Question

(1 point) A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at a rate of 4 feet per second, how fast is the circumference changing when the radius is 20 feet?

I think the answer is `8pi`, but I don't understand why I lose the variable r when differentiating.

I get.

`(dC)/(dt)=(dC)/(dr)*(dr)/(dt)`

`(dC)/(dt)=(dC)/(dr)(2pir)*(dr)/(dt)(4)`

`(dC)/(dt)=2pi*4=8pi`

So I am not making us of radius = 20 feet. I don't know how I incorporate this into the problem.

Thanks.

Answer 1

`8pi`, as suggested in the question

Explanation:

You are correct:

Using the standard formula for the circumference of a circle, we have:

`C = 2pir`

Differentiating wrt `r`, we get:

`(dC)/(dr) = 2pi`

Using the chain rule we have:

`(dC)/(dt) = (dC)/(dr) (dr)/(dt)`

Giving us:

`(dC)/(dt) = 2pi (dr)/(dt)`

Knowing (from the question) that `(dr)/(dt) = 4 \ ft \ s^(-1)`, then we get, as suggested:

`(dC)/(dt) = 2pi * 4 = 8pi`

The reason that the solution is independent of `r` is because there is a linear relationship between the perimeter of a circle and the radius, so that the derivative is constant If however we consider how fast the Area were increasing then this would be dependant upon (ie a function of) `r` as the area is a quadratic function of `r`.