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4 Steps to Solve Any Related Rates Problem – Part 1

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Are you having trouble with Related Rates problems in Calculus?
Let’s break ’em down and develop a strategy that you can use to solve them routinely for yourself.

I. What’s related about these rates? (Or, “How to recognize a Related Rates problem.”)

Related rates problems will always give you the rate of one quantity that’s changing, and ask you to find the rate of something else that’s changing as a result. Here are three common problem-scenarios to illustrate:

  1. Expanding Circle Example. Circle with radius r that increases at 1 mm per second.Given: The radius of a particular circle increases at 1 millimeter each second. As a result, its area changes. Question: How fast is its area changing [at some particular instant]?

  2. Ladder leans against a wall. Its base slides away from the wall at 2 ft/s.Sliding Ladder Example. Given: A 10-foot ladder leans against a wall. The base of the ladder slides horizontally away from the wall at 2 feet per second. As a result, the top of the ladder moves down the wall. Question: How quickly is the top of the ladder sliding down the wall [at some particular instant]?

  3. Water leaks out of a cone at 15 cubic-cm each second.Water Leaving a Cone Example. Given: A large cone of given size is being drained of water at the constant rate of 15 cm$^3$ each second. The water’s surface level in the cone falls as a result. Question: At what rate is the water level falling [at a particular instant]?    (We’ll solve this problem from start to finish in our next post.)



In each case you’re given the rate at which one quantity is changing. That is, you’re given the value of the derivative with respect to time of that quantity:

  1. “The radius . . . increases at 1 millimeter each second” means the radius changes at the rate of  $\dfrac{dr}{dt} = 1$ mm/s.

  2. “The base of the ladder slides horizontally away from the wall at 2 feet per second” means the ladder-base’s x-position changes at the rate of  $\dfrac{dx}{dt} = 2$ ft/s.


  3. When a quantity is decreasing, we have to make the rate negative.
  4. “A cone is being drained of water at the constant rate of 15 cm$^3$ each second” means the volume of water in cone changes at the rate of  $\dfrac{dV}{dt} = -15$ cm$^3$/s.
    Notice that we have to make the rate negative to capture that the water’s volume is decreasing. To do so, we insert the negative sign “by hand”—we just stick it in there.


The upshot: Related rates problems will always tell you about the rate at which one quantity is changing (or maybe the rates at which two quantities are changing), often in units of distance/time, area/time, or volume/time. The question will then be
\[ \left.
\begin{align*}
\text{how fast. . . } \\
\text{how quickly. . . } \\
\text{at what rate. . .}\\
\end{align*}
\right\} \] $$\text{…does another quantity change as a result?}$$
The rate you’re after is related to the rate(s) you’re given. Your job is to find that relationship.

II. Your job: Develop a Relationship between Quantities

This is where students often initially get stuck when doing Related Rates problems. (But after working lots of these problems you won’t anymore; they’ll be totally routine for you.) The skill you have to practice is developing a relationship between

  • the quantity you’re after, and
  • the quantity (or quantities) you’ve been given.
There’s just a tiny handful of approaches you’ll ever use.

You’ll take a derivative in a minute to get a relationship between the rates. Before then, though, we have to write a relationship between the quantities themselves. Fortunately, there’s just a tiny handful of approaches you’ll ever use. We’ll illustrate two of the most common using our first two examples above. (We’ll analyze the third example in our next post.)

  1. Expanding Circle Example. We need a relationship between the circle’s area, and its radius. This one’s easy, and just draws upon our everyday geometry-knowledge for a circle:

    $$A =\pi r^2$$
    This expression captures the relationship between

    • the quantity you’re after (something about the circle’s area, A), and
    • the quantity you’ve been given (something about the circle’s radius, r).
  2. The wall and floor form a right angle, and the ladder is the hypotenuse of the right triangle.Sliding Ladder Example. This example illustrates one of the two approaches you’ll use most often: the Pythagorean theorem.
    Let’s call the horizontal distance from the wall to the ladder’s bottom x, and the vertical distance from the ground to the ladder’s top y. Then since the ladder is 10 feet long, at any instant we know

    $$x^2 + y^2 = 10^2$$
    This expression captures the relationship between

    • the quantity you’re after (something about the ladder-top’s vertical position, y), and
    • the quantity you’ve been given (something about the ladder-bottom’s horizontal position, x).

The upshot: The new skill you’ll probably need to practice is developing the relationship between the quantity whose rate you’re after, and the quantity (or quantities) whose rate(s) you know about. There are only a few approaches you’ll typically use; we’ve listed them all in our Problem Solving Strategy box below.

III. Take the Derivative with Respect to Time

Related Rates questions always ask about how two (or more) rates are related, so you’ll always take the derivative of the equation you’ve developed with respect to time. That is, take $\dfrac{d}{dt}$ of both sides of your equation. Be sure to remember the Chain Rule!

Let’s apply this step to the equations we developed in our two examples above:

  1. Expanding Circle Example.
    \begin{align*}
    A &= \pi r^2 \\ \\
    \frac{d}{dt}[A] &= \frac{d}{dt}\left[\pi r^2 \right] \\ \\
    \frac{dA}{dt} &= 2\pi r \frac{dr}{dt}
    \end{align*}

    Why is that dr/dt there? Open for an explanation.
    Jump ahead to 4:05 in the video to see the relevant discussion.
    Are you wondering why that $\dfrac{dr}{dt}$ appears? The answer is the Chain Rule.

    While the derivative of $r^2$ with respect to r is $\dfrac{d}{dr}r^2 = 2r$, the derivative of $r^2$ with respect to time t is $\dfrac{d}{dt}r^2 = 2r\dfrac{dr}{dt}$.

    Remember that r is a function of time t: the radius grows as time passes. We could have captured this time-dependence explicitly by writing our relation as
    $$A(t) = \pi [r(t)]^2$$
    to remind ourselves that both A and r are functions of time t. Then when we take the derivative,
    \begin{align*}
    \frac{dA(t)}{dt} &= 2\pi r(t) \left[\frac{d}{dt}r(t)\right]\\ \\
    &= 2\pi r(t) \left[\frac{dr}{dt}\right] \end{align*}

    [Recall $\dfrac{dr}{dt} = 1 \,\frac{\text{mm}}{\text{s}}$ in this problem.]

    Most people find that writing the explicit time-dependence A(t) and r(t) annoying, and so just write A and r instead. Regardless, you must remember that r depends on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dr}{dt}$ term.

    That last expression gives us the relation we need between the unknown $\dfrac{dA}{dt}$ and the known $\dfrac{dr}{dt} = 1 \,\frac{\text{mm}}{\text{s}}$, and so we’re essentially finished. The problem will always specify the other conditions of the moment you’re interested in. For instance, this problem might have asked:

    Question: How fast is the circle’s area changing when its radius is 20 mm?

    Circle with r = 20 mm.
    In that case we have $r = 20$ mm and $\dfrac{dr}{dt} = 1\, \tfrac{\text{mm}}{\text{s}}$, and so
    \begin{align*}
    \frac{dA}{dt} &= 2\pi r \frac{dr}{dt}\\ \\
    &= 2\pi (20)(1) = 40\pi \, \tfrac{\text{mm}^2}{\text{s}} \quad \cmark
    \end{align*}


    The problem could have specified any value for r; it doesn’t matter. At this point, we’re just substituting in values.


  2. Sliding Ladder Example.
    \begin{align*}
    x^2 + y^2 &= 10^2 \\ \\
    \frac{d}{dt}\left[ x^2 + y^2\right] &= \frac{d}{dt}[10^2] \\ \\
    2x\frac{dx}{dt} + 2y\frac{dy}{dt} &= 0 \text{ [*]} \\
    \end{align*}
  3. The last expression gives us the relation we need between the unknown $\dfrac{dy}{dt}$ and the known $\dfrac{dx}{dt} = 2 \,\frac{\text{ft}}{\text{s}}$, so again the problem is essentially finished. If it had asked, for instance,

    Question: How quickly is the top of the ladder sliding down the wall when the base is 8 feet from the wall?

    then we would have x = 8 ft and $\dfrac{dx}{dt} = 2$ ft/s, and we’re looking for $\dfrac{dy}{dt}$. But then in this case we still have one unknown: we don’t know y.

    So we have to stop and do a quick subproblem: our original Pythaogrean statement specifies y when $x = 8$.


    Begin subproblem to find $y$ when $x = 8$:
    The wall and floor form a right angle, and the ladder is the hypotenuse of the right triangle.
    \begin{align*}
    x^2 + y^2 &= 10^2 \\
    8^2 + y^2 &= 100 \\
    y^2 &= 100 – 64 = 36 \\
    y &= 6 \text{ ft} \\
    \end{align*}
    End subproblem.


    So now we have $x =8$ ft, $y = 6$ ft, and $\dfrac{dx}{dt} = 2$ ft/s. We’re solving for $\dfrac{dy}{dt}$.


    Ladder when x = 8 ft and y = 6 ftStarting with the equation marked [*] above:
    \begin{align*}
    2x\frac{dx}{dt} + 2y\frac{dy}{dt} &= 0 \\ \\
    2y\frac{dy}{dt} &= -2x\frac{dx}{dt} \\ \\
    \frac{dy}{dt} &= -\frac{x}{y}\frac{dx}{dt} \\ \\
    &= – \frac{8}{6}(2) \\ \\
    &= -\frac{8}{3}\,\frac{\text{ft}}{\text{s}} \quad \cmark
    \end{align*}


    Again, the problem could have given us any value for x, and then we’d find the associated y and substitute those values in. (Or the problem could specify a value for y and then we’d have to find the associated x. Whatever.) At this point we’re just substituting in values.

  4. Water Leaving a Cone Example.
    To see the complete solution to this problem, please visit Part 2 of this blog post on how to solve related rates problems.

The upshot: Take the derivative with respect to time of the equation you developed earlier. Then substitute the values you’ve been given to find the quantity you’re after. (You might first have to complete a little subproblem to determine another quantity at the moment of interest, like we did in the Ladder Example.)

IV. 4-Step Problem Solving Strategy

To summarize, here are the four steps that will help you solve very-nearly any Related Rates problem (an image, so you can easily save it): Related rates problem solving strategy

As promised, in the next post we’ll complete the “Water Leaving A Cone” example, which will illustrate the common use of similar triangles in solving Related Rates problems.

If you’d like more example problems with complete solutions, please visit our Related Rates page.


Over to you:

  • 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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Anonymous

very helpful

Lucy

Thank you so much for this post. I’ve been trying EVERYTHING in order to learn more about related rates and all of them left me more confusing than before. THIS have helped me so much! Thank you!!!

Deepak

It’s a nice post about related rates calculus. I the way you have described it. It’s really helpful. Thanks for sharing it.

Anonymous

Thanks for the post, extremely helpful 🙂 Hoping I can use this info to pass my test later today!

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