# Blog

## 4 Steps to Solve Any Related Rates Problem – Part 2

### What are your thoughts or questions?

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Anonymous
2 months 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
2 months 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 2 months ago by Matheno
Jake
3 years ago

Just saved my life for a calc midterm

Editor
3 years ago

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

Anonymous
4 months ago

well, did it? @jake

Editor
4 months ago

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

Anonymous
4 months 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!

Meola
4 years ago

I’ve been trying to learn this for days now and I’ve finally got it! Thank you!!

Editor
4 years ago

You’re very welcome, Meola. We’re happy to have helped! : )

Jason Kirsch
6 years ago

I have a similar problem, but I can’t seem to get the correct answer. My cone has a height of 8 inches and a diameter of 12inches. If the liquid pours out at 4 cubic inches a second, how fast is the level of the liquid dropping when the liquid is 3 inches deep?

Barrett
7 years ago

how are you eliminating r as a variable using r/h? where does that formula come from?
just trying to figure out how r and h are related in that step, as it is not listed where that part comes from.

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