The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 40 mm?

Solution:

Using r to represent the radius and t for time, you can write the first rate as:

dr/ dt = 4mm

r = r(t) = 4t

The formula for a solid sphere's volume is:

V = V(r) = 4/3πr³

When you take the derivative of both sides with respect to time...

dVdt = 4/3π(3r²)(drdt)... Remember the Chain Rule for implicit differentiation.

The general format for this is:

dV(r)dt = dV(r)dr(t)dr(t)dt with V = V(r) and r = r(t)

So, when you take the derivative of the volume, it is with respect to its variable r (dV(r)dr(t)), but we want to do it with respect to t(dV(r)dt)

Since r = r(t)and r(t) is implicitly a function of t, to make the equality work, you have to multiply by the derivative of the function r(t) with respect to t(dr(t)dt)as well. That way, you're taking a derivative along a chain of functions, so to speak (V→r→t)

Now what you can do is simply plug in what r is (note you were given diameter) and what drdt is, because dVdt describes the rate of change of the volume over time of a sphere.

dVdt = 4/3π(3(20mm)²)(4mms)

= 6400πmm³s

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

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The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 40 mm?

Summary:

The radius of a sphere is increasing at a rate of 4 mm/s, the volume increasing when the diameter is 40 mm is 6400πmm³.