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

SectION 14.4 Tangent Planes and Linear Approximations 933
EXAMPLE 4
(a) If z − f sx, yd − x 2 1 3xy 2 y 2 , find the differential dz.
(b) If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values
of Dz and dz.
In Example 4, dz is close to Dz because
the tangent plane is a good approximation
to the surface z − x 2 1 3xy 2 y 2
near s2, 3, 13d. (See Figure 8.)
SOLUTION
(a) Definition 10 gives
dz − −z −z
dx 1 dy − s2x 1 3yd dx 1 s3x 2 2yd dy
−x −y
(b) Putting x − 2, dx − Dx − 0.05, y − 3, and dy − Dy − 20.04, we get
60
40
z 20
0
_20
5 4 3 2 1
x
0
4
2
y
0
The increment of z is
dz − f2s2d 1 3s3dg0.05 1 f3s2d 2 2s3dgs20.04d − 0.65
Dz − f s2.05, 2.96d 2 f s2, 3d
− fs2.05d 2 1 3s2.05ds2.96d 2 s2.96d 2 g 2 f2 2 1 3s2ds3d 2 3 2 g
− 0.6449
FIGURE 88
Notice that Dz < dz but dz is easier to compute.
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7et140408
05/03/10
MasterID: 01592
EXAMPLE 5 The base radius and height of a right circular cone are measured as
10 cm and 25 cm, respectively, with a possible error in measurement of as much as
0.1 cm in each. Use differentials to estimate the maximum error in the calculated
volume of the cone.
SOLUTION The volume V of a cone with base radius r and height h is V − r 2 hy3.
So the differential of V is
dV − −V
−r
dr 1
−V
−h
dh −
2rh
3
dr 1 r 2
Since each error is at most 0.1 cm, we have | Dr | < 0.1, | Dh |
3 dh
< 0.1. To estimate the
largest error in the volume we take the largest error in the measurement of r and of h.
Therefore we take dr − 0.1 and dh − 0.1 along with r − 10, h − 25. This gives
dV − 500
3
s0.1d 1 100
3
s0.1d − 20
Thus the maximum error in the calculated volume is about 20 cm 3 < 63 cm 3 .
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Functions of Three or More Variables
Linear approximations, differentiability, and differentials can be defined in a similar
manner for functions of more than two variables. A differentiable function is defined by
an expression similar to the one in Definition 7. For such functions the linear approximation
is
f sx, y, zd < f sa, b, cd 1 f x sa, b, cdsx 2 ad 1 f y sa, b, cdsy 2 bd 1 f z sa, b, cdsz 2 cd
and the linearization Lsx, y, zd is the right side of this expression.
If w − f sx, y, zd, then the increment of w is
Dw − f sx 1 Dx, y 1 Dy, z 1 Dzd 2 f sx, y, zd
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