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Rates of change — 0606

Connect two changing quantities through a shared parameter.

Notes and key methods

If \(\displaystyle A=\pi r^{2}\) and r varies with time, \(\displaystyle \frac{dA}{dt}= (\frac{dA}{dr})(\frac{dr}{dt})= \frac{2\pi rdr}{dt}\). For \(\displaystyle V= \frac{4\pi r^{3}}{3}, \frac{dV}{dt}= \frac{4\pi r^{2}dr}{dt}\). State the radius at the required instant before evaluating the rate.

Units distinguish a length rate from an area or volume rate. A negative derivative means a decrease and must not be discarded merely because the radius is positive.

Check the conditions

Three worked examples

Example 1 · Connected rates of change

A spherical bubble has radius \(\displaystyle r=3+3t \mathrm{cm}\), where t is measured in seconds. Let \(\displaystyle A=\pi r^{2}\) be its circular cross-sectional area and \(\displaystyle V= \frac{4\pi r^{3}}{3}\) its volume. Evaluate all instantaneous rates at radius 8 cm.

  1. Find the rate of change of the circular cross-sectional area at the stated radius, with units.
  2. Find the volume rate at the same instant, with units.
  3. Find the time in seconds to reach this radius.
  4. Find \(\displaystyle \frac{dV}{d}A\) at this radius.
Show the reasoning

(a)

\(\displaystyle A=\pi r^{2}\), so \(\displaystyle \frac{dA}{dt}= \frac{2\pi rdr}{dt}=\)\(48 \pi\) \(\displaystyle \frac{\mathrm{cm}^{2}}{s}\).

Answer: \(48 \pi\) \(\displaystyle \frac{\mathrm{cm}^{2}}{s}\)

(b)

Differentiate \(\displaystyle V= \frac{4\pi r^{3}}{3}\) with respect to time and use the same radius and radius rate.

Answer: \(768 \pi\) \(\displaystyle \frac{\mathrm{cm}^{3}}{s}\)

(c)

Use r=r₀+qt.

\(\displaystyle \frac{5}{3}\)

Answer: \(\displaystyle \frac{5}{3}\).

(d)

Divide the two time derivatives.

\(16\)

Answer: \(16\).

Example 2 · Connected rates of change

A spherical bubble has radius \(\displaystyle r=1+4t \mathrm{cm}\), where t is measured in seconds. Let \(\displaystyle A=\pi r^{2}\) be its circular cross-sectional area and \(\displaystyle V= \frac{4\pi r^{3}}{3}\) its volume. Evaluate all instantaneous rates at radius 6 cm.

  1. Find the rate of change of the circular cross-sectional area at the stated radius, with units.
  2. Find the volume rate at the same instant, with units.
  3. Find the time in seconds to reach this radius.
  4. Find \(\displaystyle \frac{dV}{d}A\) at this radius.
Show the reasoning

(a)

\(\displaystyle A=\pi r^{2}\), so \(\displaystyle \frac{dA}{dt}= \frac{2\pi rdr}{dt}=\)\(48 \pi\) \(\displaystyle \frac{\mathrm{cm}^{2}}{s}\).

Answer: \(48 \pi\) \(\displaystyle \frac{\mathrm{cm}^{2}}{s}\)

(b)

Differentiate \(\displaystyle V= \frac{4\pi r^{3}}{3}\) with respect to time and use the same radius and radius rate.

Answer: \(576 \pi\) \(\displaystyle \frac{\mathrm{cm}^{3}}{s}\)

(c)

Use r=r₀+qt.

\(\displaystyle \frac{5}{4}\)

Answer: \(\displaystyle \frac{5}{4}\).

(d)

Divide the two time derivatives.

\(12\)

Answer: \(12\).

Example 3 · Connected rates of change

A spherical bubble has radius \(\displaystyle r=2+1t \mathrm{cm}\), where t is measured in seconds. Let \(\displaystyle A=\pi r^{2}\) be its circular cross-sectional area and \(\displaystyle V= \frac{4\pi r^{3}}{3}\) its volume. Evaluate all instantaneous rates at radius 7 cm.

  1. Find the rate of change of the circular cross-sectional area at the stated radius, with units.
  2. Find the volume rate at the same instant, with units.
  3. Find the time in seconds to reach this radius.
  4. Find \(\displaystyle \frac{dV}{d}A\) at this radius.
Show the reasoning

(a)

\(\displaystyle A=\pi r^{2}\), so \(\displaystyle \frac{dA}{dt}= \frac{2\pi rdr}{dt}=\)\(14 \pi\) \(\displaystyle \frac{\mathrm{cm}^{2}}{s}\).

Answer: \(14 \pi\) \(\displaystyle \frac{\mathrm{cm}^{2}}{s}\)

(b)

Differentiate \(\displaystyle V= \frac{4\pi r^{3}}{3}\) with respect to time and use the same radius and radius rate.

Answer: \(196 \pi\) \(\displaystyle \frac{\mathrm{cm}^{3}}{s}\)

(c)

Use r=r₀+qt.

\(5\)

Answer: \(5\).

(d)

Divide the two time derivatives.

\(14\)

Answer: \(14\).

Common mistakes

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