Why Exponential Models Appear
A quantity follows an exponential growth or decay model when its instantaneous rate of change is proportional to the amount currently present. If a population grows faster when it is larger, or a medicine leaves the body at a rate proportional to the amount remaining, the differential equation often has the form \(y'=ky\).
In a contextual response, finding a formula is only part of the task. The parameter \(k\) communicates how the quantity changes, while the initial condition determines its scale. A complete interpretation should identify the units, describe growth or decay, and explain what a predicted value represents.
Solving the Model with an Initial Condition
For the equation \(y'=ky\), the solution satisfying \(y(t_0)=y_0\) is \(y(t)=y_0e^{k(t-t_0)}\). Here \(e\) is the base of the natural exponential function. When the initial time is zero, this becomes \(y(t)=y_0e^{kt}\).
The initial value and the rate constant have distinct roles. The value \(y_0\) fixes the amount at time \(t_0\). The value \(k\) controls the relative rate of change: at any time, the derivative is \(k\) times the current amount. The elapsed time \(t-t_0\), rather than \(t\) alone, appears in the exponent when the initial condition is specified at \(t_0\).
Why the Exponential Formula Works
A direct verification shows that the formula has the required derivative and initial value. A short argument also shows why every differentiable solution of the differential equation must have this form.
Multiply the unknown function by \(e^{-kt}\) and differentiate using the product rule.
The resulting derivative is zero, so the product is constant on the interval.
The constant is determined by \(y(t_0)=y_0\), yielding the stated formula.
For completeness, define \(F(t)=e^{-kt}y(t)\). By the product rule and \(y'=ky\),
A differentiable function whose derivative is zero throughout an interval is constant there, by the Mean Value Theorem. Thus \(F(t)=C\) for a constant \(C\), and \(y(t)=Ce^{kt}\). Applying \(y(t_0)=y_0\) gives \(y_0=Ce^{kt_0}\), so \(C=y_0e^{-kt_0}\). Substitution yields \(y(t)=y_0e^{k(t-t_0)}\). This argument also includes the zero solution when \(y_0=0\); it does not require dividing by \(y\).
Interpreting the Sign and Units of \(k\)
For a positive quantity \(y\), the equation \(y'=ky\) means that the sign of \(y'\) is the sign of \(k\). Therefore, for a positive initial amount, \(k>0\) gives growth, \(k<0\) gives decay, and \(k=0\) gives a constant amount. If the initial value is zero, the solution stays zero for every \(k\). More generally, the sign of the derivative is the sign of \(ky\); the usual growth-and-decay descriptions assume the modeled quantity is positive.
| Parameter | Behavior when \(y_0>0\) | Interpretation |
|---|---|---|
| \(k>0\) | \(y\) increases | Exponential growth |
| \(k<0\) | \(y\) decreases | Exponential decay |
| \(k=0\) | \(y\) is constant | No change in the modeled quantity |
If \(y\) is measured in bacteria, grams, or milligrams and time is measured in hours, then \(y'\) has units of the amount per hour. Since \(y'=ky\), \(k\) has units of inverse hours. This also makes the exponent \(k(t-t_0)\) dimensionless, as required for an exponent.
The parameter \(k\) is not the number of units gained or lost in one time unit. Instead, \(y'=ky\) states that the rate at a particular time depends on the amount at that time. The same \(k\) can correspond to different rates for different current values of \(y\).
Worked Examples
Worked Example: Predicting a Growing Population
A bacteria population satisfies \(P'=0.18P\), where time is measured in hours. At time \(t=0\), there are 500 bacteria. Find the model and predict the population after 4 hours.
Here \(k=0.18\) per hour, \(P(0)=500\), and the initial time is zero. The model is
Because \(k>0\) and the initial population is positive, the population increases. At 4 hours,
To the nearest whole bacterium, the model predicts approximately 1,027 bacteria after 4 hours. The rate constant has units of inverse hours, and the predicted value is a population, not a rate.
Check: \(P(0)=500e^0=500\), as required. Differentiating gives \(P'(t)=0.18(500e^{0.18t})=0.18P(t)\), so the model satisfies the differential equation as well as the initial condition.
Worked Example: Interpreting Exponential Decay
The amount \(A\) of a medicine in a patient’s body satisfies \(A'=-0.25A\), with time \(t\) in hours and amount in milligrams. Initially, \(A(0)=80\) milligrams. Find the amount after 3 hours and interpret the result.
The parameter is \(k=-0.25\) per hour, so the model is
Since \(k<0\) and \(A(0)>0\), the amount decreases. At 3 hours,
To the nearest tenth of a milligram, approximately 37.8 mg remains after 3 hours. This prediction is the amount at that time; the instantaneous rate of change at 3 hours would instead be \(A'(3)=-0.25A(3)\), measured in milligrams per hour.
Check: \(A(0)=80e^0=80\) mg. Also, \(A'(t)=80(-0.25)e^{-0.25t}=-0.25A(t)\), which verifies the equation.
Worked Example: Finding the Rate Constant from Two Measurements
A positive quantity \(Q\) follows \(Q'=kQ\). It is 12 units at day 2 and 6 units at day 5. Find \(k\), write a model based on the day-2 measurement, and interpret the result.
Using the solution with initial time \(t_0=2\), the two measurements give
Divide both sides of the measurement equation by 12:
Apply the natural logarithm to both positive sides, then solve for \(k\):
Therefore, a model is \(Q(t)=12e^{-(\ln 2)(t-2)/3}\). The negative rate constant indicates decay. The model says the quantity halves over the three days from day 2 to day 5.
Check: At day 2, \(Q(2)=12e^0=12\). At day 5, \(Q(5)=12e^{-\ln 2}=12/2=6\). Both supplied measurements are reproduced.
Worked Example: Finding When a Quantity Reaches a Threshold
A quantity starts at 240 mg and decays according to \(M'=-0.12M\), where time is measured in days. When does the model predict that it reaches 90 mg?
The initial-value model is \(M(t)=240e^{-0.12t}\). Set the model equal to the requested threshold and solve:
Taking natural logarithms gives \(\ln(3/8)=-0.12t\), so
The model predicts that the amount reaches 90 mg after approximately 8.17 days. This time is measured from the stated initial time, day 0.
Check: Since \(0.12t=\ln(8/3)\) at this time, \(M(t)=240e^{-\ln(8/3)}=240(3/8)=90\) mg.
Common Pitfall: Treating \(k\) as a Fixed Amount Change
A frequent error is to interpret \(k=-0.2\) as a loss of 0.2 units per time unit and use a linear model. For example, if \(y'=-0.2y\) and \(y(0)=100\), writing \(y=100-0.2t\) is incorrect. Its derivative is the constant \(-0.2\), but the required derivative is \(-0.2y\). At \(t=0\), the required rate is \(-0.2(100)=-20\), not \(-0.2\).
The exponential model is \(y(t)=100e^{-0.2t}\). Its derivative is \(y'(t)=-20e^{-0.2t}=-0.2y(t)\), so the rate changes with the current amount. The parameter \(k\) is a proportional rate constant, not a fixed number of units gained or lost.
Writing a Contextual Interpretation
A clear response links the mathematical result to the situation. State the model, give the sign and units of \(k\), and explain the prediction with its time and quantity units. If the result is a decimal, round only at the final step and state the precision. A prediction is an implication of the model; it does not establish that the real-world process will continue to follow the model indefinitely.
Check Your Understanding
For each question, include units and a contextual interpretation where appropriate.
- A population satisfies \(P'=0.04P\) and \(P(0)=250\). Write its model and predict \(P(5)\).
- A quantity satisfies \(y'=-0.3y\). What does the sign of \(k\) tell you when \(y\) is positive?
- A model has \(y'=ky\), with \(y(1)=8\) and \(y(4)=2\). Set up the equation that determines \(k\).
- For \(A(t)=60e^{-0.15t}\), verify the initial value and differential equation.
- Explain why \(k\) in \(y'=ky\) is not generally the amount lost per unit of time.
