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Related Rates Problem – Cylinder Drains Water

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:
Water in cylinder

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.

Hide/Show Strategy
  1. Draw a picture of the physical situation.
    Don’t stare at a blank piece of paper; instead, sketch the situation for yourself. Really.
  2. Write an equation that relates the quantities of interest.
    1. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
    2. To develop your equation, you will probably use:
      • a simple geometric fact (like the relation between a circle’s area and its radius, or the relation between the volume of a cone and its base-radius and height);
      • a trigonometric function (like $\tan{\theta}$ = opposite/adjacent);
      • the Pythagorean theorem;
      • similar triangles.

    Most frequently (> 80% of the time) you will use the Pythagorean theorem or similar triangles.

  3. Take the derivative with respect to time of both sides of your equation. Remember the Chain Rule.
  4. Solve for the quantity you’re after.
Let’s use the strategy to solve this problem.
Water surface falls as the water drains
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.
When a quantity is decreasing, we have to make the rate negative.
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.
\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} \\ \\
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:
\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
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.

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

You must practice for yourself, pencil in your hand, before your exam.

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!

Over to you:

  • What tips do you have to share about solving Related Rates problems?
  • What questions do you have?

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As of September 2022, we're using our Forum for comments and discussion of this topic, and for any math questions. We'd love to see you there and help! Please tap to visit our Forum:

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3 years ago

Honestly, I never know (in any down deep depth of understanding) why I am really doing what I am doing in math, I just know the steps someone said to do but don’t ask my why the $&@/ it’s done that way ….like at all. Sucks how school speeds through subjects like a bat outta hell and covers topics in a manner that is more shallow than a puddle.

3 years ago

I really learn from this site. I also have a problem set which im not really sure with the way i solve it.
The problem is:
An open top cylindrical water tank containing a certain volume of water with its depth and given area of its hole at the bottom of the container. Find the time rate of the cylindrical water tank will be drained.

3 years ago

Thanks, Marwyn! We’re glad to know you’re learning here at Matheno.

We’d be happy to try to help you develop the solution to your problem, but the question as stated doesn’t make sense to us. Is there other information provided, perhaps in an accompanying figure? Please let us know and we’ll do our best to assist!

Debbie Gaydos
4 years ago

The correct answer is -0.9pi ft^3/sec, yes? Even rounded to 1 decimal place, as indicated by the given answer of “-3.0”, it should be -2.8, not -3.0. I think it is very confusing to give an estimate rather than exact result, with no indication of having done so, and the answer given to 1 decimal places but rounded incorrectly, and without the simplified exact result on which the estimate is based, given first.

Reply to  Debbie Gaydos
4 years ago

Hi, Debbie,

You are absolutely correct, and thank you very much for writing to ask us about it. The error was due to some editing that happened a while back, and thanks to your alerting us we’ve now corrected it. We strive to be clear and correct, and certainly not confusing. We apologize for the error there in that last line, and for the time you must have spent trying to make sense of it.

Thank you again for bringing it to our attention. We appreciate it!

5 years ago

cancel that… I forgot to multiply by pi

5 years ago

shouldn’t it be -0.9 ft/sec?

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