Comparing Exponential Models with Different Rate Constants

Why Compare Rate Constants?

Two exponential models can start at the same amount and then make different predictions because their rate constants differ. Comparing the constants tells us which model is larger at a given positive time, whether each model grows or decays, and whether the predictions can meet again after their shared starting point.

The comparison is meaningful only when both models use the same time variable and the same units for time. A rate constant measured per year, for example, must be paired with time measured in years.

The Models and Their Rate Constants

For a constant relative rate, an exponential model with initial amount \(A\) and rate constant \(k\) has the form \(y(t)=Ae^{kt}\), where \(t\geq 0\) is elapsed time. The initial amount is \(y(0)=A\). The rate constant \(k\) has units of inverse time.

The sign of \(k\) identifies the model's direction of change: \(k>0\) gives growth, \(k<0\) gives decay, and \(k=0\) gives a constant model. The constant is a relative rate: the model satisfies \(y'(t)=ky(t)\). Thus, for example, a positive \(k\) means the amount increases at a rate proportional to its current amount.

For \(A>0\) and positive elapsed time, the ordering of predictions follows the ordering of the rate constants. If \(k_1>k_2\), then \(y_1(t)>y_2(t)\) for \(t>0\). That statement compares the amounts predicted; interpreting which model grows or decays faster also requires noticing whether the constants are positive or negative.

Worked Example: Comparing Two Growth Models

Growth and Decay Require Different Language

If both constants are positive, the larger constant gives the faster proportional growth and the larger predicted amount for every \(t>0\). If both constants are negative, the more negative constant gives the faster proportional decay and the smaller remaining amount. In the latter case, “more negative” means smaller as a number: for example, \(-0.12<-0.04\), while \(|-0.12|>|-0.04|\).

If one constant is positive and the other negative, one model grows while the other decays. For \(A>0\) and \(t>0\), the growth model is then larger than the decay model. More generally, the value comparison is given by the ordering of \(k_1\) and \(k_2\), not by comparing their absolute values.

Worked Example: Comparing Two Decay Models

Why the Rate-Constant Comparison Works

Divide one model by the other. Because the initial amount is shared and positive, it cancels, leaving a ratio that depends only on the rate-constant difference.

The exponential function is strictly increasing and \(e^0=1\). Therefore, when \(t>0\), the sign of \((k_1-k_2)t\) determines whether this ratio is greater than, equal to, or less than \(1\). In particular, if \(k_1>k_2\), then \((k_1-k_2)t>0\), so the ratio is greater than \(1\), which means \(y_1(t)>y_2(t)\). If \(k_1<k_2\), the ratio is less than \(1\), so \(y_1(t)<y_2(t)\). This proves the comparison rule.

When Do the Models Intersect?

An intersection occurs when the predictions are equal. For models with the same positive initial amount, set them equal and cancel \(A\). The exponential function is one-to-one, so equal exponential values require equal exponents:

Rearranging gives \((k_1-k_2)t=0\). If the constants differ, the only solution is \(t=0\): the models meet at their shared initial amount, but do not intersect again at any positive time. If the constants are equal, the models coincide for every \(t\geq 0\), rather than meeting at just one time.

Worked Example: Finding the Intersection

Special Cases and a Common Pitfall

The shared-initial-amount assumption matters. If \(A=0\), both models equal zero for every time, whatever their rate constants; canceling \(A\) would not be valid. If \(A>0\) but \(k_1=k_2\), they are identical at every time. These cases differ from the usual situation of a positive initial amount and distinct rate constants.

A common mistake is to say that the decay with the “larger rate constant” decays faster simply because its constant is larger. For example, comparing \(-0.12\) and \(-0.04\), that claim treats \(-0.04\) as the larger constant and incorrectly labels it the faster decay.

So the model with \(k=-0.04\) actually leaves more material at positive times. The model with \(k=-0.12\), which has the smaller and more negative constant, decays faster proportionally. Comparing the exponents at the same positive time, or computing the model ratio, prevents this sign error.

Why This Comparison Is Useful

Rate-constant comparisons help interpret competing predictions without evaluating both models at every time. The ratio \(y_1(t)/y_2(t)=e^{(k_1-k_2)t}\) also quantifies their relative separation. The shared starting amount cancels, so the ratio isolates the effect of the difference in rates.

Remember, though, what is being compared: the rate constant describes proportional change, while the model value describes the amount. For decay models, a smaller rate constant means a smaller remaining amount at positive times, but it does not mean every measure of instantaneous absolute change must have the same ordering at every time.