Me6 - Numerical solutions in context

Me6 - Numerical solutions in context

When will a generator use its planned fuel allowance, and what can a numerical answer promise? Under Me6, you will solve a contextual root problem, estimate an accumulated quantity from a rate, and evaluate the numerical result against the model's domain and data.

Turn a physical target into a root problem

A generator's cumulative fuel use is modelled by

C(t)=3t+8(1−e−0.3t),C(t)=3t+8(1-\mathrm e^{-0.3t}),

where tt is hours after starting and CC is litres, for 0≤t≤60\le t\le6. When will it use 1515 litres? Assume the operating conditions match those used to construct the model.

You already know

A root equation has zero on one side. Continuity and a sign change establish existence; strict increase establishes at most one root.

Time occurs both outside and inside an exponential. Ordinary algebra and logarithms do not isolate it in an elementary formula, so a numerical method is appropriate.

Supported interval

A supported interval is the input range for which a model is intended to apply. A numerical root outside it is not automatically a valid physical prediction.

Subtract the target from cumulative consumption to measure excess or shortfall.

Worked example 1: locating a fuel target

Question. Form and locate the equation for the 1515-litre time.

Plan. Subtract 1515, evaluate nearby times and inspect the derivative.

F(t)=3t+8(1−e−0.3t)−15F(3)=−1.252557…<0F(4)=2.590446…>0F′(t)=3+2.4e−0.3t>0.\begin{aligned} F(t)&=3t+8(1-\mathrm e^{-0.3t})-15\\ F(3)&=-1.252557\ldots<0\\ F(4)&=2.590446\ldots>0\\ F'(t)&=3+2.4\mathrm e^{-0.3t}>0. \end{aligned}

Why subtract the target? A zero then means consumption equals the specified allowance.

Answer. Solve F(t)=0F(t)=0. Continuity gives a solution in (3,4)(3,4) hours, and strict increase makes it unique there and on the supported interval.

Check. C(0)=0C(0)=0 and C(6)=24.677609…>15C(6)=24.677609\ldots>15, so the target is attainable within the model.

Change only the allowance to 1818 litres.

Worked example 2: a larger allowance

Question. Form and locate the corresponding time equation for the same generator.

Plan. Replace the target constant and test the next useful interval.

G(t)=3t+8(1−e−0.3t)−18G(4)=−0.409554…<0G(5)=3.214959…>0G′(t)=3+2.4e−0.3t>0.\begin{aligned} G(t)&=3t+8(1-\mathrm e^{-0.3t})-18\\ G(4)&=-0.409554\ldots<0\\ G(5)&=3.214959\ldots>0\\ G'(t)&=3+2.4\mathrm e^{-0.3t}>0. \end{aligned}

Why is the derivative unchanged? The allowance changes the comparison level, not the rate of fuel use.

Answer. There is exactly one supported solution, in (4,5)(4,5) hours.

Check. More allowed fuel gives a later time because cumulative use increases.

A bracket helps choose a starting value. Newton–Raphson is convenient when the derivative can be evaluated reliably; repeated interval halving retains a continuous sign bracket without a derivative. A fixed point method additionally needs a useful rearrangement and observed or justified convergence.

Watch out

A derivative sign establishes at most one solution, not existence. Also compare the target with the range attained during the supported interval.

Change only the allowance again.

Compute a time with a justified stopping decision

The calculation should remain readable in its units: F(t)F(t) is litres, F′(t)F'(t) is litres per hour, and their ratio is hours.

Newton–Raphson in context

Provided F′(tn)≠0F'(t_n)\ne0,

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

In the formula sheet.

Use stored calculator values throughout, then verify the requested accuracy in the target equation.

Worked example 3: the planned fuel time

Question. Use Newton–Raphson from t0=4t_0=4 for the 1515-litre target and justify the answer to two decimal places.

Plan. Calculate successive times, then evaluate the rounding boundaries.

tn+1=tn−3tn+8(1−e−0.3tn)−153+2.4e−0.3tnt1=3.304179568…t2=3.318662919…F(3.315)=−0.014266820…<0F(3.325)=0.024597677…>0.\begin{aligned} t_{n+1}&=t_n-\frac{3t_n+8(1-\mathrm e^{-0.3t_n})-15}{3+2.4\mathrm e^{-0.3t_n}}\\ t_1&=3.304179568\ldots\\ t_2&=3.318662919\ldots\\ F(3.315)&=-0.014266820\ldots<0\\ F(3.325)&=0.024597677\ldots>0. \end{aligned}

Why evaluate the target equation again? A numerical pattern suggests a candidate; a continuous sign bracket establishes the rounding.

Answer. The model predicts 1515 litres used after 3.323.32 hours.

Check. A further iterate is 3.318670118…3.318670118\ldots, consistent with both the bracket and supported interval.

Change only the allowance to 1818 litres, retaining the start and requested precision.

Worked example 4: adapting the allowance

Question. Find and justify the 1818-litre time to two decimal places.

Plan. Change the numerator's constant, then test the appropriate half-unit boundaries.

tn+1=tn−3tn+8(1−e−0.3tn)−183+2.4e−0.3tnt1=4.110010321…t2=4.110361170…G(4.105)=−0.019835740…<0G(4.115)=0.017158271…>0.\begin{aligned} t_{n+1}&=t_n-\frac{3t_n+8(1-\mathrm e^{-0.3t_n})-18}{3+2.4\mathrm e^{-0.3t_n}}\\ t_1&=4.110010321\ldots\\ t_2&=4.110361170\ldots\\ G(4.105)&=-0.019835740\ldots<0\\ G(4.115)&=0.017158271\ldots>0. \end{aligned}

Why does the bracket change? It must enclose the new target's root, not the previous one.

Answer. The model predicts 4.114.11 hours.

Check. It is later than the 1515-litre time and remains below the supported six-hour limit.

In a spreadsheet, retain columns for time, function value, derivative and next time. A calculation should stop for an undefined step or zero denominator and flag departure from the physical domain; a maximum iteration count prevents endless cycling.

For comparison, the delivery equation 5t+10(1−e−0.2t)=255t+10(1-\mathrm e^{-0.2t})=25, with minutes and litres, gives 3.914210120…3.914210120\ldots minutes. The numerical method transfers, but the units and model do not.

Exam tip

State the recurrence, starting value, relevant iterates and interpreted answer. Give the requested precision; extra displayed digits do not certify the physical model.

Use the final allowance with the same start.

Convert a rate estimate into a total quantity

Another generator's fuel-use rate is modelled by q(t)=6e−t2/9q(t)=6\mathrm e^{-t^2/9} litres per hour for 0≤t≤30\le t\le3, with time in hours. A rate at one instant is not the fuel used over the whole interval.

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