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A student wrote in to us (slightly edited):
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 your actual concern instead.]
Students often have some trouble with the move of going from the first to second equation there, 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:
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)]^3Then 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.
Let’s consider a different problem, to further illustrate the key issue here. 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 + 4t, 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 the last equation would look like this:
\dfrac{dA}{dt} = \pi \left[2r \dfrac{dr}{dt}\right]
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(5t). The equation immediately above is just the general way of writing the equation we need, that is correct and applies to whatever r(t) actually is.
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{dr}{dt}) term from applying the Chain Rule to the equation we had at the end of Step 2. Since the equation at the end of Step 2 was valid, and we next apply the derivative \dfrac{d}{dt} to both sides of that valid equation, 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 asking!