1.09g - Numerical solutions in context | OCR A-Level Mathematics A

When will a pump deliver its target volume, and how accurately can a calculation answer that question? Under 1.09g, you will turn a physical target into a numerical root, estimate totals from rates, and separate numerical precision from the assumptions and measurements behind a model.

Turn a physical target into a root problem

When will this pump have delivered $25$ litres? Its cumulative delivery is modelled by

V(t)=5t+10(1-\mathrm e^{-0.2t}),

where $t$ is in minutes and $V$ in litres, for $0\le t\le5$ . Assume these tested operating conditions continue and none of the delivered water leaks away.

You already know

A numerical root method solves an equation with zero on one side. A continuous sign change locates a root, while a positive derivative establishes strict increase and at most one root.

Here the time appears both linearly and inside an exponential. Ordinary algebra and logarithms do not isolate it in an elementary formula, so numerical solving is appropriate.

Model domain

The model domain is the range of inputs for which the stated assumptions are supported. An algorithm returning a time outside $0\le t\le5$ does not justify using this delivery model there.

The target becomes a volume excess or shortfall.

Worked example 1: translating a target

Question. Form and locate the $25$ -litre root.

Plan. Subtract the target, test nearby times and inspect the delivery rate.

\begin{aligned} F(t)&=5t+10(1-\mathrm e^{-0.2t})-25\\ F(3)&=-5.488116\ldots<0\\ F(4)&=0.506710\ldots>0\\ F'(t)&=5+2\mathrm e^{-0.2t}>0. \end{aligned}

Why subtract $25$ ? The required time is where modelled delivery minus target equals zero.

Answer. Solve $F(t)=0$ . Continuity gives a root in $(3,4)$ minutes, and strict increase makes it unique.

Check. $V(0)=0$ , and $V(5)=31.321206\ldots>25$ ; the target is reached within the supported interval.

Change only the target to $30$ litres.

Worked example 2: a later target

Question. Form and locate the $30$ -litre root for the same pump.

Plan. Change only the subtracted target and move the bracket as necessary.

\begin{aligned} G(t)&=5t+10(1-\mathrm e^{-0.2t})-30\\ G(4)&=-4.493290\ldots<0\\ G(5)&=1.321206\ldots>0\\ G'(t)&=5+2\mathrm e^{-0.2t}>0. \end{aligned}

Why is the derivative unchanged? Changing a target changes the constant in the root equation, not the delivery rate.

Answer. There is exactly one solution in $(4,5)$ minutes.

Check. A larger target must take longer because the delivered volume strictly increases.

Newton–Raphson is convenient when the derivative is easy and a nearby start is available. Bisection repeatedly halves a continuous sign bracket and retains a root when a reliable derivative is unavailable. A fixed point rearrangement needs a convergence check.

Watch out

A positive derivative gives at most one root; it does not alone establish that the target is reached. Compare the target with the model's attainable values too.

Change only the target once more.

Compute a time with a justified stopping decision

The physical interpretation must survive every step: inputs are minutes, function values are litres, and the Newton correction is a time.

Newton correction in context

For a target function $F$ with nonzero derivative,

t_{n+1}=t_n-\frac{F(t_n)}{F'(t_n)}.

In the formula sheet.

Use the original target equation for a final sign check, rather than trusting repeated displayed digits.

Worked example 3: the opening delivery time

Question. Use Newton–Raphson from $t_0=4$ for the $25$ -litre target, and justify the answer to two decimal places.

Plan. Iterate with retained precision, then test the rounding boundaries.

\begin{aligned} t_{n+1}&=t_n-\frac{5t_n+10(1-\mathrm e^{-0.2t_n})-25}{5+2\mathrm e^{-0.2t_n}}\\ t_1&=3.914097348\ldots\\ t_2&=3.914210120\ldots\\ F(3.905)&=-0.054478344\ldots<0\\ F(3.915)&=0.004671460\ldots>0 \end{aligned}

Why test half-unit boundaries? They verify the exact two-decimal rounding interval.

Answer. According to the model, $25$ litres have been delivered after $3.91$ minutes.

Check. The time lies in the initial bracket and the supported domain; $F(t_2)\approx-1.16\times10^{-9}$ litres.

Change only the target to $30$ , retaining $t_0=4$ , the method and requested precision.

Worked example 4: adapting the calculation

Question. Find and justify the $30$ -litre time to two decimal places.

Plan. Replace the target in the numerator and evaluate the new rounding boundaries.

\begin{aligned} t_{n+1}&=t_n-\frac{5t_n+10(1-\mathrm e^{-0.2t_n})-30}{5+2\mathrm e^{-0.2t_n}}\\ t_1&=4.761747790\ldots\\ t_2&=4.770340671\ldots\\ G(4.765)&=-0.030825390\ldots<0\\ G(4.775)&=0.026878554\ldots>0 \end{aligned}

Why evaluate $G$ , not $F$ ? The two targets have different roots.

Answer. The modelled time is $4.77$ minutes.

Check. It lies between $4$ and $5$ , as the earlier bracket required.

A spreadsheet can show time, function value, derivative and next time in separate columns. Stop for an undefined step, a zero derivative or departure from the valid domain; include a maximum iteration count. None of these safeguards proves the physical model correct.

For comparison, $H(t)=4t+6(1-\mathrm e^{-0.5t})-20$ , from $4$ , gives $3.730371$ and $3.732100$ to six decimal places. A different delivery model requires its own derivative denominator.

Exam tip

State the method, start, iterates and interpreted unit. A root accurate to $0.01$ minute for a formula is not automatically a real delivery prediction accurate to $0.01$ minute.

Use the remaining target with the same start.

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1.09g - Numerical solutions in context

Turn a physical target into a root problem

Compute a time with a justified stopping decision

Continue this lesson.

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