# Calculus AB Related Rates A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1cm^3/min. At what rate is the diameter decreasing when the diameter is 10 cm? Volume of a sphere` (V) = 4/3*pi*(d^3/8)`

Where d is the diameter.

It is given that rate of decreasing volume is 1cm^3/min.

`(dV)/dt = 4/3*pi*3*d^2/8*(dd)/dt`

`(dV)/dt = 1/2*pi*r^2*(dd)/dt`

`(dV)/dt =` rate of volume decreasing with time

`(dr)/dt =` rate of diameter decreasing with time

When the diameter is...

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Volume of a sphere` (V) = 4/3*pi*(d^3/8)`

Where d is the diameter.

It is given that rate of decreasing volume is 1cm^3/min.

`(dV)/dt = 4/3*pi*3*d^2/8*(dd)/dt`

`(dV)/dt = 1/2*pi*r^2*(dd)/dt`

`(dV)/dt =` rate of volume decreasing with time

`(dr)/dt =` rate of diameter decreasing with time

When the diameter is 10cm

`(dV)/dt = 1/2*pi*r^2*(dd)/dt`

`1 = 1/2*pi*10^2*(dd)/dt`

`(dd)/dt = 0.0064`

So the rate of decreasing diameter is 0.0064cm/min or 6.4mm/min.

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