Losing the chain in related rates of change

A rates-of-change question hands you one rate, say how fast a radius grows, $\frac{dr}{dt}$, and asks for another, such as how fast the area grows, $\frac{dA}{dt}$. The two are joined by the chain rule through the shared variable.

The tempting mistake is to differentiate $A$ with respect to $r$, get $\frac{dA}{dr}$, and then write that down as if it were $\frac{dA}{dt}$. It is inviting because $\frac{dA}{dr}$ is the derivative you know how to compute, so it feels like the finished job.

But it answers the wrong question: it is a rate per unit length, not per unit time. Losing that final multiplication by $\frac{dr}{dt}$ drops the one piece of information the question deliberately supplied, and the units alone show the answer cannot be right.

A circle's radius grows at $\frac{dr}{dt}=0.2$ cm s$^{-1}$. Find the rate of change of area when $r=5$ cm, given $A=\pi r^{2}$.

First, $\frac{dA}{dr}=2\pi r$. The tempting shortcut stops there:

The chain rule multiplies by the rate the question gave:

So the area grows at $2\pi$ cm² s$^{-1}$. The wrong version, $10\pi$, is five times too big, exactly the factor $\frac{1}{0.2}$ that was thrown away by skipping the multiplication.

1. 1

   Write down every rate

   List what you are given (for example $\frac{dr}{dt}$) and what you must find (for example $\frac{dA}{dt}$). Seeing both as "per time" tells you the chain must connect them.

2. 2

   Build the bridge

   Differentiate the formula that links the variables to get the middle derivative, here $\frac{dA}{dr}=2\pi r$.

3. 3

   Write the chain explicitly

   Put $\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}$ on its own line before substituting anything, so the missing factor cannot vanish.

4. 4

   Substitute the value of the variable

   Only now put in the given number for $r$, keeping the rate $\frac{dr}{dt}$ in the product.

5. 5

   Check the units

   The answer should read as area-per-time (cm² s⁻¹). If your units come out as area-per-length, you forgot the last factor.

Our teachers make you write the chain $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$ as a full line before any number goes in, so the given rate can never quietly disappear. In a one-to-one lesson we practise reading the question for the units of the answer first, "they want cm³ per second", which instantly tells you a second factor is needed.

Because the paper awards method marks, that explicit chain line earns credit even if the arithmetic slips. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To master related rates, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.