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.

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.

We are given that the volume of water in the cup is decreasing at the rate of 15 cm$^3$/s, so $\dfrac{dV}{dt} = -15\, \tfrac{\text{cm}^3}{\text{s}}$. Remember that we have to insert that negative sign “by hand” since the water’s volume is decreasing.


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.

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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.

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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.

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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We’d love your comments:

• What tips do you have to share about solving Related Rates problems?
• Or what questions do you have?
• Or how can we make posts such as this one more useful to you?

Please comment below!

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