Use proportional rates with a reference state

Try

A synthetic quantity doubles every three hours and begins at five units. Make a table at times zero, three, and six. The values are five, ten, and twenty. Equal time intervals multiply the amount by a fixed factor; they do not add a fixed amount. Draw a second table for an additive model that gains five units every three hours and identify the first time the predictions differ.

The continuous proportional-growth model is y′=ky, where k is a constant fractional rate with units of inverse hours. We assume the rate responds immediately and proportionally to the current amount.

Predict

If the amount doubles over three hours, is k equal to 1/3? That would confuse a finite fractional increase with an instantaneous fractional rate. Predict whether the correct value is larger or smaller than 1/3, then derive it using an exponential rather than a straight-line approximation.

Experiment

Separation gives y=y₀e^(kt). With y₀=5 and y(3)=10, division yields e^(3k)=2. Taking natural logarithms gives k=(ln 2)/3, approximately 0.231 per hour. This is smaller than 1/3. The exact formula y=5·2^(t/3) reproduces the table.

Time in hours	Proportional model	Additive comparison
0	5	5
3	10	10
6	20	15

Both models fit the first two observations. Their disagreement at six hours shows why fitting two points does not establish the mechanism. Add a hypothetical third observation and explain which model it would favor, while acknowledging measurement uncertainty.

Understand

For decay, write y′=−ky with positive k. The half-life H satisfies e^(−kH)=1/2, hence H=(ln 2)/k. If a quantity falls from eighty to twenty in four hours, it experiences two half-lives, so H=2 hours and k=(ln 2)/2. A negative rate constant in y′=ky expresses the same decay with a different sign convention; state your convention before interpreting parameters.

Cooling uses a shifted state. Let T be temperature and A a constant ambient temperature. Setting z=T−A changes T′=−k(T−A) into z′=−kz. Thus T=A+(T₀−A)e^(−kt). It is the temperature difference that has a constant fractional decay rate, not the temperature measured from an arbitrary zero.

For ambient twenty and initial thirty-six, reducing the difference by a factor of four gives T=20+16/4=24. Multiplying the whole temperature by one quarter would incorrectly predict nine. The equilibrium T=A must be preserved by the model, and a starting temperature below ambient approaches it from below.

For finite time and positive k, the exponential factor stays positive. An initially different temperature approaches ambient asymptotically rather than reaching it at a finite time in this ideal model. Real measurements may become indistinguishable from ambient because instruments have finite resolution.

Apply

Design a small model check using equally spaced observations. Exponential behavior predicts similar ratios of positive amounts, or similar differences of their logarithms. Cooling instead predicts similar ratios of deviations from the measured ambient value. Explain why a changing ambient temperature invalidates that simple test.

List one mechanism that could break proportional growth, such as limited resources, and one measurement limitation. Keep a solved differential equation distinct from an empirically validated model. Complete the practice with units and an explicit statement about what is decaying.

Continue

The logistic model introduces a state-dependent growth reduction. Compare its early behavior with exponential growth, then inspect equilibria and saturation before assuming that every positive initial population must increase.