Determine the model constants
Two rings fix two numbers
A sample of a radioactive substance has mass \(M\) milligrams after \(t\) hours. The model is \(M = M_0e^{kt}\), but the starting mass \(M_0\) and the constant \(k\) are unknown.
The chalk rings mark two measurements: 80 mg at \(t = 0\) and 50 mg at \(t = 3\).
At \(t = 0\), \(e^0 = 1\), so the model gives \(M = M_0\), and the first measurement gives \(M_0 = 80\).
Move the slider for \(k\). The curve starts at the first ring for every \(k\), and it reaches the second ring for only one value of \(k\).
At that value, \(80e^{3k} = 50\). This equation fixes \(k\).
Two measurements, two unknowns
The model \(y = y_0e^{kt}\) has two unknowns. Each measurement gives one equation, so two measurements can fix both numbers.
Move the slider for \(y_0\). The curve moves up or down, because \(y_0\) is the height where the curve crosses the vertical axis.
Move the slider for \(k\). The curve bends more or less, because each unit of time multiplies the amount by \(e^k\).
Only one value of \(y_0\) and one value of \(k\) put the curve through both rings.
Notation
Write the model with two unknowns. The number \(y_0\) is the amount at \(t = 0\), and \(k\) is the growth or decay constant.
\[y = y_0e^{kt}\]If one measurement is at \(t = 0\), it gives \(y_0\) directly, because \(e^0 = 1\).
\[y(0) = y_0\]Substitute the other measurement \((t_1, y_1)\). Divide by \(y_0\) so that the exponential is alone.
\[y_1 = y_0e^{kt_1} \;\Rightarrow\; e^{kt_1} = \frac{y_1}{y_0}\]Take the natural logarithm of both sides, because the logarithm undoes the exponential. Then divide by \(t_1\).
\[kt_1 = \ln\frac{y_1}{y_0} \;\Rightarrow\; k = \frac{1}{t_1}\ln\frac{y_1}{y_0}\]If the amount falls, then \(y_1 < y_0\), so the ratio is less than 1, its logarithm is negative and \(k < 0\).
If neither measurement is at \(t = 0\), each gives an equation with both unknowns. Divide the second equation by the first, so that the unknown \(y_0\) cancels.
\[\begin{aligned} \frac{y_2}{y_1} &= \frac{y_0e^{kt_2}}{y_0e^{kt_1}} = e^{k(t_2 - t_1)} \\ k &= \frac{\ln(y_2/y_1)}{t_2 - t_1} \end{aligned}\]Then substitute \(k\) into the first equation to find \(y_0\).
When it fails
An exponential is never 0, because \(e^{kt} > 0\) for every \(k\) and \(t\). The curve falls toward the axis but stays above it, so a measurement of 0 cannot lie on it.
In the equation for \(k\), the ratio \(y_1/y_0\) would be 0, and the logarithm of 0 is not defined. The exponential model does not fit data that reach 0.
Example 1: A growing town
A town has 12000 people at \(t = 0\) and 15000 people at \(t = 5\) years. Find \(P_0\) and \(k\) in \(P = P_0e^{kt}\).
The first measurement is at \(t = 0\), so \(P_0 = 12000\). The curve with \(k = 0\) is flat, so it passes through the first ring only.
Substitute the second measurement.
\[15000 = 12000e^{5k}\]Divide by 12000. Then take the natural logarithm of both sides and divide by 5.
\[\begin{aligned} e^{5k} &= 1.25 \\ 5k &= \ln 1.25 \\ k &= \frac{\ln 1.25}{5} \end{aligned}\]As a check, \(k \approx 0.0446\) per year. It is positive, because the town grows.
\[P = 12000e^{(\ln 1.25)t/5}\]
Example 2: Neither measurement is at the start
A deer herd has 300 animals at \(t = 2\) years and 750 animals at \(t = 5\) years after a survey began. Find \(P_0\) and \(k\) in \(P = P_0e^{kt}\).
Each measurement gives an equation with both unknowns.
\[300 = P_0e^{2k}, \qquad 750 = P_0e^{5k}\]A tempting wrong move: Use the first count as the starting amount, \(P_0 = 300\).
\[P_0 = 300\]The first ring is at \(t = 2\), not at \(t = 0\). With \(P_0 = 300\), the second ring gives \(750 = 300e^{5k}\), the red dashed curve. At \(t = 2\) it is at about 433, so it misses the first ring.
Divide the second equation by the first, so that the unknown \(P_0\) cancels.
\[\begin{aligned} \frac{750}{300} &= \frac{P_0e^{5k}}{P_0e^{2k}} = e^{3k} \\ k &= \frac{\ln 2.5}{3} \approx 0.305 \end{aligned}\]Substitute \(k\) into the first equation and divide by \(e^{2k}\) to find \(P_0\). The curve now passes through both rings.
\[P_0 = 300e^{-2k} \approx 163\]
Practice
Algae covers 4 square meters of a pond at \(t = 0\) and 9 square meters at \(t = 6\) days. Find \(A_0\) and \(k\) in \(A = A_0e^{kt}\).
Move the slider for \(k\) until the curve reaches the second ring. Then write the equation for that ring.
Solution
The first measurement is at \(t = 0\), so \(A_0 = 4\). Substitute the second measurement.
\[9 = 4e^{6k}\]Divide by 4. Then take the natural logarithm of both sides and divide by 6.
\[\begin{aligned} e^{6k} &= \frac94 \\ 6k &= \ln\frac94 \\ k &= \frac16\ln\frac94 \end{aligned}\]As a check, \(k \approx 0.135\) per day. Because \(\frac94 = \left(\frac32\right)^2\), the same number is \(\frac13\ln\frac32\).
\[k = \frac13\ln\frac32\]
Questions
Questions 1 and 2 use the curve \(y = y_0e^{kt}\) through the two chalk rings, at \(t = 1\) and \(t = 3\).
What is \(y_0\), the height of the curve at \(t = 0\)?
- 80
- 40
- 25
Answer
80
What is the height of the curve at \(t = 2\), halfway between the rings?
- 20
- 25
- 30
Answer
20
A quantity is 500 at \(t = 0\) and 800 at \(t = 4\). Which value of \(k\) fits \(y = y_0e^{kt}\)?
- \(\frac14\ln 1.6\)
- \(\ln 1.6\)
- \(\frac14\ln\frac{500}{800}\)
- 75
Answer
\(\frac14\ln 1.6\)
Summary
The measurement at \(t = 0\) gives the starting amount, and a later measurement gives \(k\) through a logarithm.
This works because the model has two unknowns, and each measurement gives one equation.
The same two steps fit the model to any exponential quantity with two known values.
This is one of four skills in topic 7.8 of Which & Why: AP Calculus.