Differentiation: Question 4

Worked solution

Part (a): Volume in terms of radius

The bubble is a sphere, so

$V = \frac{4}{3}\pi r^3$

Differentiating with respect to $r$:

$\frac{dV}{dr} = \frac{4}{3}\pi \times 3r^2 = 4\pi r^2$

Part (b): Connecting the rates with the chain rule

We are given $\dfrac{dr}{dt} = 0.3$ (constant). By the chain rule,

$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} = 4\pi r^2 \times 0.3 = 1.2\pi r^2$

Part (c): Rate of increase of volume when $r = 5$

Substituting $r = 5$ into the result from Part (b):

$\frac{dV}{dt} = 1.2\pi (5)^2 = 1.2\pi(25) = 30\pi$

As a decimal: $30\pi \approx 94.2477\ldots \approx 94.2$ cm³ per second (3 s.f.).

Independent check by direct numerical estimate. Over a short time interval $\delta t = 0.1$ s, the radius grows by $\delta r = 0.3 \times 0.1 = 0.03$ cm, from $5$ to $5.03$ cm.

$V(5) = \frac{4}{3}\pi(5)^3 = \frac{4}{3}\pi(125) = 166.6667\pi, \qquad V(5.03) = \frac{4}{3}\pi(5.03)^3 = \frac{4}{3}\pi(127.2635) = 169.6847\pi$

$\delta V = 169.6847\pi - 166.6667\pi = 3.0180\pi$

Average rate over this interval: $\dfrac{\delta V}{\delta t} = \dfrac{3.0180\pi}{0.1} = 30.180\pi$, which is very close to the exact calculus value of $30\pi$ found above (the small difference is due to using a finite, rather than infinitesimal, time step, and the volume’s convex growth). This confirms

$\frac{dV}{dt}\bigg|_{r=5} = 30\pi \approx 94.2 \text{ cm}^3\text{/s}.$

Part (d): Finding $r$ when $\dfrac{dV}{dt} = 4.8\pi$

Using $\dfrac{dV}{dt} = 1.2\pi r^2$ from Part (b):

$1.2\pi r^2 = 4.8\pi$

Divide both sides by $\pi$, then by $1.2$:

$1.2r^2 = 4.8 \implies r^2 = \frac{4.8}{1.2} = 4 \implies r = 2 \text{ (taking the positive root, since } r > 0\text{)}$

Check: substituting back, $1.2\pi(2)^2 = 1.2\pi(4) = 4.8\pi$ ✓.

Final answers

• (a) $V = \dfrac{4}{3}\pi r^3$, $\dfrac{dV}{dr} = \boxed{4\pi r^2}$
• (b) $\dfrac{dV}{dt} = \boxed{1.2\pi r^2}$
• (c) $\boxed{30\pi \approx 94.2}$ cm³ per second
• (d) $r = \boxed{2}$ cm