Oil spilling from a ruptured tanker spreads in a circle on the surface of the ocean. The area of the spill increases at a rate of 9π m²/min. How fast is the radius of the spill increasing when the radius is 10 m?

Question

Evelyn Agard · Accepted Answer

#dr|_(r=10)=0.45m//min#.

Since area of a circle is #A=pi r^2#, we may take the differential on each side to obtain : #dA=2pirdr# Hence the radius changes at the rate #dr=(dA)/(2pir)=(9pi)/(2pir)# Thus, #dr|_(r=10)=9/(2xx10)=0.45m//min#.

Christopher Agresta · Answer

undefined To find the rate at which the radius of the spill is increasing when the radius is 10 meters, use the formula for the area of a circle, A = πr^2, and differentiate it with respect to time.

Given that the area of the spill increases at a rate of 9π m²/min, substitute this into the derivative of the area formula:

dA/dt = 9π

Now, differentiate the area formula with respect to time:

dA/dt = 2πr(dr/dt)

Substitute the given value of dA/dt and the radius (r = 10m) into the equation:

9π = 2π(10)(dr/dt)

Solve for dr/dt:

9π = 20π(dr/dt)

dr/dt = 9π / 20π

dr/dt = 9 / 20

So, when the radius is 10 meters, the radius of the spill is increasing at a rate of 9/20 meters per minute.