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4 Steps to Solve Any Related Rates Problem – Part 2

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Eric
2 months ago

I genuinely was struggling so hard, thank you for making related rates so digestible. Feeling much better about attacking my final exam tomorrow!

Editor
2 months ago

We’re aiming to make things digestible, so are very glad to hear this helped! Good luck with your exams. : )

Anonymous
5 months ago

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…!

Editor
5 months ago

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. 🙂

Steve
9 months ago

We can only take the derivative with respect to one variable, so we need to eliminate one of those two.

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.

P.S. You can solve for dr/dt and r using similar triangles as well. 🙂

Last edited 9 months ago by Steve
Matheno
Editor
9 months ago

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! : )

Anonymous
1 year ago

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.

Editor
1 year ago

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.)
$A(t) = \pi \left[r(t) \right]^2$
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.

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.]

Last edited 1 year ago by Matheno
Jake
4 years ago

Just saved my life for a calc midterm

Editor
4 years ago

Thanks, Jake!! We’re happy to have helped, and hope your midterm goes well!

Anonymous
1 year ago

well, did it?

Editor
1 year ago

Alas, didn’t think to update us, so we’ll probably never know. We’d like to think so, though!

Anonymous
1 year ago

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!

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