Matheno - Learn Well and Excel

Related Rates

Calculus Related Rates Problem:
How fast is the water level falling as water drains from the cone?

Water drains from a cone at 15 cubic-cm each second.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.
To solve this problem, we will use our standard 4-step Related Rates Problem Solving Strategy.

The cone, with sizes labeled.
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 water's volume and the full cone form similar triangles.
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.

Buy us a coffee We're working to add more,
and would appreciate your help
to keep going! 😊

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*}

Why is that dh/dt there? Open for an explanation.
Are you wondering why that  $\dfrac{dh}{dt}$  appears? The answer is the Chain Rule.

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.

[collapse]

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.


Web-based homework warning icon
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: $\boxed{\dfrac{15}{16\pi}} \, \dfrac{\text{cm}}{\text{s}} \quad \checkmark$


Buy us a coffee If we've helped, please consider
giving a little something back.
Thank you! 😊

Return to Related Rates Problems


Small owl logo
Want access to all of our Calculus problems and solutions? Please visit our Calculus Home screen. It’s all free. (Why? Just because we’re educators who believe you deserve the chance to develop a better understanding of Calculus for yourself, and so we’re aiming to provide that. We hope you’ll take advantage!)

And if you have a Calculus question, please pop over to our Forum and post.  Related rates problems can be especially challenging to set up.  If you could use some help, please post and we’ll be happy to assist!



These days, we use our Forum for comments and discussion of this topic, and for any math questions. It’s also free for you to use, and if you’d like you can post anonymously with any username you choose. We’d love to see you there and help! Please tap to visit our Forum: community.matheno.com.

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

We'd appreciate your feedback! 😊
How helpful?

What are your thoughts or questions?

Subscribe
I'd like to be
8 Comments
newest
oldest most voted
Inline Feedbacks
View all comments
Anonymous
1 month ago

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

Julián Sierra
6 months ago

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”

Erin K
4 years ago

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.

Nicky G
5 years ago

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?

This site is free?!?

We don't charge for anything on this site, we don't run ads, and we will never sell your personal information.

We're passionate educators with a goal:

Provide high-quality, interactive materials to dedicated learners everywhere in the world, regardless of ability to pay,
so they (you!) can learn well and excel.

We're working to develop more.
If you'd like to give back and help support our efforts, all we ask is that you perhaps

Buy us a coffee We're working to add more,
and would appreciate your help
to keep going! 😊