Why Shift the Time Origin?
An exponential model is often written with time measured from an initial moment, \(t=0\). In practice, however, the first available measurement may be taken later. If a population was first measured on a particular date, or a substance was first sampled after an experiment had begun, it is convenient to use that observation as the starting value for predictions.
Changing the time origin is a change in how we label elapsed time. It does not change the underlying growth or decay process. In particular, the rate constant remains the same, provided the same constant-rate exponential model applies throughout the time interval being considered.
The Time-Zero Model and the Shifted Model
Suppose \(P(t)\) is described by a constant-rate exponential model referenced to time zero. Here, \(P(0)\) is the value at the original starting time, and \(k\) is the constant rate parameter. The time \(t\) and the reference time \(t_0\) must be measured in the same time unit.
The quantity \(P(t_0)\) is now the starting value in the shifted description. The elapsed time is \(t-t_0\), not \(t\). If \(t>t_0\), this elapsed time is positive; if \(t<t_0\), it is negative.
Deriving the Shifted Formula
Begin with the model referenced to time zero, \(P(t)=P(0)e^{kt}\). Evaluate it at \(t_0\) to find the value at the later reference time. Then rewrite the exponent at time \(t\) as the sum of the exponent at \(t_0\) and the additional elapsed time.
Substituting \(t_0\) for \(t\) gives \(P(t_0)=P(0)e^{kt_0}\).
Since \(t=t_0+(t-t_0)\), the exponential law gives \(e^{kt}=e^{kt_0}e^{k(t-t_0)}\).
Substitute \(P(t_0)=P(0)e^{kt_0}\) into the rewritten expression. The result is \(P(t)=P(t_0)e^{k(t-t_0)}\).
In equations, the derivation is:
This is an exact rewrite of the original model. To check the identity in the other direction, substitute the expression for \(P(t_0)\) into the shifted formula:
Thus the shifted form gives the same value at every time as the original form. Only the chosen reference time and starting value have changed.
Why the Rate Constant Does Not Change
The shifted formula contains the same \(k\) as the original model because the change from \(t_0\) to \(t\) is still governed by the same exponential factor. For a time interval of length \(\Delta\), starting at \(t_0\), the formula gives
The factor \(e^{k\Delta}\) depends on the interval length and the original rate constant. It does not depend on where that interval begins. If \(k>0\), the model grows as time advances; if \(k<0\), it decays. Shifting the origin does not reverse or otherwise alter that behavior.
The model assumes a constant rate parameter over the interval in question. If the process does not follow one constant-rate exponential model across that interval, this single shifted formula need not describe it. Also, \(t\), \(t_0\), and the time units associated with \(k\) must be consistent.
Worked Examples
Worked Example: Re-reference a Growth Model at a Later Time
Suppose a quantity follows \(P(t)=120e^{0.04t}\), with time measured in years. Rewrite the model using the value at \(t_0=5\), then find the model's value at \(t=8\).
First calculate the value at the new reference time:
Use this as the shifted starting value. The elapsed time from year \(5\) to year \(t\) is \(t-5\), so
At \(t=8\), the elapsed time is \(8-5=3\) years. Therefore,
Check against the original model: \(P(8)=120e^{0.04(8)}=120e^{0.32}\), which agrees exactly. The rate constant is \(0.04\) in both forms.
Worked Example: Re-reference a Decay Model
A decaying quantity is modeled by \(P(t)=900e^{-0.1t}\), where time is measured in days. Rewrite it using the value at day \(6\), then calculate the value at day \(9\).
At day \(6\), the model gives
Using day \(6\) as the reference time gives
From day \(6\) to day \(9\), the elapsed time is \(9-6=3\) days. Thus
Check with the original formula: \(P(9)=900e^{-0.1(9)}=900e^{-0.9}\). The two expressions agree. The negative rate constant remains \(-0.1\); the shift has not changed the decay rate.
Worked Example: Predict Backward from a Later Measurement
Suppose a quantity has rate constant \(k=-0.08\) per day and is measured as \(P(4)=250\). Use this measurement to find \(P(1)\), the value three days earlier.
Take \(t_0=4\). Since \(1-4=-3\), the shifted formula gives
For a check, first recover the time-zero value from the measurement:
Then the original time-zero model gives
This matches the shifted calculation. The negative elapsed time correctly takes the model backward from the measurement at day \(4\).
A Common Mistake: Using Absolute Time After the Shift
After choosing \(t_0\) as a new reference time, it is tempting to multiply the new starting value by \(e^{kt}\). That uses the original absolute time again, rather than the time elapsed since \(t_0\).
For the growth model \(P(t)=120e^{0.04t}\) referenced at \(t_0=5\), the new starting value is \(P(5)=120e^{0.20}\). A mistaken calculation at \(t=8\) would be
This applies growth over eight additional years, even though only three years have passed since the new starting point. The correct calculation uses the elapsed time \(8-5=3\):
The practical check is simple: at \(t=t_0\), the shifted formula must return \(P(t_0)\), because its exponent is \(k(t_0-t_0)=0\). A formula that does not have this property has not used the new time origin correctly.
When This Rewrite Is Useful
A later measurement can be more useful than an earlier starting value when predictions are made from the time of that measurement. The shifted formula puts the observed amount directly into the model, so you do not need to carry the original value \(P(0)\) through each calculation. It also makes the relevant interval explicit: a prediction at time \(t\) uses the elapsed time \(t-t_0\).
Before applying the formula, identify three quantities: the reference time \(t_0\), the value measured there \(P(t_0)\), and the unchanged rate constant \(k\). Then use \(t-t_0\) in the exponent. The same expression handles a future prediction and a backward calculation; only the sign of the elapsed time changes.
Check Your Understanding
For each question, use the shifted formula and check that the time in the exponent is elapsed time from the reference point.
- If \(P(t)=50e^{0.3t}\), write an equivalent model referenced to \(t_0=2\).
- A decay model has \(P(5)=80\) and \(k=-0.04\). Write an expression for \(P(9)\) using the measurement at time \(5\).
- In \(P(t)=P(t_0)e^{k(t-t_0)}\), what does the exponent become when \(t=t_0\), and what value must the formula return?
- A model referenced at \(t_0=7\) is used to estimate the value at \(t=4\). Is \(t-t_0\) positive or negative?
- Why must the rate constant stay the same when only the time origin is shifted?
