# A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.1 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 10 cm. (Note the answer is a positive number).

We are given the rate of decrease of the diameter `(dd)/(dt)=.1"cm"/"min" ` . We are asked to find the rate of the decrease of the volume when the diameter is 10cm:

The formula for the volume of a sphere is `V=4/3pir^3 ` .

The derivative is ` (dV)/(dt)=4pir^2(dr)/(dt) `

If...

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We are given the rate of decrease of the diameter `(dd)/(dt)=.1"cm"/"min" ` . We are asked to find the rate of the decrease of the volume when the diameter is 10cm:

The formula for the volume of a sphere is `V=4/3pir^3 ` .

The derivative is ` (dV)/(dt)=4pir^2(dr)/(dt) `

If the diameter is 10cm, the radius is 5cm. Also, if `(dd)/(dt)=.1 ` then `(dr)/(dt)=.05 `

Then substituting the given values:

`(dV)/(dt)=4pi(5)^2(.05)~~15.71 `

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We could also use `V=4/3pir^3=4/3pi(d/2)^3=(pid^3)/6 `

Then `(dV)/(dt)=(pid^2)/2(dd)/(dt) `

And `(dV)/(dt)=(100pi)/2*.1~~15.71 ` as before.

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The volume of a sphere is linked to its radius by the formula:

`V=4/3pir^3`

where V is the volume of the sphere and r is the radius.

The rate of change of one value with respect to another is found by the process of differential calculus. The derivative of a single order equation:

`y=x^a`

can be expressed as:

`dy/dx=ax^(a-1)`

thus the derivative of V with respect to r is:

`(dV)/(dr)=(4/3pi)3r^2=4pir^2`

thus when the radius is 10cm, the rate of change of the volume of the sphere with respect to the radius is `4*pi*10^2 (cm^3)/(cm)`

As the diameter is decreasing at `0.1 (cm)/(min)` , the radius is decreasing at `0.05(cm)/(min)`. Thus the rate of change of the volume of the sphere is:

`(4*pi*10^2 (cm^3)/(cm))xx(0.05(cm)/(min))=4*pi*100*0.05(cm^3/min)~~62.832(cm)^3/(min)`