Thearea of a floating circular oil spill in the ocean is increasing at a rate of  96pi m^2/sec Whenthe area=1000pi how fast is radius increasing m/sec

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The area of a circular oil spill is increasing at the rate of 96*pi m^2/s.

The area of a circle with radius r is equal to A = pi*r^2. The rate of change of the area with respect to the radius is given by `(dA)/(dt) = (2*pi*r)*(dr)/(dt) `

=> `(dr)/dt = (dA)/(dt)/(2*pi*r)`

When the area = `1000*pi` the rate at which the radius is increasing has to be determined. When A = 1000*pi, `r = sqrt(1000*pi/pi)` = `sqrt 1000`

`(dr)/dt = (96*pi)/(2*pi*sqrt1000)`

=> `48/sqrt 1000`

The radius is increasing at `48/sqrt 1000` m/s

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