4 Steps to Solve Any Related Rates Problem – Part 2
In our last post, we developed four steps to solve any related rates problem.
We introduced three examples to illustrate the basic ideas, and solved two of them there.
As promised, we’ll solve the third here.
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Water Leaving a Cone Example
Here’s the problem statement, now with some additional details about the cone itself and the moment we’re interested in:
Water in a Cone Example. Given: An inverted cone is 20 cm tall, has an opening radius of 8 cm, and was initially full of water. It is now being drained of water at the constant rate of 15 cm$^3$ each second. The water’s surface level falls as a result. Question: 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.)
Let’s use our Problem Solving Strategy to answer the question.
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.
We have a relation between the volume of water in the cup at any moment, and the water’s current height, h:
$$V = \frac{1}{3} \pi r^2h $$
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) $:
\begin{align*}
\frac{r}{8} &= \frac{h}{20} \\[8px]
r &= \frac{8}{20} h \\[8px]
&= \frac{2}{5} h
\end{align*}
End subproblem.
Then substituting the expression for r into our relation for V:
\begin{align*}
V &= \frac{1}{3} \pi \left(\frac{2}{5} h \right)^2h \\ \\
&= \frac{1}{3}\frac{4}{25} \pi h^3 \\ \\
&= \frac{4}{75} \pi h^3
\end{align*}
3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
\begin{align*}
\frac{dV}{dt} &= \frac{d}{dt}\left(\frac{4}{75} \pi h^3 \right) \\ \\
&= \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*}
4. Solve for the quantity you’re after.
At this point we’re just substituting values. 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 last expression above:
\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*}
The negative value indicates that the water’s height h is decreasing, which is correct.
Notice how our “Four Steps to Solve Any Related Rates Problem” led us straightforwardly to the solution. This is the strategy we use time and again; you can too.
Caution: IF you are using a web-based homework system and the question asks,
At what rate does the water level fall?
then the system may (depending entirely on how the question-writer entered their answer) already account for the negative sign, and so to be correct you must enter a POSITIVE VALUE: $\boxed{\frac{15}{16\pi}} \, \tfrac{\text{cm}}{\text{s}} \quad \checkmark$
That’s to say, if you think you did the problem correctly but the system tells you that your answer is wrong, try entering your value without the negative sign. Depending on how your instructor constructed your assignment, this convention may even vary from problem to problem in your homework set, just depending on which problems they chose, and the different answer writers for the various problems. (That can be really frustrating, we agree!) Let’s be clear: $\dfrac{dh}{dt}$ as we found it above is a negative value, which is a key take-away here.
Time to practice
Of course just reading our solution, or watching someone else solve problems, won’t really help you get better at solving calculus problems. Instead you need to practice for yourself, pencil in your hand, so you can get stuck and make mistakes and do all the other things people do when they’re learning something new. (And ideally do all those things before you’re taking an exam!) We have lots of problems for you to use, each with a complete step-by-step solution.
For more example problems with complete solutions, please visit our free Related Rates page!
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What are your thoughts or questions?
I genuinely was struggling so hard, thank you for making related rates so digestible. Feeling much better about attacking my final exam tomorrow!
We’re aiming to make things digestible, so are very glad to hear this helped! Good luck with your exams. : )
There´s a thing that tricks our minds: if you look down, dV/dt is an infinitésimal cone wich corresponds to if we look up, dV/dt is an infinitesimal cylinder. Volumes are different…!
Getting the visualization right is indeed important; otherwise, it’s difficult to know how to proceed for sure! In this case, the side-view is the one to have in mind. 🙂
This statement, if I understand it correctly, means that “related rate” problems are only relating two functions; i.e. Volume to Height or Area to Radius, etc…
How does that apply to related rate problems that use x^2+y^2=z^2?
We don’t eliminate any variable here, typically we have the rate of change for two of the three and are solving for the third rate at a specific x/y/z value.
My thought process, as a teacher sharing to my students, is to list out all the information given and what is being asked. IF any rate of change for any function(variable) in the equation that relates the quantities is missing, THEN you must ‘eliminate’ that variable by finding a technique (such as similar triangles) that could substitute the function for a function that is given (such as in this case replacing ‘r’ in terms of ‘h’).
Is my thought process correct?
I understand the reasoning of this cone problem, however I am trying to generalize the process of solving related rates for all scenarios and the quoted statement seems to contradicts a pythagorean related rate.
Thanks for your help.
P.S. You can solve for dr/dt and r using similar triangles as well. 🙂
This is a great question that we love! Thanks so much for writing in with it.
We’ve addressed it over on our Forum, where math displays a bit better. If you have follow-up questions or other ideas to discuss, please feel free to write us either here or there on the Forum.
For now, thanks again for asking this terrific question! : )
My English is poor, but I hope you understand what I want to say. In your example “water level falls as it drains from a cone“ Step 3, line 3, you change from the given equation to its first derivative. I consider that not precise enough for an explanation, as it is not correct also. You have to take the first derivative, but also to explain what for. As the two expressions are not the same! If I am wrong please tell me. Best regards.
First, thanks so much for writing with your question! We’d like to do our best to try to address your concern since we would like every step to make complete sense to you.
From what you wrote, I think that the specific place you’re referring to is
\[ \begin{align*}
\dfrac{dV}{dt} &= \frac{4}{75}\pi \left[ \dfrac{d}{dt}\left(h^3 \right) \right]\\[8px]
&= \frac{4}{75}\pi \left[] 3h^2 \dfrac{dh}{dt}\right]
\end{align*} \]
[If that’s NOT the mathematical move you think is incorrect, please write back and let us know and we’ll happily address you actual concern instead.]
Students often have some trouble with that move, which is why we address “Why is that dh/dt there?” in the small box at the end of Step 3. Click the small arrow to the left of that box to show the text.
—
I’ll add to that discussion a bit in case it helps.
Let’s consider a different problem, just to illustrate the key issue. Say you have a circle with a radius that changes as a function of time according to time: $r(t) = \sin(5t).$ Then we ask how quickly the circle’s area is changing as a result. We start with
\[\begin{align*}
A(t) &= \pi \left[r(t) \right]^2 \\[8px]
&= \pi \left[\sin (5t) \right]^2
\end{align*}\]
To find the rate at which the area changes, we take the derivative $\dfrac{d}{dt}$ of both sides of that equation:
\[\begin{align*}
\dfrac{d}{dt}[A(t)] &= \pi\dfrac{d}{dt}\left[\sin(5t) \right]^2 \\[8px]
&= \pi \left[2 \sin(5t) \cdot \cos(5t) \cdot 5 \right]
\end{align*} \]
Pay particular attention there to the “$\cos(5t) \cdot 5$,” which comes from the Chain Rule.
Let’s now do the calculation again, but for any $r(t)$ rather than $r(t) = \sin(5t).$ (Maybe $r(t) = t^3 + 5t,$ or $r(t) = \dfrac{1}{t+1}$, or $r(t) = e^{t^3}.$ It doesn’t matter.)
We again start with
\[ A(t) = \pi \left[r(t) \right]^2 \]
We take the derivative of both sides with respect to time:
\[\begin{align*}
\dfrac{d}{dt}[A(t)] &= \pi \dfrac{d}{dt} \left[r(t) \right]^2 \\[8px]
&= \pi \left[ 2 r(t) \dfrac{dr(t)}{dt}\right]
\end{align*} \]
We’d typically write all of those lines without the time dependence, so it’d look like this:
\[\dfrac{dA}{dt} = 2r \dfrac{dr}{dt}\]
The important thing to notice is that the $\dfrac{dr}{dt}$ term there is the same as the “$\cos(5t) \cdot 5$” in the calculation above when we had $r(t) = \sin(t)$.
That’s all to say that it’s not that “you change from the given equation to its first derivative.” Instead, we get the $\dfrac{dh}{dt}$ (or $\dfrac{d}{dt}$) term from applying the Chain Rule to the equation we had at the end of Step 2. Since we apply the derivative $\dfrac{d}{dt}$ to both sides of that valid equation at the end of Step 2, and then take correct derivatives (including using the Chain Rule!) in Step 3, the equation we have at the end of Step 3 is correct as well.
Again, I hope that I’ve addressed your actual question. But if not, please let us know and we’ll discuss that instead.
For now, thanks again for writing with your question!
[If anyone would like to continue this conversation, please do so on the Forum where we’ve cross-posted this thread. As you’ll see, math renders much better there than here in the Comments on the site.]
Just saved my life for a calc midterm
Thanks, Jake!! We’re happy to have helped, and hope your midterm goes well!
well, did it? @jake
Alas, @jake didn’t think to update us, so we’ll probably never know. We’d like to think so, though!
that’s a shame… but i do have to say you have an outstandingly fast response time, given a 9 year old post. Anyways, much thanks for the article!