Chapter 6 · Application of Derivatives

Exercise 6.1 · Q5: Ripples at 5 cm/s: rate of enclosed area at $r = 8$ cm

Exercise 6.1Rate of change of quantities · Section 6.2

The question

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Exam answer

Let $r$ cm be the radius of the wave and $A$ cm² the enclosed area at time $t$ s. Given $\frac{dr}{dt} = 5$ cm/s.

A = \pi r^2

Differentiating w.r.t. $t$ ⁠,

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

At $r = 8$ ⁠,

\frac{dA}{dt} = 2\pi(8)(5) = 80\pi

$\therefore$ the enclosed area is increasing at $80\pi$ cm²/s.

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

The one idea this problem is built on

"Waves move at 5 cm/s" is a statement about the radius of the circular wave: it grows by 5 cm each second, so $\frac{dr}{dt} = 5$ ⁠. The question asks about the area inside the wave. The chain rule links the two: $\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$ ⁠, the circumference times the speed of the wave.

Given, to find, and the plan

Given: The circular wave spreads outward at 5 cm/s, so $\dfrac{dr}{dt} = 5$ cm/s. The instant of interest: $r = 8$ cm.
To find: $\dfrac{dA}{dt}$ ⁠, the rate of increase of the enclosed area, at $r = 8$ ⁠.
Plan: Write $A = \pi r^2$ ⁠, differentiate with respect to $t$ by the chain rule, then substitute $r = 8$ and $\frac{dr}{dt} = 5$ ⁠.

Basics you need

1Speed of a circular wave is the rate of its radius

A wave that moves outward at 5 cm/s is 5 cm further from the centre each second, so its radius satisfies $\frac{dr}{dt} = 5$ ⁠. Basics: rate of change as a derivative

2Area of a circle

The area enclosed by a circle of radius $r$ is $A = \pi r^2$ ⁠. Basics: area and volume formulas

3Related rates (chain rule)

$\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt}$ (Section 6.2; Theorem 4, Section 5.3.1). Basics: related rates (chain rule)

Solution

Name the quantities and translate the speed

At time $t$ seconds after the stone is dropped, let the radius of the wave be $r$ cm and the enclosed area $A$ cm $^2$ ⁠. Then
$A = \pi r^2, \qquad \frac{dr}{dt} = 5 .$
The wave is a circle centred where the stone hit the water. Its "speed" is how fast the circle's edge moves outward, that is, how fast the distance from the centre (the radius) grows. The sign is positive because the circle expands.

A speed of a moving boundary is the rate of change of its distance from a fixed point: here $\frac{dr}{dt}$ ⁠. 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 known as a formula in $r$ ⁠, and $r$ changes with $t$ ⁠. The chain rule converts "area per cm of radius" ( $2\pi r$ ⁠) into "area per second" by multiplying by "cm of radius per second" ( $\frac{dr}{dt}$ ⁠).

Chain rule (Theorem 4, Section 5.3.1) with $f(r) = \pi r^2$ ⁠, $f'(r) = 2\pi r$ ⁠. Basics: related rates (chain rule)
Substitute the instant: $r = 8$ ⁠, $\frac{dr}{dt} = 5$
$\frac{dA}{dt} = 2\pi(8)(5) = 80\pi .$
So at the instant the radius is 8 cm, the enclosed area is increasing at $80\pi$ cm $^2$ /s, about 251.33 cm $^2$ /s.

The radius 8 cm describes one instant, so it goes in after differentiating. The positive answer confirms the area increases. Units: cm $^2$ per second.

The same steps with 4 cm/s and $r = 10$ give $80\pi$ in Example 3: a different instant and speed that happen to give the same product $2\pi r\,\frac{dr}{dt}$ ⁠. Lesson: rate of change of quantities

Ripples drawn every 0.4 s after the stone lands (radii 2, 4, 6 and 8 cm); the gold ring is the wave at $r = 8$ cm. The spacing is even because the radius grows uniformly, yet each ring encloses a larger band of extra area than the one before.
Interpret the answer

At $r = 8$ the circumference is $16\pi$ cm, and the wave's edge moves outward 5 cm each second, sweeping roughly a band of area $16\pi \times 5 = 80\pi$ cm $^2$ per second at that moment. Later, when the circle is larger, the same speed sweeps out area faster.

This reading of $2\pi r\,\frac{dr}{dt}$ as (length of the moving edge) $\times$ (speed of the edge) makes the formula easy to remember and to check for sense.

For a circle, $\frac{dA}{dr} = 2\pi r$ is the circumference (Exercise 6.1 Q1). Basics: area and volume formulas

Result

$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$ ⁠, and at $r = 8$ cm the enclosed area is increasing at

\frac{dA}{dt} = 80\pi\ \text{cm}^2/\text{s} \approx 251.33\ \text{cm}^2/\text{s}.

(Textbook answer: $80\pi$ cm $^2$ /s.)

Check it with numbers

The wave has radius $r = 5t$ at time $t$ seconds after the stone lands, so $r = 8$ at $t = 1.6$ s. Then $A(t) = \pi(5t)^2 = 25\pi t^2$ ⁠, and

A'(t) = 50\pi t, \qquad A'(1.6) = 80\pi .

This direct calculation, without the chain rule, gives the same answer.

Common mistakes

Taking 5 cm/s as the area rate. The speed of the wave is $\frac{dr}{dt}$ ⁠; its units, cm/s, are those of a length per time, not an area per time.
Answering $16\pi$ ⁠. That is $\frac{dA}{dr}$ at $r = 8$ ⁠; it still has to be multiplied by $\frac{dr}{dt} = 5$ ⁠.
Substituting $r = 8$ into the area first, giving the constant $64\pi$ and a false rate of $0$ ⁠.

What to take away

Read a speed of a circular front as $\frac{dr}{dt}$ ⁠.
$\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$ ⁠: (length of the moving edge) $\times$ (its speed).
This is Exercise 6.1 Q3 in a different setting: the routine "relation, differentiate in time, substitute" is the same.

Every fact used, and where it comes from

Fact	Used in	Source
Speed of the wave is $\frac{dr}{dt}$	Step 1	Rate of change as a derivative, Section 6.2 (Basics)
$A = \pi r^2$	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)
Positive rate means increasing	Step 3	Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
Same situation with 4 cm/s and $r = 10$	Whole solution	Example 3, Section 6.2

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Exercise 6.1 · Q5: Ripples at 5 cm/s: rate of enclosed area at r=8r = 8r=8 cm

Basics you need

Solution

Name the quantities and translate the speed

Differentiate with respect to time (chain rule)

Substitute the instant: r=8r = 8r=8⁠, drdt=5\frac{dr}{dt} = 5dtdr​=5

Interpret the answer

Every fact used, and where it comes from

Exercise 6.1 · Q5: Ripples at 5 cm/s: rate of enclosed area at $r = 8$ cm

Substitute the instant: $r = 8$ ⁠, $\frac{dr}{dt} = 5$