Example 35: Heated disc, radius growing at 0.05 cm/s: rate of increase of area at $r = 3.2$ cm

Basics you need

Solution

1. Name the quantities

   At time $t$ seconds, let $r$ be the radius of the disc (cm) and $A$ its area (cm²). We are told $\frac{dr}{dt} = 0.05$ cm/s, and we want $\frac{dA}{dt}$ at the instant when $r = 3.2$ cm.

   Writing down exactly which rate is known and which is wanted, with units, keeps the substitution in Step 4 honest. The answer will be in cm² per second.
   The starting radius of 3 cm only describes the situation; the rate is asked at $r = 3.2$ cm, so 3 cm is not used in the calculation.

2. The relation between area and radius

   $$A = \pi r^2.$$
   This holds at every instant of the heating, so we may differentiate it with respect to time.
   Area of a circle. Basics: area and volume formulas

3. Differentiate with respect to time (chain rule)

   $$\frac{dA}{dt} = \frac{d}{dr}\left(\pi r^2\right)\cdot\frac{dr}{dt} = 2\pi r\,\frac{dr}{dt}.$$

   The outer function is $r \mapsto \pi r^2$ and the inner function is $t \mapsto r(t)$⁠.

   The area depends on time only through the radius, so the chain rule is the tool. The factor $2\pi r$ is the circumference: when the radius grows by a little, a thin ring of length about $2\pi r$ is added.
   Chain rule; power rule $\frac{d}{dr}r^2 = 2r$⁠. Basics: related rates

4. Substitute the instant

   With $r = 3.2$ cm and $\frac{dr}{dt} = 0.05$ cm/s:

   $$\begin{aligned} \frac{dA}{dt} &= 2\pi(3.2)(0.05) \\ &= 0.32\pi\ \text{cm}^2/\text{s} \\ &\approx 1.0053\ \text{cm}^2/\text{s}. \end{aligned}$$

   So the area is increasing at $0.32\pi$ cm²/s, about 1.005 cm² every second, at that instant.

   We differentiate first and substitute afterwards; substituting $r = 3.2$ into $A = \pi r^2$ first would give the constant $10.24\pi$⁠, whose derivative is 0. The units multiply as cm × (cm/s) = cm²/s.
   $2 \times 3.2 \times 0.05 = 6.4 \times 0.05 = 0.32$⁠.

5. About the textbook's "approximate" wording

   The textbook writes the same calculation with a time step $\Delta t$⁠: the change in radius over a short time $\Delta t$ is about $\frac{dr}{dt}\Delta t$⁠, and the change in area is about $dA = 2\pi r\left(\frac{dr}{dt}\Delta t\right)$⁠. Taking $\Delta t = 1$ second gives $2\pi(3.2)(0.05) = 0.320\pi$ cm² in that second. The number is the same as the exact instantaneous rate found in Step 4, $\frac{dA}{dt} = 0.32\pi$ cm²/s.

   The question asks for a rate at an instant, and that is exactly $\frac{dA}{dt}$⁠; no approximation is needed. The book's $\Delta t$ version describes the approximate increase over the next second. For comparison, if the radius keeps growing at 0.05 cm/s, the actual increase in that second, from $r = 3.2$ to $r = 3.25$⁠, is $\pi(3.25^2 - 3.2^2) = 0.3225\pi$ cm², very close to $0.32\pi$⁠.
   For a small change, (change in $A$⁠) $\approx$ (rate of change of $A$⁠) $\times$ (elapsed time). Basics: rate of change as a derivative

Every fact used, and where it comes from