Calibrate exponential models from observations

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A cooling cup of tea approaches room temperature, not zero, so its model is a fixed background plus an exponential part. This lesson finds the multiplier for each time step and recovers an unknown background from evenly spaced readings, then checks which targets the model can actually reach.

The multiplier belongs to a time step

An exponential model with a fixed background has the form

Q(t) = B + Ab^t, b > 0, b ≠ 1.

Here t is the numerical time in the stated unit. The background is B; the initial deviation from it is A = Q(0)-B. It is Q-B, rather than necessarily Q, that multiplies by b each time unit.

If observations are Δ time units apart, their deviation multiplier is

r = b^Δ = (Q(Δ)-B)/(Q(0)-B), b = r^(1/Δ).

The positive root matters: a negative base does not define this real-time model. A percentage change of p% over the observed interval gives r = 1 + p/100 for an increase or r = 1-p/100 for a decrease in the positive exponential part. Keep that interval beside the multiplier.

A sensor reading follows Q(t) = 20 + Ab^t, with t in hours. It reads 100 units at t = 0 and 65 units at t = 2. Find the multiplier of Q-20 over two hours, then the positive one-hour base. Enter exact decimals.

Two-hour deviation multiplier

One-hour base

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Subtract the background from each reading: r = (65-20)/(100-20) = 45/80 = 0.5625 over two hours. Since b² = r and b > 0, the one-hour base is b = 0.75. The recovered model Q(t) = 20 + 80(0.75)^t gives Q(2) = 65.

For the sensor above, find

A, the two-hour deviation multiplier and the positive one-hour base.

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Remove the fixed part
Method
Use the initial reading to find A.
Reason
The exponential multiplier applies to the deviation from the background.
Working
Answer
A = 100-20 = 80. The two observed deviations are 80 and 45 units.
Match the observed interval
Method
Compare the deviations over two hours, then take the positive square root.
Reason
The observations span two one-hour steps; their ratio is the square of the one-hour base.
Working
Answer
r = b² = (65-20)/(100-20) = 45/80 = 0.5625,; b = square root of 0.5625 = 0.75.
State the result
Working
Answer
The initial deviation is A = 80, the two-hour multiplier is r = 0.5625 and the one-hour base is b = 0.75. Thus Q(t) = 20 + 80(0.75)^t.

Quick check

The original two-hour reading is recovered: 20 + 80(0.75)² = 65. Taking the raw ratio 65/100 would retain the added background; using 0.5625 as the one-hour base would apply a two-hour change every hour.

Q(t) = 20 + 80b^t: 65.0
Q(t) − 20: 45.0
Deviation multiplier b^t: 0.563

Predict how much of the starting gap above 20 remains after two hours, then compare. The model is Q(t) = 20 + 80bᵗ, with 0 < b < 1.

Differences remove an unknown background

Now suppose B is unknown. Use three observations at equally spaced times 0,Δ,2Δ. Write their values as Q₀,Q₁,Q₂ and put r = b^Δ:

Q₀ = B + A, Q₁ = B + Ar, Q₂ = B + Ar²,; Q₁-Q₀ = A(r-1), Q₂-Q₁ = Ar(r-1).

Taking the ratio of successive differences removes both the background and amplitude:

r = (Q₂-Q₁)/(Q₁-Q₀), B = (Q₁-rQ₀)/(1-r), A = Q₀-B.

These formulas require Q₁ ≠ Q₀, r > 0 and r ≠ 1. Flat readings do not uniquely recover a background and amplitude. Nonzero equal differences would give r = 1, which cannot produce those differences in this model. Unequally spaced readings cannot be used in this ratio without another argument.

A quantity is modelled by Q(t) = B + Ab^t, with t ≥ 0 in hours. Its readings are:

Time t (hours)	0	1	2
Quantity Q (units)	12	24	32

Find B,A,b and check the model against all three readings. Does the negative or positive sign of A agree with the observed direction of change?

Write your answer, then compare it with the solution

Your difference ratio, background, amplitude, base and three checks, or work on paper

Hint 1

Compare changes rather than raw readings

The increases are

12 and

8 units over equal one-hour steps. Their ratio describes the multiplier of the deviation from the unknown background.

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Recover the step multiplier
Method
Divide the second difference by the first.
Reason
Subtracting consecutive readings cancels the same fixed background; their changes still have the exponential multiplier.
Working
Answer
r = (32-24)/(24-12) = 8/12 = 2/3.
The interval is one hour, so b = r = 2/3.
Recover the background and amplitude
Method
Use the ratio in Q₁-rQ₀ = (1-r)B.
Reason
The terms in A cancel when the earlier reading is multiplied by the step multiplier.
Working
Answer
B = (24-(2/3)12)/(1-2/3) = 48,; A = 12-48 = -36,; Q(t) = 48-36(2/3)^t.
The deviation is negative. Its magnitude shrinks, so Q rises towards 48 from below even though b < 1.
Check every original observation
Method
Substitute t = 0,1,2 into the recovered model.
Reason
One matching observation does not check all three parameters.
Working
Answer
Q(0) = 48-36 = 12,; Q(1) = 48-24 = 24,; Q(2) = 48-16 = 32.
All three agree. In particular, using the raw ratio 24/12 = 2 would contradict the next raw ratio 32/24 = 4/3.

Quick check

Compare all three original readings with your recovered model. A negative initial deviation approaches the background from below; the base alone does not determine whether the whole quantity rises or falls.

Check which targets can be reached

Keep the recovered model exact until the final calculation. For the example,

Q(3) = 48-36(2/3)³ = 112/3 ≈ 37.3 units (3 s.f.).

For a target of 40 units, isolate the positive exponential before taking logarithms:

48-36(2/3)^t = 40,; (2/3)^t = 2/9,; t = (ln(2/9))/(ln(2/3)) ≈ 3.71 hours (3 s.f.).

Both logarithms are negative, giving a positive time. This is beyond the observed two-hour range: the calculation is a prediction assuming the same model continues.

The target 48 is different. For every finite t, (2/3)^t > 0, so Q(t) < 48. The model approaches 48 but never attains it at a finite time. A target below the initial 12 units is also unattainable for t ≥ 0, since the model is increasing over that interval. A logarithm that gives a negative time does not answer a question restricted to t ≥ 0.

A different quantity has model P(t) = 10 + 6(0.8)^t for t ≥ 0. Which description follows from the whole model?

Direction and finite-time limit

The quantity increases towards 10 because the base is below one.The quantity decreases and eventually equals 10 at a finite time.The quantity decreases from 16 towards 10, but never equals 10 at finite time.

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P decreases from 16 towards 10. Its deviation 6(0.8)^t is positive at every finite time, so P(t) > 10. This differs from the worked model’s negative deviation: reducing a positive deviation lowers the quantity, while reducing the magnitude of a negative deviation raises it.

Three fitted readings determine the parameters within this stated nonconstant model; they do not prove that an exponential family is the right physical choice. Check how observations were obtained, whether a fixed background is plausible and whether a prediction extrapolates beyond the data.

Common mistakes

Ignoring the background value. In Q(t) = B + Ab^t, the part that multiplies by b is Q-B, not Q.
Using readings at uneven times. The difference method needs equally spaced times.
Rounding the multiplier early. Keep exact values until the final answer.
Assuming a limiting value is reached. With a nonzero initial deviation and 0 < b < 1, the quantity approaches B without attaining it at a finite time.

Syllabus and review details

G3 Additional Mathematics 2027 · 2027

Last reviewed: 2026-11-07

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Guided steps

Reading settings

The multiplier belongs to a time step

Two hours is not one hour

Recover the two different multipliers

Remove the fixed part

Match the observed interval

State the result

Differences remove an unknown background

An increasing quantity can have a base below one

Recover the step multiplier

Recover the background and amplitude

Check every original observation

Check which targets can be reached

Check the amplitude as well as the base

Common mistakes

Lesson done

The multiplier belongs to a time step

Two hours is not one hour

Recover the two different multipliers

Differences remove an unknown background

An increasing quantity can have a base below one

Check which targets can be reached

Check the amplitude as well as the base

Common mistakes

Questions