Lesson 3.1: Related Rates

Related rates are simpler than their name makes them sound. Basically, you’ve got a function that depends on some variables. When one of those variables changes, how does something else in the function change?

Suppose you have a cylindrical tank of radius $ 2\textrm{ m}$ (it’s as tall as it needs to be; the height of the tank isn’t relevant to the problem) and you’re filling it with water at a rate of $ 5 \textrm{ cm}^3\cdot\textrm{s}^{-1}$. A typical related rate question would pose this situation, followed by “How fast is the height of the water changing?”

We’ll need a few equations to tackle the problem. Firstly, we know that $ V(r,h) = \pi r^2 h$—the formula for the volume of a cylinder; in this case, the cylindrical column of water. We’re given $ r=2\textrm{ m}$, and $ h$ is changing with time. Therefore $\displaystyle V=4\pi h(t)$ and since we’re concerned with an issue of time, we need to find $ \frac{dV}{dt}$, which is a good time to use the chain rule: $\displaystyle \frac{dV}{dt} = 4\pi \frac{dh}{dt}$

The other given in the problem is $ \frac{dV}{dt} = 5 \textrm{ m}^3\cdot\textrm{s}^{-1}$. We plug that in to get $\displaystyle 5 = 4\pi \frac{dh}{dt}$ and it’s then trivial to find that $\displaystyle \frac{dh}{dt} = \frac{5}{4\pi}$

But that’s a simpler case than we usually see, because the shape in question is a cylinder. Cynlinders are easy. More common is a question like that but dealing with a pyramid or a cone (ice cream cones in calculus problems are notoriously leaky; more strangely yet, they tend to leak at a very constant rate) and also often including surface area.

As a further example, suppose that you have a (right circular) cone with the annoying habit of a radius $ r$ that’s increasing by $ 2\textrm{ cm}\cdot\textrm{s}^{-1}$ from a starting point of $ 0\textrm{ cm}$. Its height is a constant $ 75\textrm{ cm}$. Here are a few questions to consider:

  1. When the cone has radius $ 10\textrm{ cm}$, how quickly is its volume increasing?
  2. When the cone has volume $ 3600\pi\textrm{ cm}^3$, how quickly is the area of its base increasing?
  3. When the cone’s radius and its volume are increasing at the same rate (neglecting units), what is the radius?

If this were an AP test, you would be given that $ V=\tfrac{1}{3}\pi r^2 h$ for a right circular cone. Knowing that $ h$ is constant, we differentiate that equation with respect to time using the chain rule: $\displaystyle \frac{dV}{dt} = \tfrac{2}{3}\pi h r(t)\frac{dr}{dt}$ Then, plugging in all the known values for the first question, $ \frac{dV}{dt}=\tfrac{2}{3}\pi (75) (10) (2)$, so $\displaystyle \frac{dV}{dt} = 1000\pi\textrm{ cm}^3\cdot\textrm{s}^{-1}$

To tackle the second problem, we first find $ r$ when $ V=3600\pi$; by solving the volume equation, we are interested in $ \frac{dA}{dt}$ when $ r=12\textrm{ cm}$. Since $ A_{\textrm{base}}=\pi r^2$, we can again use the chain rule to find $\displaystyle \frac{dA}{dt}=2\pi r \frac{dr}{dt}$ and now we plug in our known $ r$ and $ \frac{dr}{dt}$ and get $\displaystyle \frac{dA}{dt}=48\pi\textrm{ cm}^2\cdot\textrm{s}^{-1}$

The third question is irritating from the standpoint of dimensional analysis, but the College Board doesn’t care. We already have an expression for $ \frac{dV}{dt}$, so we need to set that expression equal to $ \frac{dr}{dt}$: $\displaystyle \tfrac{2}{3}\pi h r\frac{dr}{dt} = \frac{dr}{dt}$ which gives us $ r=\frac{1}{50\pi}$.

Similar questions are common, involving shapes including composites—often troughs, shaped like trapezoidal prisms—and situations as the object being filled with or drained of a liquid, or the object itself changing shape according to given equations.

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