## Related Rates

### Calculus Related Rates Problem:

How fast is the water level falling as water drains from the cone?

An inverted cone is 20 cm tall, has an opening radius of 8 cm, and was initially full of water. The water now drains from the cone at the constant rate of 15 cm$^3$ each second. The water’s surface level falls as a result. At what rate is the water level falling when the water is halfway down the cone?

(*Note*: The volume of a cone is $\dfrac{1}{3}\pi r^{2}h$. You may leave $\pi$ in your answer; do not use a calculator to find a decimal answer.)

### Calculus Solution

Let’s unpack the question statement:

- We’re told that volume of water in the cone
*V*is changing at the rate of $\dfrac{dV}{dt} = -15$ cm$^3$/s. (We must insert the negative sign “by hand” since we are told that the water is draining out, and so its volume is*decreasing*.) - As a result, the water’s height in the cone
*h*is changing at the rate $\dfrac{dh}{dt}$, which is the quantity we’re after. - The inverted cone has a radius of 8 cm at its top, and a full height of 20 cm.
- The problem is asking us about $\dfrac{dh}{dt}$ at a particular instant, when the water is halfway down the cone, and so when $h = 10$ cm. We’ll use this value toward the end of our solution.

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

See the figure.

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

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The height of the water changes as time passes, so we’re going to keep that height as a variable, *h*.

*B. To develop your equation, you will probably use . . . similar triangles.*

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.

Here we need to develop a relationship between the rate we’re given, $\dfrac{dV}{dt} = -15$ cm$^3$/s, and the rate we’re after, $\dfrac{dh}{dt}$. We thus first need to write down a relationship between the water’s volume *V* and its height-in-the-cone *h*. But we know that relationship since it was given in the problem statement:

$$V = \frac{1}{3} \pi r^2h $$

Notice that this relation expresses the water’s volume as the function of *two* variables, *r* and *h*. We can only take the derivative with respect to one variable, so we need to eliminate one of those two. Since the question asks us to find the rate at which the water is falling when its at a particular height, let’s keep *h* and eliminate *r* as a variable using similar triangles.

*Begin subproblem to eliminate r as a variable.*

The figure is the same as in Step 1, but with the rest of the cone removed for clarity. Note that there are two triangles, a small one inside a larger one. Because these are similar triangles, the ratio of the base of the small triangle to that of the big triangle $\left(\dfrac{r}{8} \right)$ must equal the ratio of the height of the small triangle to that of the big triangle $\left(\dfrac{h}{20} \right)$:

$$\frac{r}{8} = \frac{h}{20} $$

Hence

\begin{align*}

r &= \frac{8}{20} h \\[8px]
&= \frac{2}{5} h

\end{align*}

*End subproblem.*

Now let’s substitute the expression we just found for *r* into our relation for *V*:

\begin{align*}

V &= \frac{1}{3} \pi r^2h \\ \\

&= \frac{1}{3} \pi \left(\frac{2}{5} h \right)^2h \\ \\

&= \frac{1}{3}\pi \frac{4}{25} h^3 \\ \\

&= \frac{4}{75} \pi h^3

\end{align*}

That’s it. That’s the key relation we need to be able to proceed with the rest of the 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}{75} \pi h^3 \right) \\ \\

\frac{dV}{dt} &= \frac{4}{75} \pi \frac{d}{dt}\left(h^3 \right) \\ \\

&= \frac{4}{75} \pi \left(3h^2 \frac{dh}{dt} \right) \\ \\

&= \frac{4}{25} \pi h^2 \frac{dh}{dt}

\end{align*}

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

Remember that *h* is a function of time *t*: the water’s height decreases as time passes. We could have captured this time-dependence explicitly by always writing the water’s height as *h(t)*, and then explicitly showing the water’s volume in the cone as a function of time as

$$V(t) = \frac{4}{75} \pi [h(t)]^3$$

Then when we take the derivative,

\begin{align*}

\frac{dV(t)}{dt} &= \frac{4}{75} \pi \frac{d}{dt}[h(t)]^3 \\ \\

&= \frac{4}{75}\pi\,3[h(t)]^2 \frac{d}{dt}h(t)\\ \\

&= \frac{4}{25}\pi [h(t)]^2 \frac{dh(t)}{dt}

\end{align*}

[Recall that we’re looking for $\dfrac{dh(t)}{dt}$ in this problem.]

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

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

We have $\dfrac{dV}{dt} = -15 \, \tfrac{\text{cm}^3}{\text{s}}$, and want to find $\dfrac{dh}{dt}$ at the instant when *h* = 10 cm.

Starting from our preceding expression, let’s first solve for $\dfrac{dh}{dt}$ and then substitute the values we’re given:

\begin{align*}

\frac{dV}{dt} &= \frac{4}{25} \pi h^2 \frac{dh}{dt} \\ \\

\frac{dh}{dt} &= \frac{25}{4\pi h^2} \frac{dV}{dt} \\ \\

&= \frac{25}{4\pi (10)^2} (-15) \\ \\

&= \frac{25}{4\pi (100)} (-15) \\ \\

&= -\frac{15}{16\pi} \text{ cm/s} \quad \cmark

\end{align*}

That’s the answer. The negative value indicates that the water’s height *h* is decreasing, as we expect.

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

At what rate does the water’s height *decrease*?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{\dfrac{15}{16\pi}} \, \dfrac{\text{cm}}{\text{s}} \quad \checkmark$

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you need.

What if you don’t have a value for dV/dt? I have a problem where I am given the height of the water, but not given any values for the speed at which it’s draining.

Thanks for asking, Erin. You need to be given _some_ rate, for either dV/dt or dh/dt, in order to find the other one’s related rate. If you provide some more detail about the question statement, we’ll do our best to help!

why does the height change from 20 to 10? I know you got the 10 from the smaller triangle when comparing it to the larger one, but why would you use that one instead of 20 if that makes sense?

Thanks for asking, Nicky! To answer it, look back at the figure at the top where we defined the different variables: “h” is the value of the water’s height in the cone at every instant, so as time passes and the water drains, the value of h changes. The problem asks us to find dh/dt *at the instant when h = 10 cm*,” so that’s the value of h we have to use here.

A key take-away is that that figure at the top matters a LOT: in a problem like this, it does a great deal of work in defining the variables for us and acts as a guide to everything that follows. When it comes to substitute values, it’s important to look back and see what each variable means so that you don’t use a wrong value.

Your physical sense of what’s going on here might help too: if you think about water draining out of a cone, you probably have the sense that at first (when the cone is full) the water level drops slowly, but then toward the end (when the cone is almost empty) the water level drops quickly. That is, dh/dt starts small and gets larger as the water level drops — it doesn’t stay the same throughout — which means that dh/dt _somehow_ depends on the value of h at that instant. So we have to calculate the value of dh/dt for the moment the question tells us, which here happens to be when h = 10 cm. (And it’s through our calculations that we find out exactly how dh/dt depends on h: as you see above, it turns out to depend on 1/h^2.)

Does that help? We hope so!