If a snowball melts so that its surface area decreases at a rate of 7 cm2/min, find the rate at which the diameter decreases when the diameter is 8 cm. cm/min

7/16π

Step-by-step explanation:

The surface area of the snow ball = 4πr^2

Let the surface area= y

Then y= 4πr^2

The derivative can be taken with respect to time as

dy/dt= (8πr)dr/dteqn(1)

We were told that its surface area decreases at a rate of 7 cm2/min.

Then dy/dt=-7

diameter = 8 cm. cm/min, the radius r= 4cm/min

r= 4

Substitute the values of r into eqn(1)

dy/dt=(8π×4)dr/dt

-7=(32π) dr/dr

dr/dt= -7/32πeqn(**)

But we know that diameter d= 2× radius

d=2reqn(*)

We can find the derivatives with respect to t as

dd/dt=2dr/dt

Then substitute en(**) we have

dd/dt=2(7/32π)

dd/dt= -14/32π

Divide both numerator and denominator by 2

dd/dt= -7/16π

Hence, the diameter decreases by the rate of 7/16π

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step-by-step explanation:

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