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- anonymous

The radius of a circular oil slick expands at a rate of 2 m/min.
(a) How fast is the area of the oil slick increasing when the radius is 25 m
(b) If the radius is 0 at time , how fast is the area increasing after 4 mins

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- anonymous

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- helpmeplease

dr/dt = 2m/min, Make a differential equation that relates the time to the function.

- anonymous

We know that dr/dt = 2 from the question. The area of the circular slick is A = (pi)r^2. We take the derivative:
\[dA/dt=2\pi rdr/dt\]
Part a gives the radius = 25 m, so we have enough info to find how fast the area is changing.
\[dA/dt = 2\pi (25)(2) = 100\pi=314.16\]
The area is increasing by 100pi square meters per minute, or 314.16 square meters per minute.
Part b - I'm not sure at what time the radius is 0, but I'd guess it's at t = 0? That means at t = 4, the radius will have increased to 8 m. Put this value in for r and use 2 again for dr/dt.
\[dA/dt = 2\pi (8)(2)=32\pi=100.53\]
So the area is increasing at 100.53 square meters per minute at t = 4.

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