suppose we let only x vary while
keeping y fixed, say y − b, where b is a constant. Then we are really considering a func-
tion of a single variable x, namely, tsxd − f sx, bd. If t has a derivative at a, then we call
it the partial derivative of f with respect to x at sa, bd and denote it by fxsa, bd. Thus
1 fxsa, bd − t9sad where tsxd − f sx, bd
By the definition of a derivative, we have
t9sad − lim tsa 1 hd 2 tsad
hl0 h
and so Equation 1 becomes
2 fxsa, bd − lim f sa 1 h, bd 2 f sa, bd
h
hl0
Similarly, the partial derivative of f with respect to y at sa, bd, denoted by fysa, bd, is
obtained by keeping x fixed sx − ad and finding the ordinary derivative at b of the func-
tion Gsyd − f sa, yd:
3 fysa, bd − lim f sa, b 1 hd 2 f sa, bd
h
hl0
With this notation for partial derivatives, we can write the rates of change of the heat
index I with respect to the actual temperature T and relative humidity H when T − 968F
and H − 70% as follows:
fT s96, 70d < 3.75 fHs96, 70d < 0.9
If we now let the point sa, bd vary in Equations 2 and 3, fx and fy become functions of
two variables.
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914 Chapter 14 Partial Derivatives
4 If f is a function of two variables, its partial derivatives are the functions fx
and fy defined by
fxsx, yd − lim f sx 1 h, yd 2 f sx, yd
h
hl0
fysx, yd − lim f sx, y 1 hd 2 f sx, yd
h
hl0
There are many alternative notations for partial derivatives. For instance, instead of
fx we can write f1 or D1 f (to indicate differentiation with respect to the first variable) or
−fy−x. But here −fy−x can’t be interpreted as a ratio of differentials.
Notations for Partial Derivatives If z − f sx, yd, we write
fxsx, yd − fx − −f − − f sx, yd − −z − f1 − D1 f − Dx f
−x −x −x
fysx, yd − fy − −f − − f sx, yd − −z − f2 − D2 f − Dy f
−y −y −y
To compute partial derivatives, all we have to do is remember from Equation 1 that
the partial derivative with respect to x is just the ordinary derivative of the function t of
a single variable that we get by keeping y fixed. Thus we have the following rule.
Rule for Finding Partial Derivatives of z − f sx, yd
1. To find fx, regard y as a constant and differentiate f sx, yd with respect to x.
2. To find fy, regard x as a constant and differentiate f sx, yd with respect to y.
EXAMPLE 1 If f sx, yd − x3 1 x2 y3 2 2y2, find fxs2, 1d and fys2, 1d.
SOLUTION Holding y constant and differentiating with respect to x, we get
and so fxsx, yd − 3x 2 1 2 x y 3
fxs2, 1d − 3 ? 22 1 2 ? 2 ? 13 − 16
Holding x constant and differentiating with respect to y, we get ■
fysx, yd − 3x 2 y 2 2 4y
fys2, 1d − 3 ? 22 ? 12 2 4 ? 1 − 8
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.3 Partial Derivatives 915
z T¡ Interpretations of Partial Derivatives
S C¡ To give a geometric interpretation of partial derivatives, we recall that the equation
T™ z − f sx, yd represents a surface S (the graph of f ). If f sa, bd − c, then the point Psa, b, cd
lies on S. By fixing y − b, we are restricting our attention to the curve C1 in which the
P (a, b, c) C™ vertical plane y − b intersects S. (In other words, C1 is the trace of S in the plane y − b.d
Likewise, the vertical plane x − a intersects S in a curve C2. Both of the curves C1 and
0 C2 pass through the point P. (See Figure 1.)
xy Note that the curve C1 is the graph of the function tsxd − f sx, bd, so the slope of its tan-
(a, b, 0) gent T1 at P is t9sad − fxsa, bd. The curve C2 is the graph of the function Gs yd − f sa, yd,
so the slope of its tangent T2 at P is G9sbd − fysa, bd.
FFIIGGUURREE11
The partial derivatives of ff aatt(saa,,bb)daarree Thus the partial derivatives fxsa, bd and fy sa, bd can be interpreted geometrically as
the slopes of the tangents to CC¡1 and CC™2. the slopes of the tangent lines at Psa, b, cd to the traces C1 and C2 of S in the planes y − b
and x − a.
7et140301
04/29/10 As we have seen in the case of the heat index function, partial derivatives can also be
MasterID: 01569 interpreted as rates of change. If z − f sx, yd, then −zy−x represents the rate of change of
z with respect to x when y is fixed. Similarly, −zy−y represents the rate of change of z with
respect to y when x is fixed.
EXAMPLE 2 If f sx, yd − 4 2 x2 2 2y2, find fxs1, 1d and fys1, 1d and interpret these
numbers as slopes.
SOLUTION We have
fxsx, yd − 22x fysx, yd − 24y
fxs1, 1d − 22 fys1, 1d − 24
The graph of f is the paraboloid z − 4 2 x2 2 2y2 and the vertical plane y − 1 inter-
sects it in the parabola z − 2 2 x2, y − 1. (As in the preceding discussion, we label it
C1 in Figure 2.) The slope of the tangent line to this parabola at the point s1, 1, 1d is
fxs1, 1d − 22. Similarly, the curve C2 in which the plane x − 1 intersects the parabo-
loid is the parabola z − 3 2 2y2, x − 1, and the slope of the tangent line at s1, 1, 1d is
fys1, 1d − 24. (See Figure 3.)
zz zz
z=z4=-4≈--≈2-¥2¥ z=z4=-4≈--≈2-¥2¥
C¡ C¡ C™ C™
y=y1=1 x=x1=1
(1, 1(,1,1)1, 1) (1, 1(,1,1)1, 1)
x x2 2 (1, 1()1, 1) yy yy
FFIIGGFUUIGRRUEER22E 2 x x2 2 (1, 1()1, 1) ■
F IGFFUIIGRGUEUR3REE33
7e7te1t4104300320–20–303
040/42/92/91/010
MMasatestreIDrI:D0:10517507-701-71Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
916 Chapter 14 Partial Derivatives
Figure 4 is a computer-drawn counterpart to Figure 2. Part (a) shows the plane y − 1
intersecting the surface to form the curve C1 and part (b) shows C1 and T1. [We have used
the vector equations rstd − k t, 1, 2 2 t2l for C1 and rstd − k1 1 t, 1, 1 2 2t l for T1.]
Similarly, Figure 5 corresponds to Figure 3.
4 4
3
z2 3
1
00 z2
y1 0 1 y1 0
2 1x 00 2 1x
FFIGUURREE44 (a) (b)
4 4
3
z2 3
1
00 z2
FFIGUURREE55 y1 0 1 y1 0
2 1x 00 2 1x
EXAMPLE 3 In Exercise 14.1.39 we defined the body mass index of a person as
7et140304-05 Bsm, hd − m
04/29/10 h2
MasterID: 01572-7C3alculate the partial derivatives of B for a young man with m − 64 kg and h − 1.68 m
and interpret them.
Solution Regarding h as a constant, we see that the partial derivative with respect to
S Dm is −B − m 1
−m sm, hd − −m h2 − h2
so −B s64, 1.68d − 1 < 0.35 skgym2dy kg
−m s1.68d2
This is the rate at which the man’s BMI increases with respect to his weight when he
weighs 64 kg and his height is 1.68 m. So if his weight increases by a small amount,
one kilogram for instance, and his height remains unchanged, then his BMI will
increase by about 0.35.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 14.3 Partial Derivatives 917
Now we regard m as a constant. The partial derivative with respect to h is
S D S D−B
− m 2 2m
−h −h h2 2 h3 − 2 h3
sm, hd − −m
so −B s64, 1.68d − 2 ? 64 < 227 skgym2dym
−h 2 s1.68d3
This is the rate at which the man’s BMI increases with respect to his height when he
weighs 64 kg and his height is 1.68 m. So if the man is still growing and his weight
stays unchanged while his height increases by a small amount, say 1 cm, then his BMI
will decrease by about 27s0.01d − 0.27. ■
S DEXAMPLE 4 If f sx, yd − sin x , calculate −f and −f .
11y −x −y
SOLUTION Using the Chain Rule for functions of one variable, we have
S D S D S D−f
x − x x 1
−x 11y −x 11y 11y ? 11y
− cos ? − cos
S D S D S D −fcos x − x x x yd2
−y − 11y ? −y 11y − 2cos 11y ? s1 1 ■
Some computer software can plot EXAMPLE 5 Find −zy−x and −zy−y if z is defined implicitly as a function of x and y by
surfaces defined by implicit equations
in three variables. Figure 6 shows such the equation
a plot of the surface defined by the
equation in Example 5. x3 1 y3 1 z3 1 6xyz − 1
FIGURE 6 SOLUTION To find −zy−x, we differentiate implicitly with respect to x, being careful to
treat y as a constant:
3x 2 1 3z2 −z 1 6yz 1 6xy −z − 0
−x −x
Solving this equation for −zy−x, we obtain
−z − 2 x2 1 2yz
−x z2 1 2xy
Similarly, implicit differentiation with respect to y gives
−z − 2 y2 1 2xz ■
−y z2 1 2xy
Functions of More Than Two Variables
Partial derivatives can also be defined for functions of three or more variables. For
example, if f is a function of three variables x, y, and z, then its partial derivative with
respect to x is defined as
fxsx, y, zd − lim f sx 1 h, y, zd 2 f sx, y, zd
h
hl0
and it is found by regarding y and z as constants and differentiating f sx, y, zd with respect
to x. If w − f sx, y, zd, then fx − −wy−x can be interpreted as the rate of change of w with
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
918 Chapter 14 Partial Derivatives
respect to x when y and z are held fixed. But we can’t interpret it geometrically because
the graph of f lies in four-dimensional space.
In general, if u is a function of n variables, u − f sx1, x2, . . . , xn d, its partial derivative
with respect to the ith variable xi is
−u − lim f sx1, . . . , xi21, xi 1 h, xi11, . . . , xn d 2 f sx1, . . . , xi , . . . , xnd
−xi h
hl0
and we also write −u −f
−xi −xi
− − fx i − fi − Di f
EXAMPLE 6 Find fx, fy, and fz if f sx, y, zd − ex y ln z.
SOLUTION Holding y and z constant and differentiating with respect to x, we have
fx − ye x y ln z
ex y
z
Similarly, fy − xe x y ln z and fz − ■
Higher Derivatives
If f is a function of two variables, then its partial derivatives fx and fy are also functions
of two variables, so we can consider their partial derivatives s fx dx, s fx dy, s fy dx, and s fy dy,
which are called the second partial derivatives of f. If z − f sx, yd, we use the following
notation: S Ds fxdx
− fx x − f11 − − −f − −2f − −2z
−x −x −x 2 −x 2
S Ds fxdy
− fx y − f12 − − −f − −2f − −2z
−y −x −y −x −y −x
S Ds fydx
− fy x − f21 − − −f − −2f − −2z
−x −y −x −y −x −y
S Ds fydy
− fy y − f22 − − −f − −2f − −2z
−y −y −y 2 −y 2
Thus the notation fx y (or −2fy−y −x) means that we first differentiate with respect to x and
then with respect to y, whereas in computing fyx the order is reversed.
EXAMPLE 7 Find the second partial derivatives of
f sx, yd − x 3 1 x 2 y 3 2 2y 2
SOLUTION In Example 1 we found that
fxsx, yd − 3x 2 1 2xy 3 fysx, yd − 3x 2 y 2 2 4y
Therefore
fxx − − s3x 2 1 2xy3d − 6x 1 2y 3 fx y − − s3x 2 1 2xy 3 d − 6xy2
−x −y
fyx − − s3x 2 y 2 2 4yd − 6xy2 fyy − − s3x 2 y 2 2 4yd − 6x2y 2 4 ■
−x −y
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