I just finished teaching a freshman calculus course (at an American state university), and one standard topic in the curriculum is related rates. I taught my students to answer questions such as the following (taken, more or less, from the textbook):

  • A man starts walking due north at 5 ft/sec from a Point A. Ten seconds later, a woman starts walking south at 4 ft/sec from a point 20 ft due east of Point A. How fast are they moving apart when the woman has been walking for ten seconds?

  • A 6' man walks away from a 20' lamppost at a speed of 5 ft/sec. How fast is the distance between the tip of his shadow and the top of the post changing when he is 40' away?

  • A baseball player runs from first base to second at 20 ft/sec, and simultaneously another baseball player runs from third base to home at the same speed. How fast are they approaching each other after one second?

To put my question bluntly: Who cares?

My students do, but only because they know these questions will appear on their exams. The baseball question (or something very similar) is actually an exercise in Stewart, and I struggled in vain to imagine a situation in which the manager of a baseball team would need to know the answer.

This is in stark contrast to many other topics addressed in first-year calculus -- optimization, basic differential equations, etc. -- which are realistic models of questions of natural interest in business, biology, etc. Basically, all the related rates questions seemed to be cooked up in response to the fact that calculus students now knew a method to solve them.

My question is in the title. Can anyone share any related rates questions which don't seem quite as contrived, and which might naturally seem interesting and motivated to a typical class of college freshmen?

Thank you!

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    $\begingroup$ How about predator-prey or host-parasite systems? There you have two things naturally in opposition, and it's natural to think about one rate of growth pitted against the other. I don't have any sample questions to hand, but I guess you could find some in introductory math bio books. $\endgroup$ Dec 10, 2011 at 3:04
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    $\begingroup$ Excellent question, and I think it applies to a lot of other stuff in the calculus curriculum as well. Teaching students to differentiate every function that you can write down in closed form should incite them to ask "who cares?". I would think one could do better, but many mathematicians don't seem to want to. $\endgroup$ Dec 10, 2011 at 3:20
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    $\begingroup$ Michael: I actually think it requires a fair amount of sophistication to find differentiation techniques unmotivated. To a typical calculus student, "mathematics" has always been more or less synonymous with "calculation," and computing derivatives probably seems more natural than defining them. $\endgroup$ Dec 10, 2011 at 3:43
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    $\begingroup$ I like Boyle's Law and total resistance of two resistors in parallel. Both are found in almost any textbook. The value of Boyle's Law examples in particular is that simple related rates calculations confirm something they already know (increasing pressure of a gas decreases its volume and vice versa). There's a lot of value of relating a new concept you are learning to things you already know, though most students fail to grasp this metaprinciple. $\endgroup$ Dec 10, 2011 at 4:01
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    $\begingroup$ Michael: I think the value of Boyle's law with calculus is not that it confirms what students already know but that it shows them something they don't know. Boyle's law is not about calculus, as it just says $PV$ is constant under suitable conditions. There are no related rates there, so the intuition that $P$ going up makes $V$ go down has no calculus in it. Let's include calculus: students are likely to think that if $dP/dt$ is constant then $dV/dt$ (up to a sign) is the reciprocal constant, which is wrong if you use the product rule. See my answer below $\endgroup$
    – KConrad
    Dec 10, 2011 at 4:43

5 Answers 5


The skills that students are practicing in related rates problems are:

  1. Differentiating a known equation implicitly with respect to time.

  2. Interpreting the time derivative of a quantity as a rate of change.

The main reason that related rates problems feel so contrived is that calculus books do not want to assume that the students are familiar with any of the equations of science or economics. Every related rates problem inherently involves differentiating a known equation, and the only equations that the calculus book assumes are the equations of geometry.

Thus, you can find related rates problems involving various area and volume formulas, related rates problems involving the Pythagorean Theorem or similar triangles, related rates problems involving triangle trigonometry, and so forth. A few of these problems are compelling -- for example, computing the speed of an airplane based on ground observations of its altitude and apparent angular velocity -- but most of them do feel a bit contrived.

The reality, of course, is that students are familiar with many of the basic equations and concepts of science and economics, and there's no rule against using these in problems. For example, you can make up all sorts of compelling related rates problems by starting with any physics or chemistry equation and imagining a situation where you might want to take its derivative:

  1. The kinetic energy of an object is $K = \frac{1}{2}mv^2$. If the object is accelerating at a rate of $9.8 \text{m}/\text{s}^2$, how fast is the kinetic energy increasing when the speed is $30 \;\text{m}/\text{s}$?

  2. An ideal gas satisfies $PV = nRT$, where $n$ is the number of moles and $R \approx 8.314\;\; \text{J}\; \text{mol}^{-1} \text{K}^{-1}$. Give the rate at which the temperature and volume of the gas are increasing, and then ask about the rate of change in pressure when the volume and temperature reach certain amounts.

  3. The total energy stored in a capacitor is $\frac{1}{2} Q^2 / C$, where $Q$ is the amount of charge stored in the capacitor and $C$ is the capacitance. Give the value of $C$ and the rate at which $Q$ is decreasing, and ask about the rate at which the capacitor is losing energy when the energy is a certain amount.

  4. In astronomy, the absolute magnitude $M$ of a star is related to its luminosity $L$ by the formula $$ M \;=\; M_{\text{sun}} -\; 2.5\; \log_{10}(L/L_{\text{sun}}). $$ where $M_{\text{sun}} = 4.75$ and $L_{\text{sun}} = 3.839 \times 10^{26} \text{watts}$. (Note that, by convention, brighter stars have lower magnitude.) If the absolute magnitude of a variable star is decreasing at a rate of $0.09 / \text{week}$, how quickly is the luminosity of the star increasing when the magnitude is $3.8$?

It's easy to make these up: just think of any equation in science or economics whose derivative might be interesting. Wikipedia and/or science textbooks can be helpful for finding equations from a wide variety of fields.

Edit: I have compiled a list of these problems in the form that I use them in my classes, and posted them on my professional web page.

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    $\begingroup$ And you can always briefly explain the relevant formula if it is something you suspect they might not know. I find it important to make sure to introduce the relevant background as part of the problem (as opposed to explaining while you go through the solution), to make it clear that you don't necessarily expect students to know the relevant physical law (though it's great if they do!). And it is certainly worth telling students that the more physics, chemistry, biology, economics, etc. that they learn, the more applications of calculus they will discover. $\endgroup$ Dec 10, 2011 at 18:53
  • $\begingroup$ This is extremely insightful. Thank you very much! $\endgroup$ Dec 11, 2011 at 14:17
  • $\begingroup$ Wonderful (Michael Joyce's last sentence too). $\endgroup$ Dec 15, 2011 at 15:46
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    $\begingroup$ I find your first example especially compelling when it's turned the other way. Your 2000-ton rocket is moving at half a kilometer per second, and you are bored. To boost your speed by 10 m/s every second, how much power will it take? $\endgroup$
    – Vectornaut
    Nov 6, 2014 at 1:24
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    $\begingroup$ These are good questions to discuss, but they are slightly less compelling for me from an exam standpoint because I think the other thing related rates problems do is teach students to read a paragraph in English and work from there to set up an equation relevant to a situation. For these kinds of physics equations either you know (or are given) them or you don't. So great problems to illustrate "who cares" but I'm still looking for good test questions! $\endgroup$ Dec 13, 2014 at 1:01

Here's one that may be quite relevant to their lives.

Doppler radar measures the rate of change of the distance from an object to the observer. A police officer $a$ metres from a straight road points a radar gun at a car travelling along the road, $c$ metres away, and measures a speed of $v$. What is the car's actual speed?

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    $\begingroup$ It is hard not to include this practical application: arxiv.org/abs/1204.0162 (The Proof of Innocence), "The paper was awarded a special prize of $400 that the author did not have to pay to the state of California. " $\endgroup$ Jul 16, 2017 at 17:27

Here are two examples I think are interesting. (Update: I have added a third example at the end, more intricate than the other two.)

  1. A ladder that is leaning against a wall starts slipping down. If the point where the ladder touches the ground (draw your own picture) is moving away from the wall at a constant rate, is the point where the ladder touches the wall falling at a constant rate? People may reasonably think that because the ladder is "rigid" the answer should be yes.

Differentiating $x^2 + y^2 = L$ (where $L$ is the length of the ladder, a constant) shows $dx/dt$ being constant certainly doesn't make $dy/dt$ constant, and in fact since $dy/dt = -x(dx/dt)/\sqrt{L-x^2}$ we see that as $x$ increases (while being less than $L$) and $dx/dt$ is fixed the point where the ladder touches the wall is dropping faster and faster. Some students may even say that from their experience or physical intuition this actually makes sense, which raises the question of whether this is truly a physical phenomenon or a purely mathematical one that has been revealed from calculus.

  1. There is a gas is in a chamber with a flexible wall (so the chamber can expand or contract, e.g., a piston is at one end). According to the chemists, if we maintain the gas at a constant temperature while increasing or decreasing the size of the chamber then the pressure $P$ and volume $V$ satisfy the relation $PV$ = constant. (I am thinking of the ideal gas law $PV = nRT$, where $n$ and $T$ don't change.) One aspect of this equation which is obvious is that as the volume goes up/down, the pressure goes down/up. Question: If we decrease the volume at a constant rate, does the pressure increase at a constant rate, and more precisely at the rate which is the reciprocal of the rate at which the volume is going down? I think it's quite natural for people to make a snap judgement that if $dV/dt = -4$ then $dP/dt = 1/4$ because $PV$ is constant, but of course the product rule shows this is wrong, and moreover if $dV/dt$ is constant then $dP/dt$ is definitely not constant.

It's perhaps worth first discussing a situation where such intuition is right, namely where the sum of two variables is fixed, rather than the product. Find your own example where two variables $x$ and $y$ satisfy $x+y$ = constant. Then $dx/dt = -dy/dt$, which makes a lot of sense: the rate at which one goes up is exactly "opposite" to the rate at which the other goes down. But when the product is constant this is completely incorrect.

  1. Relate the rates between angular velocity and linear velocity of a crank-and-piston system. It would be too involved to present the details here, and I don't have a good image to include, so see the YouTube video here or read about it here. I find this far more interesting than the previous two examples, and covered it in a calculus lecture yesterday as the last example.
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    $\begingroup$ I concur with your first sentence! $\endgroup$ Dec 11, 2011 at 14:14
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    $\begingroup$ So in fact the ( ideal) ladder hits the ground with infinite velocity! I guess that in the real world the top of the ladder slightly pulls away from the wall. $\endgroup$ May 22, 2013 at 13:45
  • $\begingroup$ @AaronMeyerowitz googling "falling ladder paradox" leads to some articles that reach the conclusion you indicate. Steer clear of any hits on the similarly titled "ladder paradox" of special relativity. $\endgroup$
    – KConrad
    Jan 16, 2017 at 6:32
  • $\begingroup$ I really hate the ladder problem $\endgroup$
    – Lenny
    Mar 10, 2022 at 1:23

Here's something you could try, based on a passage from Richard Feynman's "Surely You're Joking, Mr. Feynman!":

When I was in high school, I'd see water running out of a faucet growing narrower, and wonder if I could figure out what determines that curve. I found it was rather easy to do. I didn't have to do it; it wasn't important for the future of science; somebody else had already done it. That didn't make any difference. I'd invent things and play with things for my own entertainment.

Assume a constant flow rate, and see whether your students can be guided to a solution. You might try a warm-up problem, where you pour water into a non-cylindrical glass (a martini glass, or something more weirdly shaped) and try to determine how quickly the level rises, as a function of cross-sectional area.

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    $\begingroup$ I was actually looking at that same quote quite recently, and musing that I actually don't know what makes water running out of a faucet grow narrower. What is the physical principle I should be using to guide the math? $\endgroup$ Dec 10, 2011 at 18:31
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    $\begingroup$ Think of the flow as measured at a horizontal cross-section as the amount of water that passes through the cross-section per second. If the flow is uniform, this is the same at every cross-section, and in that case the area of water at that cross-section is inversely proportional to the velocity at that cross-section. $\endgroup$
    – Todd Trimble
    Dec 10, 2011 at 19:17
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    $\begingroup$ Hm... why is the "flow", in this particular sense, uniform? I mean, surely if I were to keep pumping a load of solid bricks into a pipe at a constant rate, ending in a vertical drop, the bricks, when dropping, would simply fall straight down, not attempting to contract inwards to maintain a uniform "flow" as they accelerate. (I may be completely missing something still) $\endgroup$ Dec 11, 2011 at 9:59
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    $\begingroup$ Sorry, Sridhar, I didn't mean to be glib. I'm afraid I don't have a really good answer, but I suspect the answer has something to do with the near-incompressibility of water: it maintains a very nearly constant density when in its liquid state. For a stream of bricks falling, bricks that were initially close together begin to separate: the velocity vector field has non-zero divergence. Whereas the velocity vector field of an incompressible fluid in steady flow has zero divergence. See e.g. Feynman Lectures in Physics, vol. II, section 40. But I don't know why water is incompressible! $\endgroup$
    – Todd Trimble
    Dec 11, 2011 at 21:41
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    $\begingroup$ And the surface tension of water keeps the cross section (approximately) circular. $\endgroup$
    – Lubin
    May 22, 2013 at 13:24

The following example is contrived, but I created it with the same frustration at the boring and repetitive nature of most related rates problems. Since I found this MO question by doing a google search for a "really interesting related rates problem," I'm recording this here for other instructors looking for juicier problems.

Question: What is the rate of change of the width of a shadow as you walk away from a street lamp considered as a point light source which is higher than your head? You can model yourself as a rectangle is you'd like!

The answer to this question was surprising to me. You can also ask for the rate of change of the trapezoidal shadow's area.

  • $\begingroup$ I don't know if it's obvious or nobody was motivated to work though the example, but the surprising thing to me is that the width of the shadow doesn't change! $\endgroup$
    – Jim Conant
    Oct 29, 2014 at 11:55
  • $\begingroup$ Anyone ever work this one out? $\endgroup$
    – JackOfAll
    Nov 21, 2016 at 14:26
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    $\begingroup$ @JackOfAll you mean besides me? $\endgroup$
    – Jim Conant
    Nov 21, 2016 at 14:54

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