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?

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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?

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Answer 1 · 2015-06-22T00:03:43

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

$\frac{d r}{d t} = 4 \frac{mm}{s}$ $(dr)/(dt) = 4 "mm"/"s"$

or

$r = r (t) = 4 t$ $r = r(t) = 4t$

The formula for a solid sphere's volume is:

$V = V (r) = \frac{4}{3} π r^{3}$ $V = V(r) = 4/3pir^3$

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

$\frac{d V}{d t} = \frac{4}{3} π (3 r^{2}) (\frac{d r}{d t})$ $(dV)/(dt) = 4/3pi(3r^2)((dr)/(dt))$

...remember the Chain Rule for implicit differentiation. The general format for this is:

$\frac{d V (r)}{d t} = \frac{d V (r)}{d r (t)} \cdot \frac{d r (t)}{d t}$ $(dV(r))/(dt) = (dV(r))/(dr(t))*(dr(t))/(dt)$

with $V = V (r)$ $V = V(r)$ and $r = r (t)$ $r = r(t)$ .

So, when you take the derivative of the volume, it is with respect to its variable $r$ $r$ $(\frac{d V (r)}{d r (t)})$ $((dV(r))/(dr(t)))$ , but we want to do it with respect to $t$ $t$ $(\frac{d V (r)}{d t})$ $((dV(r))/(dt))$ . Since $r = r (t)$ $r = r(t)$ and $r (t)$ $r(t)$ is implicitly a function of $t$ $t$ , to make the equality work, you have to multiply by the derivative of the function $r (t)$ $r(t)$ with respect to $t$ $t$ $(\frac{d r (t)}{d t})$ $((dr(t))/(dt))$ as well. That way, you're taking a derivative along a chain of functions, so to speak ( $V \to r \to t$ $V -> r -> t$ ).

Now what you can do is simply plug in what $r$ $r$ is (note you were given diameter) and what $\frac{d r}{d t}$ $(dr)/(dt)$ is, because $\frac{d V}{d t}$ $(dV)/(dt)$ describes the rate of change of the volume over time, of a sphere.

$\frac{d V}{d t} = \frac{4}{3} π (3 {(20 mm)}^{2}) (4 \frac{mm}{s})$ $(dV)/(dt) = 4/3pi(3(20 "mm")^2)(4 "mm"/"s")$

$= 6400 π \frac{{mm}^{3}}{s}$ $= 6400pi "mm"^3/"s"$

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a 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?

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