# help with rate of change qn please?

A circular cylinder is expanding in such a way that at a time t seconds, the length of the cylinder is 20x cm and the area of the cross section is x cm^2. Given that when x=5, the area of the cross section is increasing at a rate of 0.025 cm^2 s^-1, find the length of the cylinder.

### 1 Answer

- MaryLv 78 years agoFavorite Answer
As we know that x is expanding with respect to time, x can be given by the equation,

x = at + b

The derivative of x with respect to t is

x' = a = 0.025 cm^2/s

Since the value of t is not given when x=5, b can take on any value. Let's say that when t=0, x=0 then b=0. We have

x = 0.025t

When x=5

5 = 0.025t

t = 5/0.025 = 200

However, this only tells us that area of the cross section of the cylinder is 5 at 200 seconds. This fact has nothing to do with what we want, namely the length of the cylinder, nor other values like the volume. This is a trick question!

Since x=5 at whatever the time is, the length is 20x, so,

length of the cylinder

= 20*5 = 100 cm