Suppose a pebble has been dropped into a pond forming a circular ripple. If the rate of change of the circumference of the circle is 4π in/s, what is the rate of change

Question

Suppose a pebble has been dropped into a pond forming a circular ripple. If the rate of change of the

circumference of the circle is 4π in/s, what is the rate of change of the area of the circle when the radius is

10 inches?

Answer 1

To find the rate of change of the area start with the formula for area of a circle

A = pi(r^2) now take the derivative of both sides which gives you da/dt = 2(pi)(r)(dr/dt)

Now we need to use the info about change in circumference to find dr/dt

C = 2(pi)r. Take derivative of both sides

dc/dt = 2(pi)dr/dt. We are told dc/dt is 4pi so

4pi= 2pi(dr/dt) so dr/dt = 2

Going back to our first equation da/dt= 2(pi)(r)(dr/dt) we have solved for dr/dt and r=10 so

da/dt= 2(pi)(10)(2)= 40pi inches/ sec