Example 1: Rate of change of a circle's area with its radius at $r = 5$ cm

Basics you need

Solution

1. Name the quantities and write the relation

   Let the radius be $r$ cm ($r > 0$⁠) and the area $A$ cm$^2$⁠. Then

   $$A = \pi r^2 .$$

   We want the rate of change of $A$ with respect to $r$⁠, which is $\dfrac{dA}{dr}$⁠, at the value $r = 5$⁠.

   "With respect to its radius" names the variable we differentiate with respect to. Writing the area as a formula in $r$ turns the word question into a calculus question.
   $\pi$ is a fixed number (about 3.14159), not a variable. Basics: area and volume formulas

2. Differentiate with respect to $r$

   By the constant multiple rule and the power rule (Basic 3):

   $$\frac{dA}{dr} = \frac{d}{dr}\left(\pi r^2\right) = \pi \cdot 2r = 2\pi r .$$
   This formula gives the rate for every radius at once. We differentiate before choosing the radius: if we put $r = 5$ first, $A$ becomes the constant $25\pi$⁠, and the derivative of a constant is 0, which would be wrong.
   $\frac{d}{dr}(r^2) = 2r$ is the power rule with $n = 2$⁠. Basics: rate of change as a derivative

3. Put in $r = 5$

   $$\left.\frac{dA}{dr}\right|_{r=5} = 2\pi(5) = 10\pi \approx 31.42 .$$

   So at the moment the radius is 5 cm, the area changes at the rate $10\pi$ with respect to the radius: about 31.42 cm$^2$ of area for each centimetre of radius.

   $10\pi = 31.4159\ldots$⁠, so 31.42 to two decimal places. The value is positive: as the radius increases, the area increases (Note after Example 3, Section 6.2).
   The value of a derivative at a point is written $\left.\frac{dA}{dr}\right|_{r=5}$⁠, "evaluated at $r = 5$". Lesson: rate of change of quantities

4. What the number means: the thin ring

   Let the radius grow from 5 cm to 5.01 cm. The new area is a thin ring:

   $$\begin{aligned} A(5.01) - A(5) &= \pi(5.01^2 - 5^2) \\ &= \pi(0.1001) \approx 0.3145 . \end{aligned}$$

   Dividing by the change in radius, 0.01 cm, gives $10.01\pi \approx 31.45$⁠, very close to $10\pi \approx 31.42$⁠. The ring is almost a strip of length $2\pi r$ (the circumference) and width $\Delta r$⁠, so

   $$\Delta A \approx 2\pi r \cdot \Delta r, \quad\text{that is}\quad \frac{\Delta A}{\Delta r} \approx 2\pi r .$$

   As $\Delta r \to 0$ the approximation becomes exact; that limit is the derivative.

   This explains why the derivative of the area is the circumference: the area grows at the edge, and the edge has length $2\pi r$⁠. The quotient 31.45 is a little more than 31.42 because the outer edge of the ring is slightly longer than the inner edge.
   A derivative is the limit of the average rate $\frac{\Delta A}{\Delta r}$ as $\Delta r \to 0$⁠. Basics: rate of change as a derivative

5. The units, and the words "per second"

   $A$ is measured in cm$^2$ and $r$ in cm, so $\frac{dA}{dr}$ is measured in cm$^2$ per cm. The book's solution writes the answer as "$10\pi$ cm$^2$/s", and the question says "per second". No time and no speed appear anywhere in the data, so the quantity actually computed, by the book as well as here, is the rate with respect to the radius, $\frac{dA}{dr} = 10\pi$⁠.

   The unit cm$^2$/s would belong to $\frac{dA}{dt}$⁠, the rate with respect to time. By the chain rule $\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}$⁠, so the two agree in value only if the radius happens to grow at 1 cm/s; then $\frac{dA}{dt} = 10\pi \times 1 = 10\pi$ cm$^2$/s.

   Units are a free check on any rate: a rate "of A with respect to r" always has the unit of $A$ divided by the unit of $r$⁠. This is why we read the answer as $10\pi$ cm$^2$ per cm, and note that the book's "/s" fits only the extra assumption of a radius growing at 1 cm/s.
   Rates with respect to time come from the chain rule (Section 6.2), as in Example 3. Basics: related rates (chain rule)

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