01 · Form a different target
Write the equation to find when this tank reaches 30 m³.
Hint
Subtract the new target from V(t).
Worked solution
6t + 8(1 − e⁻ᵗ) − 30 = 0, with 0 ≤ t ≤ 6.
Understand · explore · practise
Numerical methods in modelling
Apply numerical methods to original volume and temperature models. Form a target equation, choose a physical interval, compare Newton and fixed-point estimates, verify rounding and interpret answers with units and model limitations.
Before you startNewton–Raphson; fixed-point iteration; root accuracy
01 / Translate the target into an equation
Identify the quantity, variable and valid interval first.
An illustrative tank model gives V(t) = 6t + 8(1 − e⁻ᵗ), where V is volume in cubic metres and t is time in hours. The model is intended only for 0 ≤ t ≤ 6.
To find when the volume reaches T cubic metres, solve f(t) = V(t) − T = 0. The unknown is time, not volume. A useful final answer must therefore include hours and remain inside the stated interval.
When does the tank reach its target?Explore
Start at t₀ = target/6 hours.
Compare the volume at the current estimate with the target.
Each selection performs the chosen number of substitutions with retained precision.
The time must remain within the model interval 0 ≤ t ≤ 6.
There is no automatic advance. These calculations use an invented illustrative model; a numerical root describes the model and is not a guarantee about a real tank.
02 / Establish that the required solution exists
A numerical method should target a physically relevant root.
Find when the illustrative tank reaches 20 m³.Worked example
f(t) = 6t + 8(1 − e⁻ᵗ) − 20
Subtract the target volume.
f(2) ≈ −1.082682266; f(3) ≈ 5.601703453
The target is crossed between 2 and 3 hours.
The function is continuous on [2, 3]
The opposite signs guarantee a root there.
f′(t) = 6 + 8e⁻ᵗ > 0
The volume is strictly increasing, so the root is unique.
The bracket lies inside 0 ≤ t ≤ 6
It is relevant to the model’s intended time interval.
02 · Interpret the derivative
Give V′(0) and its units.
Hint
Differentiate with respect to hours.
Worked solution
V′(t) = 6 + 8e⁻ᵗ, so V′(0) = 14 m³ per hour. It is the initial filling rate.
03 / Apply Newton to the target equation
The derivative of the constant target is zero.
Use the 20 m³ target with t₀ = 20/6 hours.Worked example
tₙ₊₁ = tₙ − [6tₙ + 8(1 − exp(−tₙ)) − 20]/[6 + 8exp(−tₙ)]
Evaluate the target function and its derivative at the same input.
t₁ ≈ 2.105946382997
The first tangent estimate is within the physical interval.
t₂ ≈ 2.154436865963
Retain stored precision.
t₃ ≈ 2.154599491459
The estimates suggest about 2.15 hours; verify the rounding separately.
See a volume target become a time estimate
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Does the target affect the derivative?
If the target changes from 20 to 30 m³, what changes in the Newton formula?
Hint
A constant has zero derivative.
Worked solution
The numerator uses −30 instead of −20. The denominator remains 6 + 8exp(−tₙ).
04 · Check an output against the model
A numerical procedure returns t = −0.2 hours. May this be accepted for the tank question as stated?
Hint
Look at the intended interval.
Worked solution
No. The model is intended for 0 ≤ t ≤ 6. A negative time is outside the stated situation even if it solves some algebraic continuation.
04 / Compare an alternative rearrangement
Use the same equation and the same target.
Rearrange the tank target equation for fixed-point iteration.Worked example
6t = T − 8 + 8e⁻ᵗ
Isolate the linear term.
tₙ₊₁ = [T − 8 + 8exp(−tₙ)]/6
This uses no derivative during the numerical update.
For T = 20, start with t₀ = 20/6
Compare the same starting value in the model above.
The two methods need not give the same intermediate estimates
They are different procedures for the same root.
Agreement is useful, but does not replace a proof of accuracy
Check the final answer with the original continuous target function.
05 · A fixed point satisfies the target
Show that a fixed point of this recurrence satisfies V(t) = T.
Hint
Replace tₙ and tₙ₊₁ by the same value α and rearrange.
Worked solution
6α = T − 8 + 8e⁻ᵅ, so 6α + 8(1 − e⁻ᵅ) = T, which is V(α) = T.
06 · Method agreement
Both methods display 2.1546 hours. Does agreement alone prove that the true model root rounds to this value?
Hint
The two computations could share rounding or convergence limitations.
Worked solution
No. Use a valid sign bracket within the corresponding halfway boundaries, or another justified error bound.
05 / Verify accuracy in the original model equation
Use rounding boundaries for the quantity you will report.
Prove the 20 m³ time is 2.15 hours to 2 d.p.Worked example
The halfway boundaries are 2.145 and 2.155 hours
These surround 2.15 at two decimal places.
f(2.145) ≈ −0.06654429635
The model volume is below 20.
f(2.155) ≈ 0.002774475097
The model volume is above 20.
Continuity and strict increase give one root between them
The entire bracket rounds to 2.15 hours.
State: according to the model, the target is reached after 2.15 hours
This is a verified numerical result within the model.
07 · Different requested precision
To verify 2.155 hours to 3 d.p., which two time values should you test?
Hint
Use half a thousandth.
Worked solution
2.1545 and 2.1555 hours. Check the signs of the original target function at these boundaries before making the accuracy claim.
06 / Convert the time without changing its meaning
Decimal hours are not clock notation.
Interpret the estimate t ≈ 2.15459949 hours.Worked example
The fractional part is about 0.15459949 hours
Subtract the two whole hours.
Multiply by 60 to obtain about 9.2759694 minutes
This is 9 whole minutes plus a fraction.
Multiply the remaining fraction by 60 to obtain about 16.558 seconds
The estimate is about 2 h 9 min 17 s.
2.15 hours is 2 h 9 min
It does not mean 2 h 15 min. Match the precision to the question and data.
08 · Convert decimal hours
Convert 1.4 hours into hours and minutes.
Hint
Multiply 0.4 by 60.
Worked solution
1 hour 24 minutes.
09 · Change the variable unit
If u is measured in minutes, how should V(t) be rewritten?
Hint
Use t = u/60 everywhere, including the exponent.
Worked solution
V(u/60) = 6(u/60) + 8[1 − exp(−u/60)], for 0 ≤ u ≤ 360. Replacing only the linear term would give a different model.
07 / Check whether the target is reached in the valid interval
An extrapolated root may not answer the question.
Can this model justify a time for reaching 50 m³?Worked example
V(0) = 0 and V(6) ≈ 43.98016998 m³
These are the endpoints of the intended interval.
V′(t) > 0 throughout the interval
The maximum on [0, 6] is its endpoint value at 6.
50 m³ is not reached during the modelled six hours
There is no relevant root of V(t) − 50 = 0 in this interval.
A root obtained by extending the formula beyond six hours would be an extrapolation
It is not justified by the stated scope without further evidence.
10 · Boundary target
What if the requested target is exactly V(6)?
Hint
The allowed interval includes its endpoints.
Worked solution
Then t = 6 hours is a valid endpoint solution. A strict interior sign change is not needed when an endpoint already solves the equation.
08 / Apply the same reasoning to another physical quantity
The root is where the model reaches the requested value.
A separate illustrative temperature model is θ(t) = 15 + 60e⁻⁰·²ᵗ + 0.5t °C, valid for 0 ≤ t ≤ 10 minutes. Form a Newton method for θ = 30°C.Worked example
f(t) = 15 + 60exp(−0.2t) + 0.5t − 30
The unknown is time in minutes.
f′(t) = −12exp(−0.2t) + 0.5
The chain-rule factor −0.2 is essential.
tₙ₊₁ = tₙ − f(tₙ)/f′(tₙ)
Use a start in a justified physical bracket.
f(0) = 45; f(10) = 60e⁻² − 10 < 0
Continuity gives a root during the ten-minute interval.
f′(t) < 0 on [0, 10]
The temperature is strictly decreasing there, so the root is unique.
11 · Temperature derivative units
What are the units of θ′(t) for this second model?
Hint
The independent variable is measured in minutes.
Worked solution
Degrees Celsius per minute.
12 · Use exact methods when available
If the model were instead θ(t) = 15 + 60e⁻⁰·²ᵗ, solve θ = 30 exactly.
Hint
Isolate the exponential and take logarithms.
Worked solution
e⁻⁰·²ᵗ = 1/4, so t = 5ln4 minutes. Numerical iteration is unnecessary when the equation can be solved directly.
09 / Separate numerical accuracy from model accuracy
More decimal places do not make assumptions more realistic.
A root may be computed to many digits even though the coefficients were estimated roughly or the physical conditions vary. Numerical error concerns solving the chosen equation. Model error concerns whether that equation represents the situation well.
State assumptions, the time interval, units and relevant constraints. A filling model may omit leakage or changing supply; a temperature model may cease to apply if its surroundings change. Do not claim a prediction is exact merely because a numerical calculation is stable.
13 · Evaluate a claim
A tank’s model coefficients are approximate, but a root is printed to ten decimal places. Is the actual filling time known to that precision?
Hint
Identify which uncertainty the calculation addresses.
Worked solution
No. The calculation may locate the model’s root very accurately, while uncertain coefficients and omitted effects limit the accuracy of the real prediction.
10 / Give a mathematical answer in its physical context
The final sentence should answer the original question.
- Define the variable and units.
- Subtract the target to form f = 0.
- Use the intended physical interval.
- Check existence and uniqueness where needed.
- Differentiate correctly for Newton or state the chosen recurrence.
- Keep working precision and verify final rounding.
- Convert units carefully.
- Distinguish an accurate numerical root from a reliable real-world prediction.
14 · A complete interpretation
How should the verified tank result 2.15 be stated?
Hint
Include the quantity, units and model qualification.
Worked solution
According to the model, the tank reaches 20 m³ after 2.15 hours, correct to 2 decimal places in hours.
Section 1 of 10 · Translate the target into an equation