The circumference of the circle is increasing at a rate of 0.5 meters per minute. What's the rate of change of the area of the circle when the radius is 4 meters?

1: 3 meters per minute

2: 4 meters squared per minute

3: 4 meters per minute

4: 2 meters squared per minute

5: 7 meters per minute

This is actually for a game but I'm really bad at math.

The rate of change of the area of the circle when the radius is 4 meters = 2 meters²/minute ⇒ answer 4

Step-by-step explanation:

* Lets revise the chain rule in the derivative

- If dy/da = m and dx/da = n, and you want to find dy/dx

∴ dy/dx = dy/da : dx/da = m : n = m/n

* In our problem we have

- The rate of increasing of the circumference dC/dt = 0.5 meters/minute

- We need the find the rate of change of the area of the circle

when the radius is 4 meters

- The common element between the circumference and the area

of the circle is the radius of the circle

* We must to find dC/dr and dA/dr and use the chain rule to

find dA/dr

- Find the rate of change of the radius dr/dt

∵ C = 2πr

- Find the derivative of C with respect to r

∴ dC/dr = 2π ⇒ (1)

∵ dC/dt = 0.5 meters/minute ⇒ (2)

- Divide (1) by (2) to get dr/dt by using chain rule

∵ dC/dt : dC/dr = 0.5 : 2π

∴ dC/dt * dr/dC = 0.5 * 1/2π ⇒ cancel dC together and change

0.5 to 1/2

∴ dr/dt = 1/2 * 1/2π = 1/4π ⇒ (3)

- Find the rate of change of the area dA/dt

∵ A = πr²

- Find the derivative of A with respect to r

∴ dA/dr = 2πr

∵ r = 4

∴ dA/dr = 2π (4) = 8π ⇒ (4)

- Multiply (4) by (3) to get dA/dt by using chain rule

∵ dA/dr * dr/dt = 8π * 1/4π ⇒ divide 8 by 4 and cancel π

∴ dA/dt = 2 meters²/minute

* The rate of change of the area of the circle when the radius is

4 meters = 2 meters²/minute