Worked Solutions
Rates of Change — Worked Solutions (Preliminary Maths Extension 1)
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Worked examples for Preliminary Maths Extension 1 rates of change. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
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Example 1 — Related rates
Question
Air is pumped into a spherical balloon at a rate of $50 \text{ cm}^3/\text{s}$. Find the rate at which the radius is increasing at the instant the radius is $5 \text{ cm}$. (Volume of a sphere: $V = \tfrac{4}{3}\pi r^3$.)
Solution
Related rates: link the rates with the chain rule. We want $\dfrac{dr}{dt}$ and we're given $\dfrac{dV}{dt} = 50$.
Differentiate $V = \tfrac{4}{3}\pi r^3$: $\dfrac{dV}{dr} = 4\pi r^2$.
Chain rule: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt} \Rightarrow 50 = 4\pi r^2 \cdot \dfrac{dr}{dt}$.
At $r = 5$: $50 = 4\pi(25)\dfrac{dr}{dt} = 100\pi \dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{50}{100\pi} = \dfrac{1}{2\pi} \approx 0.159 \text{ cm/s}$.
Substitute the radius only after building the chain-rule equation — never before.
"Related rates" means two quantities are changing together, linked by an equation, and the chain rule ties their rates. We know how fast volume grows, $\dfrac{dV}{dt} = 50$, and we want how fast the radius grows, $\dfrac{dr}{dt}$.
Start from the volume formula and differentiate with respect to $r$: $\dfrac{dV}{dr} = 4\pi r^2$. The chain rule says $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$, which connects everything.
So $50 = 4\pi r^2 \cdot \dfrac{dr}{dt}$. Now — and only now — substitute the instant $r = 5$: $50 = 100\pi \cdot \dfrac{dr}{dt}$, giving $\dfrac{dr}{dt} = \dfrac{1}{2\pi} \approx 0.159 \text{ cm/s}$.
We wait to substitute $r = 5$ because it's a snapshot in time; the equation must hold for all $r$ before we freeze it at one moment.
Chain rule links the rates.
- $\dfrac{dV}{dr} = 4\pi r^2$
- $\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt} \Rightarrow 50 = 4\pi r^2 \dfrac{dr}{dt}$
- At $r = 5$: $50 = 100\pi \dfrac{dr}{dt}$
- $\dfrac{dr}{dt} = \dfrac{1}{2\pi} \approx 0.159 \text{ cm/s}$
Where the marks go
- 1 mark: Differentiates to get $\dfrac{dV}{dr} = 4\pi r^2$
- 1 mark: Sets up the chain rule $\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$
- 1 mark: Substitutes $r = 5$ and $\dfrac{dV}{dt} = 50$ correctly
- 1 mark: Correct answer $\dfrac{dr}{dt} = \dfrac{1}{2\pi} \approx 0.159$ cm/s
Key idea
Related rates connect $\dfrac{dV}{dt}$ and $\dfrac{dr}{dt}$ through the chain rule; build the equation first, substitute the instant last.
Example 2 — Exponential growth and decay
Question
A radioactive substance decays according to $M = M_0 e^{-kt}$, where $M$ is the mass in grams and $t$ is in years. An initial mass of $80 \text{ g}$ decays to $50 \text{ g}$ after $10$ years. Find the value of $k$, correct to four decimal places.
Solution
Use the data point to pin down $k$. Here $M_0 = 80$, and $M = 50$ when $t = 10$.
$50 = 80 e^{-10k} \Rightarrow e^{-10k} = \dfrac{50}{80} = \dfrac{5}{8}$.
Take logs: $-10k = \ln\tfrac{5}{8} \Rightarrow k = -\dfrac{1}{10}\ln\tfrac{5}{8} = \dfrac{1}{10}\ln\tfrac{8}{5}$.
$\ln\tfrac{8}{5} = \ln 1.6 \approx 0.4700$, so $k \approx 0.0470$.
Keep $k$ positive for decay — the negative sign already lives in the exponent.
The model $M = M_0 e^{-kt}$ has two unknowns, but the initial mass tells us $M_0 = 80$ straight away. The decay data — $50$ g after $10$ years — gives the equation we need.
$50 = 80 e^{-10k}$. Dividing both sides by $80$: $e^{-10k} = \tfrac{5}{8}$.
To free $k$ from the exponent we take the natural log of both sides: $-10k = \ln\tfrac{5}{8}$. Since $\tfrac{5}{8}$ is less than $1$ its log is negative, and dividing by $-10$ makes $k$ come out positive — exactly what we expect for decay: $k = -\tfrac{1}{10}\ln\tfrac{5}{8} = \tfrac{1}{10}\ln\tfrac{8}{5} \approx 0.0470$.
The natural log is the tool that undoes $e$, which is why it's always the step that isolates the rate constant.
Solve for $k$ from the data.
- $M_0 = 80$; at $t = 10$, $M = 50$
- $50 = 80 e^{-10k} \Rightarrow e^{-10k} = \tfrac{5}{8}$
- $-10k = \ln\tfrac{5}{8} \Rightarrow k = \tfrac{1}{10}\ln\tfrac{8}{5}$
- $k \approx 0.0470$
Where the marks go
- 1 mark: Substitutes the data to get $50 = 80 e^{-10k}$
- 1 mark: Takes logs to isolate $k$
- 1 mark: Correct value $k \approx 0.0470$
Key idea
Exponential growth/decay $M = M_0 e^{-kt}$ is solved by substituting a known data point and taking natural logs to release $k$ from the exponent.