The following is the problem at hand:

The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately ten times the radius. At what rate is the height of the oil changing when the oil is 35 inches high? (Hint: The formula for the volume of a cylinder is $V=\pi r^2 h$).

I have come to the following:

  1. $\frac{dv}{dt}=150\frac{in^3}{min}$
  2. $r=3.5$ when h=35 since $h=10r$.

I believe once you take the first derivative of $V=\pi r^2 h$ with respect to time (t) and solve for $\frac{dh}{dt}$ using the known variables listed above that you come to $\frac{dh}{dt}=\frac{600}{49pi}\frac{in}{sec}$. Though I lack the ability to test this answer to know if it's correct.


  • $\begingroup$ Not enough information. Clearly, the height of the cylinder must be greater than the oil height, $h>35''$. If we know that $h\approx 10r$, this implies only that $r$ is "greater than approximately" $3.4''$ - a totally useless description $\endgroup$ – Hagen von Eitzen Feb 27 at 21:51
  • $\begingroup$ @HagenvonEitzen, this is ultimatly what my professor and I came to also. This is a direct and complete quote from a well-known calculus textbook. Just not enough information. Thanks for the comment. $\endgroup$ – Optimizationally Feb 28 at 0:00
  • $\begingroup$ @SawyerBenson Is it possible for you to check the answer for this problem somewhere? $\endgroup$ – Michael Rybkin Feb 28 at 0:12

Physicists obtain the chain rule (calculus) by extending a fraction and swapping the denominators. In your scenario, this would look like this: $$ \frac{dV}{dt} = \frac{dV}{dt} \cdot \frac{dh}{dh} = \frac{dV}{dh} \cdot \frac{dh}{dt} $$ You already know that $\frac{dV}{dt} = 150\frac{in^3}{sec}$, and we can derive $\frac{dV}{dh} = \pi r^2$. So, with the formula from above: $$ 150\frac{in^3}{sec} = \pi r^2 \cdot \frac{dh}{dt} $$ Solving for $\frac{dh}{dt}$ (which is the function we are looking for), we get $$\frac{dh}{dt} = \frac{150}{\pi r^2}\frac{in^3}{sec}$$ We see that this doesn't depend on the time $t$, which is expected, as volume and height are proportional (in a cylinder, the radius is the same at any hight).

As we don't know anything about the relationship of the hight of the cylinder to the hight in question (which doesn't have any influence on the result), the problem can't be solved. The information "hight of the cylinder is approximately ten times the radius" must be chosen to confuse you. We can't use a formula like $h=10r$, because that $h$ is the hight of the cylinder. If $h=10r$ at any hight $h$, the cylinder would be an inverted cone.

  • $\begingroup$ Great explanation and simplification method, Wolfgang! Much appreciated, my friend. $\endgroup$ – Optimizationally Feb 28 at 14:07

First of all, each quantity there is a function of time. Secondly, I believe the idea here is that the radius and height are both growing in direct proportion to each other which means that the relation between the radius and height can be expressed the following way: $$r(t)=\frac{1}{10}h(t)$$ Using this fact, we can now express the volume of the cylinder only in terms of the height: $$ V(t)=\pi\left[r(t)\right]^2 h(t)= \pi\left[\frac{1}{10}h(t)\right]^2 h(t)= \frac{\pi}{100}\left[h(t)\right]^3 $$ Let's now differentiate both sides and express the rate at which the height is growing with respect to time in terms of the height function, which is unknown, and the rate at which the volume is growing: $$ \frac{dV}{dt}=\frac{3\pi}{100}\left[h(t)\right]^2\frac{dh}{dt}\implies\\ \frac{dh}{dt}=\frac{100}{3\pi}\frac{1}{\left[h(t)\right]^2}\frac{dV}{dt} $$ And what we do know is that at a particular point in time, $h(t)$ is going to be equal to $35$ in: $$ \frac{dh}{dt}=\frac{100}{3\pi}\frac{1}{\left[h(t)\right]^2}\frac{dV}{dt}= \frac{100}{3\pi}\frac{1}{35^2}150\approx 1.3\ in/s. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.