Question

Let a(t)a(t) be the area of a circle with radius r(t)r(t), at time tt in min. suppose the radius is changing at the rate of drdt=5drdt=5 ft/min. find the rate of change of the area at the moment in time when r=6r=6 ft

Answer 1

The relationship between the rates of change can be found by taking the derivative of area with respect to time.

... a(t) = π·r(t)²

... a'(t) = 2π·r(t)·r'(t)

At the time of interest, we have

... a'(t) = 2π·(6 ft)·(5 ft/min) = 60π ft²/min

The rate of change of area at that time is 60π ≈ 188.5 ft²/min.

Answer 2

We have the area of the circle $a(t)=\pi r(t)^{2}$. Differentiating with respect to time t,

$\frac{d}{dt} a(t)=\frac{d}{dt} \pir(t)^2=2\pi r(t)\frac{dr(t)}{dt}$

Give that $\frac{dr(t)}{dt}=5, r(t)=6$.

Substituting the numerical values, the rate of change of area at $r(t)=6$ is

$\frac{d}{dt} a(t)=2\pi (6)(5)=60 ft/min \pi$