Question

Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 mi²/h. How fast is the radius of the spill increasing when the area is 9 mi²?

a. 2 mi/h
b. 3mi/h
c. 4 mi/h
d. 6mi/h

Answer 1

The radius of the oil spill is increasing at a rate of 3/π mi/h when the area is 9 mi².

Step-by-step explanation:

To find the rate at which the radius of the spilled oil is increasing, we can use the formula for the area of a circle, A = πr2. We are given that the area is increasing at a constant rate of 6 mi²/h. So, ΔA/Δt = 6 mi²/h, where ΔA is the change in area and Δt is the change in time. We can differentiate both sides of the equation with respect to time to find the rate at which the area is changing with respect to time: dA/dt = 6 mi²/h. Since the area is given by A = πr2, we can differentiate both sides with respect to time to find the rate at which the radius is changing with respect to time: dA/dt = 2πr(dr/dt). Plugging in the values, we have 6 = 2πr(dr/dt). We are trying to find the rate at which the radius is changing when the area is 9 mi². So, plugging in A = 9 and solving for dr/dt, we get dr/dt = (6/2πr)(1/r) = 3/π mi/h. Therefore, the radius of the spill is increasing at a rate of 3/π mi/h when the area is 9 mi².