A cylindrical rubber cord is streched at a constant rate of 2 cm per second.?

Question

Assuming its volume does not change, how fast does is its radius shrinking when its length 100 cm and its radius is 1 cm?
A) 0 cm per sec
B) .01cm/sec
C) .02 cm/sec
D) 2 cm/sec
E) 3.979 cm/sec

Answer 1

First off, the volume of a cylinder is

V = pi(r^2)h

We are given:
(1) V is a constant
(2) dh/dt = 2
And we want to find dr/dt when h = 100 and r = 1.

Differentiating implicitly (keep in mind that the derivative of V is 0, since it is constant):

V = pi(r^2)h. Using the product rule and chain rule,

0 = pi [2r [dr/dt] * h + r^2 [dh/dt] ]

Plugging in our known rate dh/dt = 2,

0 = pi [2r [dr/dt] * h + r^2 (2) ]
0 = pi [2rh [dr/dt] + 2r^2]

When its length is 100 cm and its radius is 1 cm, h = 100 and r = 1. Now we plug in these values.

0 = pi [2(1)(100) [dr/dt] + 2[1]^2]
0 = pi [200 [dr/dt] + 2]
0 = 200pi [dr/dt] + 2pi
-2pi = 200pi [dr/dt]

Therefore,

dr/dt = -1 / 100

So the cord's radius is decreasing at a rate of 1/100 cm per second, or 0.01 cm/sec.

The answer is b.

Answer 2

This problem can be done in a much simpler way.

First, since the volume is a constant, we have

r^2 h = constant

Differentiate both sides wrt time,

2rr' h + r^2 h' = 0

Therefore,

-r' = r h'/(2h) = 1(2)/(200) = 0.01 cm/sec