Chapter 6 · Application of Derivatives

Example 3: Ripples on a lake: how fast the enclosed area grows at $r = 10$ cm

ExamplesRate of change of quantities · Section 6.2

The question

A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm per second. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?

Exam answer

Let the radius be $r$ cm and the enclosed area $A$ cm $^2$ at time $t$ s. Given $\frac{dr}{dt} = 4$ 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)(4) = 80\pi

Hence the enclosed area is increasing at the rate of $80\pi$ cm $^2$ /s.

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

The one idea this problem is built on

The wave is a circle whose radius grows with time, and the enclosed area depends on the radius through $A = \pi r^2$ ⁠. The chain rule links the two rates: $\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}$ ⁠. The "speed of the wave" is the rate at which the radius grows, $\frac{dr}{dt}$ ⁠.

Given, to find, and the plan

Given: The circular wave moves outward at 4 cm/s, so $\frac{dr}{dt} = 4$ cm/s. The instant of interest is when $r = 10$ cm.
To find: $\dfrac{dA}{dt}$ ⁠, the rate of increase of the area inside the wave, at that instant.
Plan: Write $A = \pi r^2$ ⁠; differentiate with respect to time by the chain rule; substitute $\frac{dr}{dt} = 4$ and $r = 10$ ⁠; state the units.

Basics you need

1Area of a circle

A circle of radius $r$ encloses area $\pi r^2$ ⁠. Basics: area and volume formulas

2Related rates (chain rule)

If $A$ depends on $r$ and $r$ depends on time $t$ ⁠, 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)

3Speed as a rate

The crest of the wave is at distance $r$ from the point where the stone fell. "The wave moves at 4 cm per second" means this distance grows by 4 cm each second: $\frac{dr}{dt} = 4$ ⁠. Basics: rate of change as a derivative

Solution

Name the quantities and write the relation

At time $t$ seconds after the stone falls, let the radius of the circular wave be $r$ cm and the area it encloses be $A$ cm $^2$ ⁠. Then
$A = \pi r^2 ,$
with $r$ ⁠, and so $A$ ⁠, functions of $t$ ⁠. We know $\frac{dr}{dt} = 4$ and want $\frac{dA}{dt}$ when $r = 10$ ⁠.

The lake seen from above, drawn to scale. The crest is at radius 10 cm and moves outward at 4 cm/s; the shaded disc is the enclosed area.

The crest of the wave is a circle centred where the stone fell, so the area inside it is the area of a disc. The radius is the only thing that changes.

Basics: area and volume formulas
Differentiate with respect to time (chain rule)

Outer function $\pi r^2$ ⁠, inner function $r(t)$ ⁠:
$\frac{dA}{dt} = \frac{d}{dr}\left(\pi r^2\right)\cdot\frac{dr}{dt} = 2\pi r\,\frac{dr}{dt} .$
We know how fast $r$ changes, and $A$ is a function of $r$ ⁠. The chain rule multiplies "how fast $A$ changes per unit of $r$ " (that is $2\pi r$ ⁠) by "how fast $r$ changes per second" (that is $\frac{dr}{dt}$ ⁠), giving how fast $A$ changes per second.

$\frac{d}{dr}(\pi r^2) = 2\pi r$ ⁠, the circumference, as in Example 1. Basics: related rates (chain rule)
Substitute the instant

At the instant when $r = 10$ ⁠, with $\frac{dr}{dt} = 4$ ⁠:
$\frac{dA}{dt} = 2\pi(10)(4) = 80\pi \approx 251.33 .$
So the enclosed area is increasing at $80\pi$ cm $^2$ /s, about 251.33 cm $^2$ /s.

The radius is substituted only after differentiating, because $r = 10$ holds at one instant only. The rate is positive, so the area is increasing, as the question expects. Units: cm $^2$ of area per second.

$80\pi = 251.327\ldots$ Lesson: rate of change of quantities
Why the area rate keeps growing although the wave speed is steady

The formula $\frac{dA}{dt} = 2\pi r \times 4 = 8\pi r$ grows with $r$ ⁠. Each second the crest moves out the same 4 cm, but a bigger circle has a longer edge, so the strip of new area added each second is longer. At $r = 10$ the edge has length $20\pi$ cm and moves out at 4 cm/s, sweeping $20\pi \times 4 = 80\pi$ cm $^2$ each second.

This is the ring picture of Example 1 in motion: area is added along the circumference, at the speed of the crest.

Result

$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$ ⁠, so when $r = 10$ 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}.

This agrees with the book's solution.

Check it with numbers

If the wave starts at the point where the stone falls and moves at a steady 4 cm/s, then $r = 4t$ ⁠, and $r = 10$ when $t = 2.5$ s. Then $A(t) = \pi(4t)^2 = 16\pi t^2$ ⁠, and

A(2.5) \approx 314.16, \qquad A(2.6) \approx 339.79 .

The average rate over $[2.5, 2.6]$ is about 256.4 cm $^2$ /s, a little above 251.33 because the rate is rising. The balanced quotient over $[2.4, 2.6]$ is

\frac{\pi(10.4^2 - 9.6^2)}{0.2} = \frac{16\pi}{0.2} = 80\pi ,

exactly the answer.

Common mistakes

Answering $\frac{dA}{dr} = 20\pi$ ⁠. That is the rate with respect to the radius; the question asks "how fast", which means per second, so multiply by $\frac{dr}{dt}$ ⁠.
Substituting $r = 10$ into $A$ first, which gives the constant $100\pi$ and the false rate 0.
Treating the area rate as constant. The radius grows steadily, but $\frac{dA}{dt} = 8\pi r$ grows with $r$ ⁠.

What to take away

Speed of a spreading circle means $\frac{dr}{dt}$ ⁠.
$\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}$ ⁠: a rate in space times a rate in time.
Write the relation, differentiate in $t$ ⁠, and only then substitute the instant.

Every fact used, and where it comes from

Fact	Used in	Source
Area of a circle $\pi r^2$	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	Power and constant multiple rules, Section 5.3; Example 1
Positive rate means increasing	Step 3	Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
$80\pi \approx 251.33$ ⁠; values of $A(t)$	Step 3; Check	Arithmetic, recomputed in the site's verification script

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Example 3: Ripples on a lake: how fast the enclosed area grows 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

Why the area rate keeps growing although the wave speed is steady

Every fact used, and where it comes from

Example 3: Ripples on a lake: how fast the enclosed area grows at $r = 10$ cm