Calculus question on rates

Question

The circumference of a circle is increasing at a rate of 1 ft/sec. At the instant the radius has a length of 19.5 feet, answer the following:

 

(A) At what rate, in ft/sec, is the radius changing?

(B) At what rate, in ft²/sec is the area changing?

Answer 1

It is easy to get "FREAKED OUT" by a question like this but really this is nothing more then doing a little calc then using algebra and substituting stuff.

We know circumference of a circle:

C = 2 π r ok so take the derivative and we get dC/dt = 2 π dr/dt

Question A asks "At what rate, in ft/sec, is the radius changing?" In mathematical language that is asking:

What is dr/dt?

Let's rearrange what we figured out before.

dC/dt = 2π dr/dt arranges to:

dr/dt = 1/2π * dC/dt but we are given dC/dt. Again we have to turn words into mathematical language. "The circumference of a circle is increasing at a rate of 1 ft/sec." is really saying:

dC/dt = 1 ft/sec

SO sub time dr/dt = 1/2π * 1 ft/sec YEAH we have an answer

Question A. At what rate, in ft/sec, is the radius changing? dr/dt = 1/2π ft/sec

NOW for B.

At what rate, in ft²/sec is the area changing? Again that is really asking: what is dA/dt. (A is area)

ok so let's start from what we know: A = π r ² let's take a derivative dA = 2πr dr/dt r is in ft

WELL we have all of that so we just need to plug a chug the stuff in for dr/dt (from question A).

dA = 2 π r(ft)* 1/2π ft/sec remember r is in ft

dA = 2π * 1/2π r ft/sec OR dA = r ft / sec

PLUG in info r = 19.5 ft

dA = 19.5 ft * ft/sec or 19.5 ft² / sec

the big hint here use the basic things you know how to find area, circumference; take derivatives; KEEP UNITS THERE. Think of units (ft, sec, whatever) as variables.

Answer 2

A) We have C = 2πr. We differentiate this equation: C' = 2πr'. From here rate of change of radius r' = C'/(2π) =

= 1/(2π)ft/sec ≈ 0.159 ft/sec

B) Area of sircle A = πr². Then rate of change of area is A' = 2πr·r' = 2πr·(1/2π) = 19.5 ft²/sec