calculus math 202

Question

The area A of a disk of radius r is given by the formula A= πr². Assume that the area increases at a rate of 3ft²/sec. Find the rate with which the radius increases at the moment when the diameter d=14ft. (Enter your answer correct to at least five decimal places, you may use the value of π provided by your calculator, or alternatively use  π ≈ 3.1415926535 )

Answer 1

This is a Related Rates question, and involves implicit differentiation.

The first step in all questions of this topic is to take the derivative of the main equation, A = π*r², with respect to time. In other words, the equation is actually A(t) = π[r(t)]². The area and radius are now functions of time, and when taking the derivative we have to use the chain rule. I find that forgetting the chain rule is the most common error that students makes with related rates.

A(t) = π[r(t)]²

A'(t) = π*2*r(t)*r'(t)

Notice the derivative of [r(t)]² is 2*r(t)*r'(t), rather than simply 2*r(t). According to the chain rule you must also multiply by the derivative of the inner function.

It can be cumbersome to always write A(t) and r(t), so they are usually left off, but it is a useful way to understand where the r'(t) comes from. Normally it would be written simply as such:

A = π*r²

A' = π*2*r*r'

After taking the derivative, I like to write down our givens, and what we are asked to find. From the question, we are told A' = 3 ft²/sec, and are asked to find r' when d = 14 ft, which is the same as r = 7 ft. From here it's just a matter of plugging in and solving.

A' = π*2*r*r'

3 = π*2*7*r'

3 = 14π*r'

r' = 3/(14π)

We are asked to give a decimal answer, which comes out to r' = 0.06821. Put into words, the radius increases by 0.06821 ft/sec.