Exercise 6.1 · Q1: Circle: $dA/dr$ at $r = 3$ cm and $r = 4$ cm

Basics you need

Part (a): the rate when $r = 3$ cm

1. Name the quantities and write the relation

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

   $$A = \pi r^2 .$$
   "Rate of change of the area with respect to the radius" asks how $A$ responds to changes in $r$⁠. Before we can differentiate, $A$ must be written as a formula in $r$ alone. The area formula does exactly that.
   Every rate-of-change problem starts the same way: name the variables with their units, then write the equation that links them. Basics: area and volume formulas

2. Differentiate with respect to $r$

   $$\frac{dA}{dr} = \frac{d}{dr}\left(\pi r^2\right) = \pi \cdot 2r = 2\pi r .$$
   The question says "with respect to its radius", so $r$ is the variable we differentiate with respect to. There is no time variable in this question, so no chain rule is needed. $\pi$ is a constant, so it stays as a factor while $r^2$ is differentiated by the power rule.
   Power rule with $n = 2$⁠: $\frac{d}{dr}(r^2) = 2r^{1} = 2r$⁠. Constant multiple rule: $\frac{d}{dr}(\pi r^2) = \pi\frac{d}{dr}(r^2)$⁠. Basics: rate of change as a derivative

3. Substitute $r = 3$

   $$\left.\frac{dA}{dr}\right|_{r=3} = 2\pi(3) = 6\pi .$$

   So when the radius is 3 cm, the area is changing at $6\pi$ cm$^2$ per cm of radius, which is about 18.85 cm$^2$/cm.

   We put in $r = 3$ only after differentiating. If we substituted first we would get $A = 9\pi$⁠, a constant, whose derivative is $0$⁠: that would wrongly say the area does not change at all.
   The value of $\frac{dA}{dr}$ at $r = r_0$ is written $\left.\frac{dA}{dr}\right|_{r=r_0}$ and is the rate of change at that particular radius (Section 6.2). Lesson: rate of change of quantities

4. See why the answer is the circumference

   Increase the radius from $r$ to $r + \Delta r$⁠. The area added is a thin ring:

   $$\begin{aligned} \Delta A &= \pi(r+\Delta r)^2 - \pi r^2 \\ &= 2\pi r\,\Delta r + \pi(\Delta r)^2 . \end{aligned}$$

   Dividing by $\Delta r$ gives $\dfrac{\Delta A}{\Delta r} = 2\pi r + \pi\,\Delta r$⁠, and as $\Delta r \to 0$ this tends to $2\pi r$⁠. At $r = 3$ with $\Delta r = 0.1$⁠: the ring has area $\pi(3.1^2 - 3^2) = 0.61\pi$⁠, so $\dfrac{\Delta A}{\Delta r} = 6.1\pi$⁠, already close to $6\pi$⁠.

   This explains the formula rather than just computing it. A thin ring is almost a long thin strip whose length is the circumference $2\pi r$ and whose width is $\Delta r$⁠. The leftover term $\pi(\Delta r)^2$ becomes negligible compared with $\Delta r$ as $\Delta r$ shrinks, and the derivative is the limit of $\frac{\Delta A}{\Delta r}$⁠.
   The derivative is the limit of the average rate $\frac{\Delta A}{\Delta r}$ as $\Delta r \to 0$ (definition of the derivative, Section 5.3). Basics: rate of change as a derivative

Part (b): the rate when $r = 4$ cm

1. Reuse the general rate

   Part (a), Step 2 found the rate for every radius:

   $$\frac{dA}{dr} = 2\pi r .$$
   The formula $2\pi r$ was derived for a general $r$⁠, before any value was substituted. That is exactly why it can be reused: only the substitution changes.
   Differentiate once in general, then evaluate at as many points as the question asks. Lesson: rate of change of quantities

2. Substitute $r = 4$

   $$\left.\frac{dA}{dr}\right|_{r=4} = 2\pi(4) = 8\pi .$$

   So when the radius is 4 cm, the area is changing at $8\pi$ cm$^2$ per cm of radius, about 25.13 cm$^2$/cm.

   The rate is positive, so the area increases as the radius increases (the Note after Example 3 in Section 6.2: a positive $\frac{dy}{dx}$ means $y$ increases as $x$ increases).
   Units: area (cm$^2$⁠) per radius (cm), so cm$^2$/cm. This is not cm$^2$/s: no time is involved in this question. (Example 1 of Section 6.2 words the same calculation at $r = 5$ "per second" and states its answer $10\pi$ in cm$^2$/s, but the quantity it computes is $\frac{dA}{dr}$⁠, area per unit of radius; the answer key gives this question's answers in cm$^2$/cm.)

3. Compare the two answers

   The rate grew from $6\pi$ to $8\pi$ when the radius grew from 3 cm to 4 cm. Because $\frac{dA}{dr} = 2\pi r$ is proportional to $r$⁠, a bigger circle gains more area for each extra centimetre of radius.

   The rate of change of the area is not a fixed number: it depends on the radius at which we measure it. That is why the question names a radius in each part. In the figure, the ring around the larger circle is longer, so the same small width adds more area.
   A rate of change is a function of the point where it is measured; a single number is that function evaluated at one point. Basics: rate of change as a derivative

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