Chapter 6 · Application of Derivatives

Exercise 6.1 · Q3: Radius grows at 3 cm/s: rate of area at $r = 10$ cm

Exercise 6.1Rate of change of quantities · Section 6.2

The question

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Exam answer

Let $r$ cm be the radius and $A$ cm² the area at time $t$ s. Given $\frac{dr}{dt} = 3$ cm/s.

A = \pi r^2

Differentiating w.r.t. $t$ ⁠,

\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}

At $r = 10$ ⁠,

\frac{dA}{dt} = 2\pi(10)(3) = 60\pi

$\therefore$ the area is increasing at $60\pi$ cm²/s.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The area depends on the radius, and the radius depends on time. So the area's rate with respect to time is a product of two rates (the chain rule): $\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt}$ ⁠. The first factor, $2\pi r$ ⁠, says how much area each centimetre of radius brings at the current size; the second, 3 cm/s, says how many centimetres of radius arrive each second.

Given, to find, and the plan

Given: $\dfrac{dr}{dt} = 3$ cm/s at all times ("uniformly"). The instant of interest: $r = 10$ cm.
To find: $\dfrac{dA}{dt}$ at $r = 10$ ⁠.
Plan: Write $A = \pi r^2$ ⁠, differentiate with respect to $t$ by the chain rule, then substitute $r = 10$ and $\frac{dr}{dt} = 3$ ⁠.

Basics you need

1Area of a circle

$A = \pi r^2$ ⁠, with $\pi$ a constant. Basics: area and volume formulas

2Related rates (chain rule)

If $A$ depends on $r$ and $r$ depends on $t$ ⁠, then $\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt}$ (Section 6.2; Theorem 4, Section 5.3.1). Basics: related rates (chain rule)

3"Increasing uniformly at 3 cm/s"

The radius grows by the same 3 cm every second, so $\frac{dr}{dt} = 3$ cm/s at every instant, and it is positive because $r$ increases. Basics: rate of change as a derivative

Solution

Name the quantities and write the relation

At time $t$ seconds let the radius be $r$ cm ( $r > 0$ ⁠) and the area $A$ cm $^2$ ⁠. Then
$A = \pi r^2, \qquad \frac{dr}{dt} = 3 .$
The relation $A = \pi r^2$ holds at every instant, so it can be differentiated with respect to time. The given rate is translated into the derivative $\frac{dr}{dt}$ with a positive sign because the radius is increasing.

Translate each rate in words into a derivative with its sign and units before calculating. Basics: rate of change as a derivative
Differentiate with respect to time (chain rule)

Outer function $\pi r^2$ ⁠, inner function $r(t)$ ⁠:
$\frac{dA}{dt} = \frac{d}{dr}\left(\pi r^2\right)\cdot\frac{dr}{dt} = 2\pi r\,\frac{dr}{dt} .$
The area is not given as a formula in $t$ ⁠, only in $r$ ⁠. The chain rule lets us differentiate with respect to $t$ anyway: differentiate with respect to $r$ ⁠, then multiply by how fast $r$ changes with $t$ ⁠.

Chain rule: $\frac{d}{dt}\big(f(r)\big) = f'(r)\,\frac{dr}{dt}$ ⁠, with $f(r) = \pi r^2$ and $f'(r) = 2\pi r$ (Theorem 4, Section 5.3.1). Basics: related rates (chain rule)
Substitute the instant: $r = 10$ ⁠, $\frac{dr}{dt} = 3$
$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi .$
So when the radius is 10 cm, the area is increasing at $60\pi$ cm $^2$ /s, about 188.50 cm $^2$ /s.

Only now do we use the particular instant. The answer is positive, confirming that the area increases. Units: cm $^2$ (area) per s (time).

The same computation with 4 cm/s and $r = 10$ gives $80\pi$ in Example 3. Lesson: rate of change of quantities

The circle at the instant $r = 10$ cm (solid) and one second later, when $r = 13$ cm (dashed). The rate $60\pi$ cm $^2$ /s is the rate at the solid circle; the area grows faster and faster as the circle gets bigger.
Notice that the rate itself changes

Although $\frac{dr}{dt}$ is constant, $\frac{dA}{dt} = 2\pi r \cdot 3 = 6\pi r$ grows with $r$ ⁠. For example, at $r = 13$ (one second later) it is $78\pi$ cm $^2$ /s. That is why the question names an instant.

"Uniformly" describes the radius, not the area. Each new centimetre of radius adds a ring whose length is the circumference $2\pi r$ ⁠, and that ring is longer when the circle is bigger.

A rate of change is a function of time (or of the current size); a single answer is its value at one instant. Basics: rate of change as a derivative

Result

$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$ ⁠, and at $r = 10$ cm,

\frac{dA}{dt} = 60\pi\ \text{cm}^2/\text{s} \approx 188.50\ \text{cm}^2/\text{s}.

(Textbook answer: $60\pi$ cm $^2$ /s.)

Check it with numbers

Because the radius grows uniformly, measure time from the instant $r = 10$ ⁠: then $r = 10 + 3t$ and

\begin{aligned} A(t) &= \pi(10 + 3t)^2 \\ &= \pi\left(100 + 60t + 9t^2\right). \end{aligned}

So $A'(t) = \pi(60 + 18t)$ ⁠, and at $t = 0$ this is $60\pi$ ⁠. It agrees with the chain-rule answer.

Common mistakes

Leaving out $\frac{dr}{dt}$ ⁠. $2\pi r = 20\pi$ is the rate with respect to the radius (cm $^2$ /cm), not with respect to time.
Thinking "uniformly" makes the area rate constant. Only $\frac{dr}{dt}$ is constant; $\frac{dA}{dt} = 6\pi r$ changes with $r$ ⁠.
Substituting $r = 10$ into $A$ first, which gives the constant $100\pi$ and a false rate of $0$ ⁠.

What to take away

Related rates: $\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}$ ⁠. Find the "geometric" factor $\frac{dA}{dr}$ ⁠, then multiply by the given rate.
For a circle, $\frac{dA}{dt} = (\text{circumference})\times\frac{dr}{dt}$ ⁠.
Exercise 6.1 Q5 (ripples on a lake) is this same problem with different numbers.

Every fact used, and where it comes from

Fact	Used in	Source
$A = \pi r^2$	Step 1	Mensuration (Basics: area and volume formulas)
Chain rule for rates with respect to time	Step 2	Section 6.2; Theorem 4, Section 5.3.1 (Basics: related rates)
$\frac{d}{dr}(\pi r^2) = 2\pi r$	Step 2	Table 5.3 and algebra of derivatives, Section 5.3
Positive rate means increasing	Step 3	Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
Same method with 4 cm/s	Whole solution	Example 3, Section 6.2

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Exercise 6.1 · Q3: Radius grows at 3 cm/s: rate of area at r=10r = 10r=10 cm

Basics you need

Solution

Name the quantities and write the relation

Differentiate with respect to time (chain rule)

Substitute the instant: r=10r = 10r=10⁠, drdt=3\frac{dr}{dt} = 3dtdr​=3

Notice that the rate itself changes

Every fact used, and where it comes from

Exercise 6.1 · Q3: Radius grows at 3 cm/s: rate of area at $r = 10$ cm

Substitute the instant: $r = 10$ ⁠, $\frac{dr}{dt} = 3$