## Related Rates

### Calculus Related Rates Problem:

As a snowball melts, how fast is its radius changing?

A spherical snowball melts at the rate of $2 \pi$ cm$^3$/hr. It melts symmetrically such that it is always a sphere. How fast is its radius changing at the instant $r = 10$ cm?

*Hint*: The volume of a sphere is related to its radius according to $V = \dfrac{4}{3} \pi r^3$.

### Calculus Solution

Let’s unpack the question statement:

- We’re told that the snowball’s volume
*V*is changing at the rate of $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr. (We must insert the negative sign “by hand” since we are told that the snowball is melting, and hence its volume is*decreasing*.) - As a result, its radius is changing, at the rate $\dfrac{dr}{dt}$, which is the quantity we’re after.
- The snowball always remains a sphere.
- Toward the end of our solution, we’ll need to remember that the problem is asking us about $\dfrac{dr}{dt}$ at a particular instant, when $r = 10$ cm.

**1. Draw a picture of the physical situation.**

See the figure.

**2. Write an equation that relates the quantities of interest.**

*B. To develop your equation, you will probably use. . . a simple geometric fact. *

This is the hardest part of Related Rates problem for most students initially: you have to know how to develop the equation you need, how to pull that “out of thin air.” By working through these problems you’ll develop this skill. The key is to recognize which of the few sub-types of problem it is; we’ve listed each on our Related Rates page. In this problem, the diagram above reminds us that the snowball always remains a sphere, which is a Big Clue.

We need to develop a relationship between the rate we’re given, $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr, and the rate we’re after, $\dfrac{dr}{dt}$. We thus first need to write down a relationship between the sphere’s volume *V* and its radius *r*. But we know that relationship since it’s a simple geometric fact that you could look up, and here it was given in the hint:

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

That’s it — that’s the key relationship we need to be able to proceed with our solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*}

\frac{d}{dt}V & = \frac{d}{dt} \left( \frac{4}{3}\pi r^3\right) \\[12px]
\frac{dV}{dt} &= \frac{4}{3}\pi \frac{d}{dt} \left( r^3\right) \\[12px]
&= \frac{4}{3}\pi \left( 3r^2 \frac{dr}{dt} \right) \\[12px]
&= 4 \pi\, r^2 \frac{dr}{dt}

\end{align*}

Are you wondering why that $\dfrac{dr}{dt}$ appears? The answer is the Chain Rule.

While the derivative of $r^3$ with respect to *r* is $\dfrac{d}{dr}r^3 = 3r^2$, the derivative of $r^3$ with respect to *time t* is $\dfrac{d}{dt}r^3 = 3r^2\dfrac{dr}{dt}$.

Remember that *r* is a function of time *t*: the radius *changes* as time passes and the snowball melts. We could have captured this time-dependence explicitly by writing our relation as

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

to remind ourselves that both *V* and *r* are functions of time *t*. Then when we take the derivative,

\begin{align*}

\frac{d}{dt}V(t) &= \frac{d}{dt}\left[ \frac{4}{3}\pi [r(t)]^3\right] \\ \\

\frac{dV(t)}{dt} &= \frac{4}{3} \pi\, 3[r(t)]^2 \left[\frac{d}{dt}r(t)\right]\\ \\

&= 4\pi [r(t)]^2 \left[\frac{dr(t)}{dt}\right]

\end{align*}

[Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr in this problem, and we’re looking for $\dfrac{dr}{dt}$.]

Most people find that writing the explicit time-dependence *V(t)* and *r(t)* annoying, and so just write *V* and *r* instead. Regardless, you *must* remember that *r* depends on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dr}{dt}$ term.

**4. Solve for the quantity you’re after.**

Solving the equation above for $\dfrac{dr}{dt}$:

\begin{align*}

\frac{dV}{dt} &= 4 \pi r^2 \frac{dr}{dt} \\[12px]
\frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px]
\end{align*}

Now we just have to substitute values. Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr,

and the problem asks about when $r=10$ cm:

\begin{align*}

\frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px]
&= \frac{1}{4 \pi (10)^2} (-2 \pi) \\[12px]
&= -\frac{1}{200} \text{ cm/hr} \quad \cmark

\end{align*}

That’s the answer. The negative value indicates that the radius is decreasing as the snowball melts, as we expect.

**Caution**: IF you are using a web-based homework system and the question asks,

At what rate does the radius *decrease*?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{\dfrac{1}{200}} \, \dfrac{\text{cm}}{\text{hr}} \quad \checkmark$

Return to Related Rates Problems

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tbh…I still don’t really get the concept

Thanks for letting us know. Our overall discussion of the topic might be more helpful: 4 Steps to Solve Any Related Rates Problem. We hope that will be more help you understand the general idea more clearly!