A student recently wrote to ask if we’d help solve a common related rates problem about water draining from a cylindrical tank. The problem was something like this:

Cylinder Drains Water.

A cylinder filled with water has a 3.0-foot radius and 10-foot height. It is drained such that the depth of the water is decreasing at 0.1 feet per second. How fast is the water draining from the tank?

Remember that related rates problems always give you the rate of one quantity that’s changing, and ask you to find the rate of something else that changes as a result. Here, the problem tells you that the water level is falling at 0.1 ft/s, and asks for the rate at which the volume of water in the tank is decreasing. That is, if we call the volume of water in the tank at any moment V, then what is dV/dt?

In an earlier post, we developed a 4-Step Strategy to solve almost any Related Rates problem.

Let’s use the strategy to solve this problem.

1. Draw a picture of the physical situation.
See the figure. Let’s call the height (or depth) of the water at any given moment y, as shown.
We are told that the water level in the cup is decreasing at the rate of $0.1\, \tfrac{\text{ft}}{\text{s}}$, so $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$. Remember that we have to insert that negative sign “by hand” since the water’s height 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 calling that the variable y.

B. To develop your equation, you will probably use . . . a simple geometric fact.
The volume V of any cylinder is its circular cross-sectional area $\left(\pi r^2 \right)$ times its height. Here, at any moment the water’s height is y, and so the volume of water in the cylinder is:
$$V = \pi r^2y$$

3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Note that the cylinder’s radius, r, is constant (r = 3.0 ft), so we’ll treat it as a constant when we take the derivative. By contrast, the water’s height y is not constant; instead it changes, and indeed, it changes at the rate dy/dt.
\begin{align*}
\frac{dV}{dt} &= \frac{d}{dt}\left(\pi r^2 y \right) \\ \\
&= \pi r^2 \frac{d}{dt}(y) \\ \\
&= \pi r^2 \frac{dy}{dt} \\ \\
\end{align*}
4. Solve for the quantity you’re after.
At this point we’re just substituting values: the problem told us $r = 3.0 \, \text{ft}$ and $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$, and we’re looking for dV/dt. Starting from our last expression above:
\begin{align*}
\frac{dV}{dt} &= \pi r^2 \frac{dy}{dt} \\ \\
&= \pi (3.0\, \text{ft})^2 \left(-0.1\, \tfrac{\text{ft}}{\text{s}}\right) \\ \\
&= -2.8 \, \tfrac{\text{ft}^3}{\text{s}} \quad \cmark
\end{align*}
That is, water is draining from the tank at the rate $\dfrac{dV}{dt} = -2.8\, \tfrac{\text{ft}^3}{\text{s}}$. The negative value indicates that the water’s volume in the tank V is decreasing, which is correct.

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Caution: IF you are using a web-based homework system and the question asks,

At what rate does the water drain?

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:   $\boxed{2.8} \, \tfrac{\text{ft}^3}{\text{s}} \quad \checkmark$

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: $\dfrac{dV}{dt}$ as we found it above is a negative value, which is a key take-away here.

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A related, harder problem that’s common on exams

Another very common Related Rates problem examines water draining from a cone, instead of from a cylinder. While the idea is very much the same, that problem is a little more challenging because of a sub-problem required to deal with the cone’s geometry. You can learn how to solve that in this blog post.

Time to practice

Reading our solution may be a great start to learn how to solve Related Rates problems, but to be ready for your test you need to practice for yourself, pencil in your hand. That’s when you’ll get stuck and make mistakes and do all the other things people do when they’re learning something new. . . and you want to do that before your exam! We have lots of problems for you to try, 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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Over to you:

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