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Chapter 6 · Application of Derivatives

Exercise 6.1 · Q12: Air bubble growing at 12\tfrac12 cm/s: volume rate at r=1r = 1 cm

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

The radius of an air bubble is increasing at the rate of 12\dfrac{1}{2} cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Exam answer

Let rr cm be the radius and VV cm³ the volume at time tt s. Given drdt=12\frac{dr}{dt} = \frac12 cm/s.

V=43πr3V = \frac43\pi r^3

Differentiating w.r.t. tt⁠,

dVdt=4πr2 drdt\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}

At r=1r = 1⁠,

dVdt=4π(1)2⋅12=2π\frac{dV}{dt} = 4\pi(1)^2\cdot\frac12 = 2\pi

∴\therefore the volume is increasing at 2π2\pi cm³/s.

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

The one idea this problem is built on

An air bubble is a sphere, so its volume is V=43πr3V = \frac{4}{3}\pi r^3 at every moment. Differentiating with respect to time by the chain rule gives dVdt=4πr2 drdt\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}⁠: the volume's rate is the radius's rate multiplied by 4πr24\pi r^2⁠, which is the bubble's surface area. Put in r=1r = 1 and drdt=12\frac{dr}{dt} = \frac12⁠.

Given, to find, and the plan
Given
The bubble is a sphere of radius rr cm at time tt s, with drdt=12\frac{dr}{dt} = \frac12 cm/s.
To find
dVdt\frac{dV}{dt} at the instant when r=1r = 1 cm.
Plan
(1) Write the volume formula. (2) Differentiate with respect to tt (chain rule). (3) Substitute r=1r = 1⁠, drdt=12\frac{dr}{dt} = \frac12⁠. (4) State the answer with units and direction.

Basics you need

1Volume of a sphere

A sphere of radius rr has volume V=43πr3V = \frac{4}{3}\pi r^3 and surface area 4πr24\pi r^2⁠. Basics: area and volume formulas

2Related rates

If rr depends on tt⁠, then ddt(r3)=3r2 drdt\frac{d}{dt}\left(r^3\right) = 3r^2\,\frac{dr}{dt} (chain rule). Basics: related rates (chain rule)

3Rate of change and its unit

dVdt\frac{dV}{dt} is measured in units of volume per unit of time, here cm3^3/s. Basics: rate of change as a derivative

Solution

  1. Name the quantities and the relation

    At time tt seconds let the radius be rr cm and the volume VV cm3^3⁠. The bubble stays spherical, so at every moment

    V=43πr3.V = \frac{4}{3}\pi r^3.

    We are told drdt=12\frac{dr}{dt} = \frac12 cm/s (positive, since the radius is increasing).

    A related-rates problem needs an equation between the quantities that holds at all times. The shape is fixed (a sphere), so the volume is completely determined by the radius through this formula.
    Volume of a sphere: 43πr3\frac{4}{3}\pi r^3⁠. Basics: area and volume formulas

  2. Differentiate with respect to time

    VV depends on tt through rr⁠. By the chain rule,

    dVdt=dVdr⋅drdt=43π⋅3r2⋅drdt=4πr2 drdt.\begin{aligned} \frac{dV}{dt} &= \frac{dV}{dr}\cdot\frac{dr}{dt} \\ &= \frac{4}{3}\pi\cdot 3r^2\cdot\frac{dr}{dt} \\ &= 4\pi r^2\,\frac{dr}{dt}. \end{aligned}
    The constant 43π\frac43\pi stays in front, r3r^3 differentiates to 3r23r^2 with respect to rr⁠, and the chain rule multiplies by drdt\frac{dr}{dt} because rr itself is changing with time. The 3's cancel, leaving 4πr24\pi r^2⁠, the surface area of the sphere: a thin new layer of thickness drdr spread over the surface adds about 4πr2 dr4\pi r^2\,dr of volume.
    Chain rule (Theorem 4, Section 5.3.1) and ddr(r3)=3r2\frac{d}{dr}\left(r^3\right) = 3r^2 (Table 5.3). This is the same pattern as Example 3 for the area of a circle. Basics: related rates (chain rule)

  3. Substitute the values at the instant

    Now put r=1r = 1 and drdt=12\frac{dr}{dt} = \frac12⁠:

    dVdt=4π(1)2⋅12=2π cm3/s.\begin{aligned} \frac{dV}{dt} &= 4\pi(1)^2\cdot\frac{1}{2} \\ &= 2\pi \text{ cm}^3/\text{s}. \end{aligned}

    As a decimal, 2π≈6.282\pi \approx 6.28 cm3^3/s.

    We substitute only after differentiating: the radius is 1 cm only at that one instant, not at all times. Units: 4πr24\pi r^2 is in cm2^2 and drdt\frac{dr}{dt} in cm/s, so the product is in cm3^3/s, a volume per second as it should be.

  4. State the direction

    dVdt=2π>0\frac{dV}{dt} = 2\pi > 0⁠, so the volume is increasing, at 2π2\pi cm3^3/s at that instant.

    The sign agrees with common sense: a growing radius means a growing bubble. The rate is not constant: since dVdt=4πr2⋅12=2πr2\frac{dV}{dt} = 4\pi r^2\cdot\frac12 = 2\pi r^2⁠, it gets larger as the bubble grows, even though the radius grows at a steady 12\frac12 cm/s.
    A positive rate means an increasing quantity (Note after Example 3, Section 6.2). Lesson: rate of change of quantities
Result

When the radius is 1 cm, the volume of the bubble is increasing at 2π2\pi cm3^3/s (about 6.28 cm3^3/s). (Textbook answer: 2π2\pi cm3^3/s.)

Check it with numbers

Let the radius be exactly 1 cm at t=0t = 0⁠. After a further 0.001 s it has grown by 12(0.001)=0.0005\frac12(0.001) = 0.0005 cm, to 1.0005 cm. The volume goes from 43π(1)3≈4.188790\frac43\pi(1)^3 \approx 4.188790 to 43π(1.0005)3≈4.195077\frac43\pi(1.0005)^3 \approx 4.195077 cm3^3⁠, a gain of about 0.006286 cm3^3 in 0.001 s. The average rate over that short time is about 6.29 cm3^3/s, very close to 2π≈6.28322\pi \approx 6.2832⁠. (It is a little above, because the rate itself is rising as the bubble grows.)

Common mistakes
  1. Forgetting drdt\frac{dr}{dt}⁠. 4πr24\pi r^2 at r=1r = 1 is 4π≈12.574\pi \approx 12.57⁠; that is dVdr\frac{dV}{dr}⁠, the rate per cm of radius, not per second.
  2. Substituting r=1r = 1 before differentiating. Then V=43πV = \frac43\pi looks constant and its rate wrongly becomes 0.
  3. Using the surface-area or circle formula for the volume. A bubble is a sphere: V=43πr3V = \frac43\pi r^3⁠.
  4. Wrong units. A volume rate is in cm3^3/s, not cm/s or cm2^2/s.
What to take away
  1. For a sphere, dVdt=4πr2 drdt\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}⁠: the surface area times the rate of the radius.
  2. Write the formula valid at all times, differentiate with respect to tt⁠, and substitute the instant's values last.
  3. A constant drdt\frac{dr}{dt} does not mean a constant dVdt\frac{dV}{dt}⁠: the volume rate grows with r2r^2⁠.

Every fact used, and where it comes from

FactUsed inSource
Volume of a sphere 43πr3\frac43\pi r^3⁠; surface area 4πr24\pi r^2Steps 1, 2Mensuration (Classes IX and X, Surface Areas and Volumes)
Chain ruleStep 2Theorem 4, Section 5.3.1; used the same way in Examples 2 and 3, Section 6.2
ddr(r3)=3r2\frac{d}{dr}\left(r^3\right) = 3r^2Step 2Table 5.3, Section 5.3
Positive rate means increasingStep 4Note after Example 3, Section 6.2 (Lesson: rate of change of quantities)
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