James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

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 C 1 and part (b) shows C 1 and T 1 . [We have used
the vector equations rstd − kt, 1, 2 2 t 2 l for C 1 and rstd − k1 1 t, 1, 1 2 2t l for T 1 .]
Similarly, Figure 5 corresponds to Figure 3.
4
4
3
3
z 2
z 2
1
1
0
0
y
1
2
1 x
0
0
0
y
1
2
1 x
0
FIGURE 4
(a)
(b)
4
4
3
3
z 2
z 2
1
1
FIGURE 5
0
0
y
1
2
1 x
0
0
0
y
1
2
1 x
0
EXAMPLE 3 In Exercise 14.1.39 we defined the body mass index of a person as
7et140304-05
Bsm, hd − m h
04/29/10
2
MasterID: 01572-73 Calculate 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
m is
−B
sm, hd −
−
−m
−m
S m h 2D − 1 h 2
so
−B
1
s64, 1.68d −
−m s1.68d < 0.35 2 skgym2 dykg
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.
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