11DIFFERENTIATION - Department of Mathematics

774 11 DIFFERENTIATION
EXAMPLE 4
SOLUTION ✔
EXAMPLE 5
b. Here, x 2 and dx 2.01 2 0.01. Therefore,
dy 3x 2 dx 3(2) 2 (0.01) 0.12
c. Here, x 2 and dx 1.98 2 0.02. Therefore,
dy 3x 2 dx 3(2) 2 (0.02) 0.24
d. As you can see, both approximations 0.12 and 0.24 are quite close to the
actual changes of y obtained in Example 2: 0.120601 and 0.237608.
Observe how much easier it is to find an approximation to the exact
change in a function with the help of the differential, rather than calculating
the exact change in the function itself. In the following examples, we take
advantage of this fact.
Approximate the value of 26.5 using differentials. Verify your result using
the key on your calculator.
Since we want to compute the square root of a number, let’s consider the
function y f(x) x. Since 25 is the number nearest 26.5 whose square
root is readily recognized, let’s take x 25. We want to know the change in
y, y, asx changes from x 25 to x 26.5, an increase of x 1.5 units.
Using Equation (11), we find
Therefore,
y dy f(x)x
1
2x x25 (1.5) 1
10(1.5) 0.15
26.5 25 y 0.15
26.5 25 0.15 5.15
The exact value of 26.5, rounded off to five decimal places, is 5.14782. Thus,
the error incurred in the approximation is 0.00218.
APPLICATIONS
The total cost incurred in operating a certain type of truck on a 500-mile trip,
traveling at an average speed of v mph, is estimated to be
C(v) 125 v 4500
v
dollars. Find the approximate change in the total operating cost when the
average speed is increased from 55 mph to 58 mph.