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I'm doing related rates calculus problems. For example the volume of a sphere:

$V (t) = \frac {4}{3} \pi r(t)^3 $

I know the derivative is:

$V' (t) = 4 \pi r(t)^2 r'(t)$.

But isn't volume a function of its radius?

$V (r (t)) = \frac {4}{3} \pi r(t)^3 $

So why isn't the derivative:

$V' (r (t)) r'(t) = 4 \pi r(t)^2 r'(t)$.

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    They are same.. – 2017-02-03
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    The last equation has r'(t) on both sides, and thus dividing both sides by r' will cancel of r' entirely. – 2017-02-03
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    no problem. just canceled. – 2017-02-03

2 Answers 2

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I know you've already accepted an answer, but I want to go ahead and give an alternative one here:

The problem here is not the chain rule; the problem is having a single letter denote a single thing.

Some folks might use $V(r)$ to denote the volume of a sphere of radius $r$, but $V(t)$ to denote the volume of a sphere whose radius is changing with time. That's (i) a really bad idea when you're starting out, and (ii) really convenient once you get used to it. But in that preliminary phase, it's the worst possible idea.

Let me be clear. Suppose you're using $V$ in both those ways. What does $V(1.5)$ denote? Is it the volume when the radius is $1.5$? Or is the volume at time $1.5$, when the radius might be, say, 14? You cannot tell when you use this kind of notation!

So to resolve this, I like to write the following sort of thing whenever I start a problem:

\begin{align} V(r) &= \text{the volume of a sphere of radius $r = \frac{4}{3} \pi r^3$}\\ h(t) &= 11 \sin (t) = \text{the radius of a pulsing sphere at time $t$} \\ W(t) &= V(h(t)) = \text{The volume of the pulsing sphere at time $t$} \end{align} where I've made up the function $h$ just to have something to demonstrate with.

Having written out those three things, it's clear what $W(1.5)$ means: it's the volume of the sphere at time $1.5$; since the radius then is $11 \sin(1.5)$, this volume is $$ W(1.5) = \frac{4}{3} \pi \left(11 \sin(1.5) \right)^3. $$ And now it's also clear how to apply the chain rule: $$ W'(t) = V'(h(t)) \cdot h'(t), $$ so that \begin{align} W'(t) &= 4 \pi (h(t))^2 \cdot (11 \cos(t))\\ &= 4 \pi (11 \sin(t))^2 \cdot 11 \cos(t). \end{align}

The "physics" way of writing this (to slightly malign physicists) is to say that

$$ V(t) = V(r(t)) $$ so that $$ \frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} $$ and then possibly amplifying by saying $$ \frac{dV}{dt}(t) = \frac{dV}{dr}(r(t)) \cdot \frac{dr}{dt}(t) $$ so that (in the example above) \begin{align} \frac{dV}{dt}(t) &= \left.4\pi r^2 \right|_{r = r(t)} \cdot 11 \cos (t)\\ &= 4\pi (11 \sin(t))^2\cdot 11 \cos (t). \end{align} with the only problem being that $V$ means two different things, $r$ is used once as a variable and once as a function (in the same term!), and only with practice can you make any sense of it.

As they used to say on the old world maps: here be dragons.

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    Thank you. This is a very rigorous explanation. I'm going to spend some time analyzing your example. I really appreciate you taking the time to write it up for me. – 2017-02-04
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From the chain rule, when you differentiate $\frac{d}{dt}[V(r(t))]$ you get $\frac{dV}{dr}\frac{dr}{dt} = \frac{dV}{dt}$. If you notice your right hand side, you have precisely this expression, where $\frac{dV}{dr} = 4\pi r^2$ and $\frac{dr}{dt} = r'(t)$

The ambiguity of $V'(r(t))$ appears to be causing the confusion, as you may be interpreting it as $\frac{dV}{dt}$.

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    I understand what you are saying about my "right hand side". And I think it is the "ambiguity" that is confusing me. I thought this is an implicit derivative, and both sides get their derivative taken with respect to t. So how is V'(r (t)) different from dV/dt? – 2017-02-03
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    Implicit differentiation is an application for the chain rule. If you have $V(r(t))$ and you wish to find $\frac{dV}{dt}$ you get $\frac{dV}{dr}\frac{dr}{dt}$. This is not so obvious with the common chain rule problems like $\frac{d}{dx} [(\sin(x))^2]$. When you take the derivative of the outside function ($(f)^2$), you are deriving with respect to the $\sin(x)$ which is why you get $2\sin(x)*(4x)$. In this sense, $V'(r(t))$ is not $\frac{dV}{dt}$ but instead $\frac{dV}{dr}$. The prime notation misses this. – 2017-02-03
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    A kmeis's comment is exactly why you should use one letter to denote one thing, in which case the prime notation captures things perfectly. – 2017-02-05