Answer to Question #276235 in Calculus for Emmy

Question

The area A of a circle is increasing at a constant rate of 2cm2s−1. If the area of the circle is given by A = πr2, what is the rate of change of the radius when the radius is 4cm?

Answer 1

$A=\pi r^2$

Differentiate both sides with respect to $t$

$\dfrac{dA}{dt}=\dfrac{d}{dt}(\pi r^2)$

$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$

Solve for $\dfrac{dr}{dt}$

$\dfrac{dr}{dt}=\dfrac{\dfrac{dA}{dt}}{2\pi r}$

Given $\dfrac{dA}{dt}=2 {cm}^2/s, r=4cm$

$\dfrac{dr}{dt}|_{r=4}=\dfrac{2{cm}^2/s}{2\pi (4cm)}=\dfrac{1}{8\pi}cm/s\approx0.040cm/s$

The rate of change of the radius is $\dfrac{1}{8\pi}cm/s\approx0.040cm/s$ when the radius is 4cm.