Further Modelling with Functions (Edexcel A-Level Mathematics): Revision Notes

Example: A taxi company charges $£3$ for the first mile and $£2$ for each additional mile. Model the cost $C(x)$ as a function of the distance $x$ (in miles) travelled. Solution:

• For $\ 0 < x \leq 1$, the cost is $£3$.
• For $\ x > 1 ,$ the cost is $\ 3 + 2(x - 1) .$ The piecewise function is: $C(x) = \begin{cases} 3 & \text{if } 0 < x \leq 1 \\ 3 + 2(x - 1) & \text{if } x > 1 \end{cases}$

Interpretation: This function reflects how the cost changes after the first mile, combining a flat rate with a linear increase.

Example: The population of a town grows at a rate of $5$% per year. If the current population is $20,000$, model the population $P(t)$ as a function of time $\ t$ (in years) and predict the population after $10$ years. Solution: The model follows the exponential growth function: $P(t) = P_0 \cdot e^{rt}$ Where:

• $\ P_0 = 20000$ (initial population)
• $\ r = 0.05$ (growth rate)
• $\ t = 10$ years Substituting the values: $P(10) = 20000 \cdot e^{0.05 \times 10} \approx 20000 \cdot e^{0.5} \approx 20000 \cdot 1.6487 = 32974$

Interpretation: After $10$ years, the population is predicted to be approximately $32,974$.

Example: A ball is thrown upwards from a height of $5$ meters with an initial velocity of $10$ $m/s$. Model the height $\ h(t)$ of the ball as a function of time $\ t ,$ considering the acceleration due to gravity $\ g = 9.8 \, \text{m/s}^2 .$ Solution: The height function is quadratic, given by: $h(t) = h_0 + v_0t - \frac{1}{2}gt^2$ Where:

• $\ h_0 = 5$ meters (initial height)
• $\ v_0 = 10$ $m/s$ (initial velocity)
• $g = 9.8 m/s^2$ Substitute the values: $h(t) = 5 + 10t - 4.9t^2$

To find when the ball hits the ground $( h(t) = 0 )$: $0 = 5 + 10t - 4.9t^2$

This is a quadratic equation: $4.9t^2 - 10t - 5 = 0$ Use the quadratic formula: $t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(4.9)(-5)}}{2(4.9)}$

Simplifying: $t = \frac{10 \pm \sqrt{100 + 98}}{9.8} = \frac{10 \pm \sqrt{198}}{9.8} \approx \frac{10 \pm 14.07}{9.8}$

This gives two solutions: $\ t = \frac{24.07}{9.8} \approx 2.46 \text{ seconds} \quad \text{(valid)}$ $\ t = \frac{-4.07}{9.8} \quad \text{(not valid, as time cannot be negative)}$

Interpretation: The ball hits the ground approximately $2.46$ seconds after it is thrown.

Example: The $pH$ of a solution is defined as $\ pH = -\log[H^+]$, where $\ [H^+]$is the hydrogen ion concentration in moles per litre. If the concentration decreases exponentially, model the pH as a function of time $\ t .$ Solution: Given $\ [H^+] = [H^+]_0 e^{-kt}$, the pH function is: $pH(t) = -\log([H^+]_0 e^{-kt}) = -\log[H^+]_0 + kt\log e$

This logarithmic model shows how pH increases as hydrogen ion concentration decreases over time.

A company manufactures electronic devices. The cost $\ C(x)$ to produce $\ x$ units is given by $\ C(x) = 500 + 20x$, and the revenue $\ R(x)$ from selling $\ x$ units is $\ R(x) = 50x - 0.1x^2 .$Determine the number of units the company must produce and sell to maximize profit.

Solution: Profit function $\ P(x) = R(x) - C(x) :$

$P(x) = (50x - 0.1x^2) - (500 + 20x)$

Simplify: $P(x) = -0.1x^2 + 30x - 500$

To maximize profit, differentiate and set to zero: $P'(x) = -0.2x + 30 = 0 \quad \Rightarrow \quad x = 150$

Verify with the second derivative: $P''(x) = -0.2 \quad (\text{Negative, so a maximum})$

Answer: The company should produce and sell $150$ $units$ to maximize profit.