11DIFFERENTIATION - Department of Mathematics

776 11 DIFFERENTIATION
SOLUTION ✔
EXAMPLE 8
EXAMPLE 9
SOLUTION ✔
outermost ring, 1989N1R, has an inner radius of approximately 62,900 kilometers
(measured from the center of the planet), and a radial width of approximately
50 kilometers. Using these data, estimate the area of the ring.
a. Using the fact that the area of a circle of radius x is A f(x) x 2 , we find
R 2 r 2 f(R) f(r)
A (Remember, A change in f when
dA
f(r)dr
x changes from x r to x R.)
where dr R r. So, we see that the area of the ring is approximately
2r(R r) square units. In words, the area of the ring is approximately equal to
Circumference of the inner circle Thickness of the ring
b. Applying the results of part (a) with r 62,900 and dr 50, we find that
the area of the ring is approximately 2(62,900)(50), or 19,760,618 square
kilometers, which is roughly 4% of Earth’s surface.
Before looking at the next example, we need to familiarize ourselves with
some terminology. If a quantity with exact value q is measured or calculated
with an error of q, then the quantity q/q is called the relative error in
the measurement or calculation of q. If the quantity q/q is expressed as a
percentage, it is then called the percentage error. Because q is approximated
by dq, we normally approximate the relative error q/q by dq/q.
Suppose the radius of a ball-bearing is measured to be 0.5 inch, with a maximum
error of 0.0002 inch. Then, the relative error in r is
dr 0.0002
r 0.5 0.0004
and the percentage error is 0.04%.
Suppose the side of a cube is measured with a maximum percentage error of
2%. Use differentials to estimate the maximum percentage error in the calculated
volume of the cube.
Suppose the side of the cube is x, so its volume is
We are given that dx
0.02. Now,
x
and so
V x 3
dV 3x 2 dx
dV
V 3x2dx x3 dx
3
x