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I was doing this simple Calc 1 problem and it took me forever to get it right and it was embarrassing. I could see that the problem was easy but I just couldn't 'see' what I was doing. I couldn't find the path. I solved it, and then I tried to figure out how I had done it and why it took so long. The problem was this:

There is a circular cone-shaped tank. It's pointed upward. It's filling at a rate of $12 m^3/s$. The radius of the base is $26m$. The height of the tank is $8m$. The water's surface is circular and has radius $r$. The height of the water is $h$. What is $\frac {\delta h}{\delta t}$ when $r=10$?

I know that $v=\frac{\pi r^2h}{3}$. I also knew that $\frac{26}{8}=\frac{r}{h}$.

After I solved it, I came up with a 'map' for the work I had done. This is what I did:

given: $v', r$

know: {r,h} {v,r,h}

need: {$h'$, {r,h}}

The know line reads: I know a function connecting r and h. I know a function connecting v,r and h.

The need line reads: I need a function connecting $h'$ and r. r is in bold because I can use the {r,h} function that I know, and the r value I was given.

The process was then:

{v,r,h} start with volume function.

{v,{r:h}, h} replace r with h using my known {r,h} function.

{v', h', h} differentiate to get these variables. v' I have, h' I want, h I don't need.

{v', h', {r,h}} replace h with r using known {r,h} function.

Now the problem is solved: I had v', I needed h', and I had r.

SO, my question is this: is there a system somewhere comparable to this process? Does anyone else do this?

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To solve for unknowns I need to know: what already-knowns could possibly derive this result. Are there sufficient knowns? Are the linear equation system closed? etc. This process is very natural, IMO but good for very complicated tasks. – Frenzy Li Aug 15 '12 at 6:32
up vote 0 down vote accepted

I have usually taken a quite similar approach when explaining to students how to start solving a related rates problems! Related rates problems are the main calculus-type problems for which I have found this really useful. In a more primitive way it comes up elsewhere in teaching calculus. Something similar can also be useful when students first meet "abstract algebra" problems where the solution just depends on using the definition correctly.

In the case under discussion, we know $\frac{dV}{dt}$ and we want $\frac{dh}{dt}$. So we need a relationship between $V$ and $r$. It is immediate that $V=\frac{\pi}{3}r^2h$. Perhaps it would be even nicer to use $$3V=\pi r^2 h.$$ Once one has a nice simple relationship, I advise differentiating right away. Since $r$ and $h$ are related linearly, we could (as you did) substitute for $h$ in terms of $r$. Probably in this case that is a good idea, but often it is not. So differentiate. We get $$3\frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)\tag{$1$}.$$ In other situations, we often need to use implicit differentiation. (In many ways the current problem is too simple to illustrate the technique.)

Ultimately, we will need information about $\frac{dr}{dt}$. Use the relationship $8r=26h$ to find that $$8\frac{dr}{dt}=26\frac{dh}{dt}\tag{$2$}.$$ In a more complicated situation, the relationship between these two derivatives might depend on $h$, $r$, and $V$.

Now freeze the situation at the moment when $r=10$. Calculate the appropriate $V$ and $h$ if needed. (In this case they are not.) Use $(2)$ to replace $\frac{dr}{dt}$ in $(1)$, and use the known value of $\frac{dV}{dt}$ to find $\frac{dh}{dt}$ at this particular moment.

Remark: Here is another too simple example. A particle is travelling along the parabola $y=x^2-5$. The $x$-coordinate is increasing at a steady $5$ units per second. How fast is the particle's distance from the origin increasing at the instant the particle reaches the point $(3,4)$?

We have $D^2=x^2+y^2$. Differentiate immediately. We get $$2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.$$ from $y=x^2-5$, we get $\frac{dy}{dt}=2x\frac{dx}{dt}$. Freeze the situation at the instant when $x=3$. The rest is simple calculation, everything needed is at hand.

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I don’t think that I’ve ever before seen anyone write out the process in so much detail and with a notation specially invented for the purpose, but anyone solving the problem is going to use pretty much the same set of ideas. Many people, though, will prefer to start at the other end:

  1. I want $\dfrac{dh}{dt}$, so I need to express $h$ in terms of $t$.

  2. I don’t have a direct dependence of $h$ on $t$, but I do know how $v$ depends on $t$; can I calculate $h$ from $v$?

  3. Well, I can calculate $v$ from $h$: $v=\dfrac13\pi r^2h$, so $h=\dfrac{3v}{\pi r^2}$. What do I do with $r$?

  4. I can’t easily express $r$ in terms of $t$. The next best thing would be to express $r$ in terms of $h$; then I’d have an equation relating $v$ to $h$, and I could solve for $h$ just in terms of $v$. Can I find a relationship between $r$ and $h$?

  5. Looking at a sketch, I see that I can use similar triangles to get such a relationship: $\dfrac{r}{8-h}=\dfrac{26}8$.

Now all of the pieces are in place, and the rest is just a bit of algebra.

The more definite and detailed the method, the less generally applicable it will be. One could probably set out a pretty detailed recipe for solving routine related rates problems, for instance, but it wouldn’t be of much use for other kinds of problems. The best general advice that I can give, I think, is to alternate between working forward from what you have and backward from what you want. If a problem isn’t too complicated, you may be able to solve it by working in just one direction. It may not even matter much which direction you try: in this problem, either approach works. In more complicated problems you may have to work in both directions, trying to make the two approaches meet somewhere in the middle.

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