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Chapter 6 · Application of Derivatives

Exercise 6.1 · Q6: Radius grows at 0.7 cm/s: rate of the circumference

Exercise 6.1Rate of change of quantities · Section 6.2
The question

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Exam answer

Let rr cm be the radius and CC cm the circumference at time tt s. Given drdt=0.7\frac{dr}{dt} = 0.7 cm/s.

C=2πrC = 2\pi r

Differentiating w.r.t. tt⁠,

dCdt=2π drdt=2π(0.7)=1.4π\frac{dC}{dt} = 2\pi\,\frac{dr}{dt} = 2\pi(0.7) = 1.4\pi

∴\therefore the circumference is increasing at 1.4π1.4\pi cm/s.

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

The one idea this problem is built on

The circumference C=2πrC = 2\pi r is a fixed multiple of the radius. When one quantity is a constant multiple of another, their rates are in the same ratio: dCdt=2π drdt\frac{dC}{dt} = 2\pi\,\frac{dr}{dt}⁠. No radius is given because none is needed: the answer is the same at every instant.

Given, to find, and the plan
Given
drdt=0.7\dfrac{dr}{dt} = 0.7 cm/s.
To find
dCdt\dfrac{dC}{dt}⁠, the rate of increase of the circumference.
Plan
Write C=2πrC = 2\pi r⁠, differentiate with respect to tt⁠, substitute the given rate, and explain why the answer does not depend on rr⁠.

Basics you need

1Circumference of a circle

A circle of radius rr has circumference C=2πrC = 2\pi r⁠. Basics: area and volume formulas

2Related rates (chain rule)

dCdt=dCdr⋅drdt\dfrac{dC}{dt} = \dfrac{dC}{dr}\cdot\dfrac{dr}{dt} (Section 6.2; Theorem 4, Section 5.3.1). Basics: related rates (chain rule)

3Derivative of a linear function

ddr(kr)=k\dfrac{d}{dr}(kr) = k for a constant kk⁠: a straight line has the same slope everywhere. Basics: rate of change as a derivative

Solution

  1. Name the quantities and write the relation

    At time tt seconds let the radius be rr cm and the circumference CC cm. Then

    C=2πr,drdt=0.7.C = 2\pi r, \qquad \frac{dr}{dt} = 0.7 .
    The circumference is a length measured around the circle, and it is tied to the radius by C=2πrC = 2\pi r at every instant. The given rate is positive because the radius increases.
    Translate "increasing at 0.7 cm/s" into drdt=+0.7\frac{dr}{dt} = +0.7 with units cm/s. Basics: rate of change as a derivative

  2. Differentiate with respect to time (chain rule)

    dCdt=ddr(2πr)⋅drdt=2π drdt.\frac{dC}{dt} = \frac{d}{dr}(2\pi r)\cdot\frac{dr}{dt} = 2\pi\,\frac{dr}{dt} .
    CC is a formula in rr⁠, and rr changes with time, so the chain rule applies. The derivative of 2πr2\pi r with respect to rr is the constant 2π2\pi⁠: each extra centimetre of radius always adds exactly 2π2\pi cm of circumference.
    Chain rule (Theorem 4, Section 5.3.1), with f(r)=2πrf(r) = 2\pi r and f′(r)=2πf'(r) = 2\pi (constant multiple of ddr(r)=1\frac{d}{dr}(r) = 1⁠). Basics: related rates (chain rule)

  3. Substitute the given rate

    dCdt=2π(0.7)=1.4π.\frac{dC}{dt} = 2\pi(0.7) = 1.4\pi .

    So the circumference is increasing at 1.4π1.4\pi cm/s, about 4.40 cm/s.

    The answer is positive, so the circumference increases. Units: the circumference is a length (cm) and time is in seconds, so the rate is in cm/s. This is a length rate, not an area rate.
    A rate of change of a length with respect to time is measured in length per time. Lesson: rate of change of quantities

  4. Why no radius is needed

    The formula dCdt=2π drdt\frac{dC}{dt} = 2\pi\,\frac{dr}{dt} contains no rr⁠. Whatever the size of the circle, if the radius grows at 0.7 cm/s the circumference grows at 1.4π1.4\pi cm/s. Compare the area, whose rate 2πr drdt2\pi r\,\frac{dr}{dt} does depend on rr (Exercise 6.1 Q3).

    r = 2 C: 4π → 5.4π cm r = 5 C: 10π → 11.4π cm dashed: radius + 0.7 cm gain in C: 1.4π cm both times
    Two circles of radius 2 cm and 5 cm, each shown with its radius growing by 0.7 cm (dashed). In both cases the circumference grows by the same amount, 2π×0.7=1.4π2\pi \times 0.7 = 1.4\pi cm, because CC is proportional to rr⁠.
    C=2πrC = 2\pi r is a straight-line relation (a constant multiple), so its rate of change with respect to rr is the same at every radius. The area πr2\pi r^2 is curved in rr⁠, so its rate does change with rr⁠. This is why the question can leave out the radius.
    If y=kxy = kx with kk constant, then dydt=k dxdt\frac{dy}{dt} = k\,\frac{dx}{dt} at every instant. Basics: related rates (chain rule)
Result

dCdt=2π drdt\dfrac{dC}{dt} = 2\pi\,\dfrac{dr}{dt}⁠, so the circumference is increasing at

dCdt=1.4π cm/s≈4.40 cm/s,\frac{dC}{dt} = 1.4\pi\ \text{cm/s} \approx 4.40\ \text{cm/s},

whatever the radius is at that moment (the formula contains no rr⁠). (Textbook answer: 1.4π1.4\pi cm/s.)

Check it with numbers

Start from any radius r0r_0 and let the radius grow at 0.7 cm/s for one second. The circumference changes from 2πr02\pi r_0 to 2π(r0+0.7)2\pi(r_0 + 0.7)⁠, an increase of exactly 2π×0.7=1.4π≈4.3982\pi \times 0.7 = 1.4\pi \approx 4.398 cm, whatever r0r_0 is. With r0=2r_0 = 2⁠: from 4π4\pi to 5.4π5.4\pi⁠; with r0=5r_0 = 5⁠: from 10π10\pi to 11.4π11.4\pi⁠. Both increases are 1.4π1.4\pi⁠.

Common mistakes
  1. Looking for a missing radius. None is needed: dCdt\frac{dC}{dt} does not depend on rr⁠.
  2. Using the area formula. The circumference is 2πr2\pi r⁠, not πr2\pi r^2⁠; the question is about a length.
  3. Wrong units. The answer is in cm/s, not cm2^2/s.
What to take away
  1. When y=kxy = kx with kk constant, dydt=k dxdt\frac{dy}{dt} = k\,\frac{dx}{dt}⁠: the rates have the same ratio as the quantities.
  2. A rate that depends on the current size (like an area rate) needs an instant; a rate of a linear quantity does not.
  3. Always read the units of the asked quantity: circumference is a length.

Every fact used, and where it comes from

FactUsed inSource
C=2πrC = 2\pi rStep 1Mensuration (Basics: area and volume formulas)
Chain rule for rates with respect to timeStep 2Section 6.2; Theorem 4, Section 5.3.1 (Basics: related rates)
ddr(kr)=k\frac{d}{dr}(kr) = kStep 2Table 5.3 and algebra of derivatives, Section 5.3
Positive rate means increasingStep 3Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
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