# At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?

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To find the rate at which the volume of a sphere is changing, given that the surface area is 4 square meters and the radius is increasing at a rate of ( \frac{1}{6} ) meters per minute, we can use related rates.

Given: Surface area, ( A = 4 ) square meters Radius, ( r = \frac{1}{6} ) meters per minute

We know that for a sphere:

[ A = 4\pi r^2 ]

Taking the derivative with respect to time, ( t ):

[ \frac{dA}{dt} = 8\pi r \frac{dr}{dt} ]

We are given ( \frac{dr}{dt} = \frac{1}{6} ) meters per minute, and we can solve for ( r ) using the formula for surface area:

[ 4 = 4\pi r^2 ] [ r = \sqrt{\frac{1}{\pi}} ]

Now, plug in the values into the rate formula:

[ \frac{dA}{dt} = 8\pi \left(\sqrt{\frac{1}{\pi}}\right) \left(\frac{1}{6}\right) ]

[ \frac{dA}{dt} = \frac{4}{3\sqrt{\pi}} ]

To find the rate at which the volume is changing, we use the formula for the volume of a sphere:

[ V = \frac{4}{3}\pi r^3 ]

[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} ]

[ \frac{dV}{dt} = 4\pi \left(\frac{1}{\pi}\right)^{\frac{3}{2}} \left(\frac{1}{6}\right) ]

[ \frac{dV}{dt} = \frac{1}{3\sqrt{\pi}} ]

So, at the given instant, the volume of the sphere is changing at approximately ( \frac{1}{3\sqrt{\pi}} ) cubic meters per minute.

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