The radius of a circle is increasing at the rate of 2 feet per minute.?

Question

What is the rate at which the area is increasing when the radius is 7 feet?

Answer 1

A = (pi) r^2 or 0.78539 d^2

dA/dt = d/dt (0.78539 d^2) = (pi) d/dt 0.78539 d^2 = 2(pi)r dr/dt

25.13248 sq feet per min

Answer 2

Related rates question:
Express the area of the circle as a function of the radius:
A = pi * r^2
Derivative of area (speed at which it is changing) = pi * 2r * r', where r' equals the speed at which r is changing.
So when r = 7, A' = pi * 2 * 7 * 2 = 28 pi feet^2 per minute

Answer 3

This is a related rate problem. You need to get the area of a circle as a function of the radius, then take the time rate of change of both sides. Specifically:

A = (pi) r^2

dA/dt = d/dt ((pi) r^2) = (pi) d/dt r^2 = 2(pi)r dr/dt

You have r and dr/dt, so plug them in and you're good to go.

Answer 4

dr/dt = 2 [feet/minute]
r = 7 [feet]
dA/dt = ? [feet^2 /minute]



Area = pi * r^2 [feet^2 / minute]

dA/dt = 2 * pi * r [feet] *(dr/dt) [feet/minute]

dA/dt = 2 * pi * 7 [feet] * 2 [feet/minute]
=88 [feet^2 / minute]

Answer 5

87.92 squarefeet/minute