Related Rates problem The radius of a sphere is increasing at a rate of 4mm/s.  How fast is the volume increasing when the diameter is 80 mm?

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The radius of the sphere increases over time at a rate:

vr = dr/dt = 4 mm/s

The radius of the sphere is:

r = D/2 = 40 mm

To find the rate of variation of volume of the sphere, we derive the expression of volume with respect to time:

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The radius of the sphere increases over time at a rate:

vr = dr/dt = 4 mm/s

The radius of the sphere is:

r = D/2 = 40 mm

To find the rate of variation of volume of the sphere, we derive the expression of volume with respect to time:

The volume of a sphere is:

V = 4/3 Π r3

Differentiating with respect to time, we have:

dV/dt = 4/3 Π(3 r2)(dr/dt)

Now we substitute the given values

 dr/dt = vr = 4 mm/s        and      r = 40 mm

dV/dt = 4Π(40)^2(4) = 80,425 mm3/s

So, in this case, the volume of the sphere when the radius is 4 mm, increasing at a rate of 80,425 mm3/s

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The volume of a sphere is given by `V=4/3 pi r^3 `

where r is the radius.

To find the rate of change of the volume, differentiate with respect to t:

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

Substitute the known values: r=40 (the diameter is 80 so the radius is 40), `(dr)/(dt)=4`

so we get:

`(dV)/(dt)=(4pi)(40)^2(4)=25600 pi`

~~80424.77

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The rate of change of the volume at the given point is 25600 pi cubic mm per second, or approximately 80424.77 cubic mm per second.

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Approved by eNotes Editorial Team