## 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 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.

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**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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Return to Related Rates Problems

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(Archived comments from before we started our Forum are below.)

## What are your thoughts or questions?

when you turn the equation of the volume of a cone into rate in terms of time why wouldn’t v= turn into V(dv/dt).

Thanks for writing in to ask!

When we take the derivitive of

Vwith respect to time, \(\dfrac{d}{dt}V\), we obtain\[\dfrac{d}{dt}V = \dfrac{dV}{dt}\]

where the quantity on the right hand side of the equation is the rate at which the water’s volume changes with respect to time. In this problem, that rate is specified to be constant, \(\dfrac{dV}{dt} = -15 \, \, \tfrac{\text{cm}^3}{\text{s}}\).

I think you’re asking why

\[\dfrac{d}{dt}V \overbrace{=}^{?} V\left(\dfrac{dV}{dt}\right) \]

which is incorrect. The quantity on the right there isn’t a rate itself; instead it says that the volume

Vis a function of the rate \(\dfrac{dV}{dt}\). (If you meant instead that the parenthese indicate multiplication, please let us know since that would have a different explanation for why it’s not a correct statement.)The key point you should make sure you understand is that when we take the derivative with respect to time of the water’s volume

V,\(\dfrac{d}{dt}V,\) we obtain the rate at which that volume is changing, \(\dfrac{dV}{dt}.\) If that doesn’t make sense, please let us know and we’ll do our best to explain further.For now, thank you again for writing in to ask, and we hope what we’ve written above helps!

I would like to ask, what if the rate is not constant. What if when the depth of the water was 6ft, the water surface is changing at a rate of 15cm3/s? And the water drained completely

Nice question — thanks for asking! The formula we found in Step 4,

dh/dt= [25/(4 pih^2)] (dV/dt), holds at every instant in time without the restriction thatdV/dtbe constant. So at any timeT, if you have the values ofhanddV/dt,then you can correctly calculatedh/dtat that moment.We don’t quite understand your additional specification there at the end, “And the water drained completely.” If you’d like to explain more about the question, we’d be happy to try to answer.

For now, we hope that helps!

-15/16Pi cubic cm? not just cm in the answer?

Thanks for asking! The units for dh/dt are cm/s, as we’d expect: the water’s

height(measured in cm) is changing at the rate of -15/(16pi) cm per second. By contrast, thevolumeof water in the cone, measured in cm^3, is changing at the rate of -15 cm^3/s as stated in the problem.Does that address your question, or was there something else you were curious about? If so, please let us know and we’ll do our best to answer.

For now, thanks for writing in!

The statement says: NOW the flow rate is 15cm3 in present.

Then what is the height loss rate WHEN THE HEIGHT IS HALFWAY 10cm, it sounds like it will be 10 cm in the future, which could lead to a misunderstanding. It would help to state that it is the same moment, “now the height is halfway 10cm”

Thanks for your comment and suggestion, Julián! We have modified the text, removing the word “now,” and hope that eliminates any potential misunderstanding.

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!