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

Section 3.10 Linear Approximations and Differentials 255
Figure 6 shows the function in Example
3 and a comparison of dy and Dy
when a − 2. The viewing rectangle is
f1.8, 2.5g by f6, 18g.
y=˛+≈-2x+1
In general,
dy − f 9sxd dx − s3x 2 1 2x 2 2d dx
When x − 2 and dx − Dx − 0.05, this becomes
dy − f3s2d 2 1 2s2d 2 2g0.05 − 0.7
(b) f s2.01d − s2.01d 3 1 s2.01d 2 2 2s2.01d 1 1 − 9.140701
dy
Îy
Dy − f s2.01d 2 f s2d − 0.140701
(2, 9)
When dx − Dx − 0.01,
FIGURE 6
dy − f3s2d 2 1 2s2d 2 2g0.01 − 0.14
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Notice that the approximation Dy < dy becomes better as Dx becomes smaller in
Example 3. Notice also that dy was easier to compute than Dy. For more complicated
functions it may be impossible to compute Dy exactly. In such cases the approximation
by differentials is especially useful.
In the notation of differentials, the linear approximation (1) can be written as
f sa 1 dxd < f sad 1 dy
For instance, for the function f sxd − sx 1 3 in Example 1, we have
dy − f 9sxd dx −
dx
2sx 1 3
If a − 1 and dx − Dx − 0.05, then
dy − 0.05
2s1 1 3 − 0.0125
and s4.05 − f s1.05d < f s1d 1 dy − 2.0125
just as we found in Example 1.
Our final example illustrates the use of differentials in estimating the errors that occur
because of approximate measurements.
ExamplE 4 The radius of a sphere was measured and found to be 21 cm with a possible
error in measurement of at most 0.05 cm. What is the maximum error in using this
value of the radius to compute the volume of the sphere?
SOLUtion If the radius of the sphere is r, then its volume is V − 4 3 r 3 . If the error
in the measured value of r is denoted by dr − Dr, then the corresponding error in the
calculated value of V is DV, which can be approximated by the differential
When r − 21 and dr − 0.05, this becomes
dV − 4r 2 dr
dV − 4s21d 2 0.05 < 277
The maximum error in the calculated volume is about 277 cm 3 .
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