Subject: Related Rates
Name: Karli
Who are you: Student

The question is:
Find the rate of change of the perimeter of a square with respect to its area.

Hi Karli.

When you send a question, it helps if you can tell us what you've tried and where you are stuck.

To solve this problem you are dealing with perimeter (P) and area (A) and time (t) is implied.

So the change in perimeter with respect to time is dP/dt and the change in area with respect to time is dA/dt.

How does the perimeter relate to the area? If R is the radius, the perimeter is 2 π R and the area is π R², so

P = 2 π R
A = π R²

re-arranging the second equation to solve for R we get:

R = (A/π)^1/2

and putting this into the first equation, we get:

P = 2 π (A/π)^1/2

which reduces to

P = 2 π^1/2 A^1/2

Now if you take the derivative of both sides with respect to time (and apply the chain rule, because A is not constant), then you get dP/dt on one side (that's what you are asked for) and an expression involving dA/dt on the other side (which is the other rate you are relating it to).

Hope this helps,
Stephen La Rocque.