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Course Notes for Calculus I

Remkes Kooistra

Section 3.4 Week 3 Activity

Subsection 3.4.1 Exponential Growth

Activity 3.4.1.

Here is a dataset. Just by looking at the dataset, guess the doubling period and write a guess of the appropriate exponential function. Evaluate that exponential function (using a calculator or computer) and compare the values with the data.
Table 3.4.1. Data Set 1 for Exponential Growth
t 0 1 2 3 4 5 6 7 8 9 10
p 1.0 1.1 1.2 1.5 1.8 2.0 2.3 2.7 3.1 3.5 4.1
Solution.
It looks like this data doubles about every five years. The starting value is 1. From the notes, I can use the provided structure for this exponential function: if \(a\) is the starting value and \(d\) is the doubling period, the function should be \(a2^{\frac{t}{d}}\text{.}\) Using the values from this data, here is the exponential function.
\begin{equation*} p(t) = 2^{\frac{t}{5}} \end{equation*}
If I wanted, I could change base to write this as an exponential function with base \(e\text{.}\) (This is useful to do in many cases, since the base \(e\) is the most common and most useful base for exponential functions in calculus.)
\begin{equation*} p(t) = e^{\frac{t \ln 2}{5}} \end{equation*}
The following table shows the values of this function to compare with the data set.
Table 3.4.2. Data Set 1 for Exponential Growth - Values
t 0 1 2 3 4 5 6 7 8 9 10
p 1.0 1.1 1.2 1.5 1.8 2.0 2.3 2.7 3.1 3.5 4.1
\(2^\frac{t}{5}\) 1.00 1.14 1.32 1.51 1.74 2.00 2.30 2.64 3.03 3.48 4.00
The function seems reasonably close to the given data.

Activity 3.4.2.

Here is a dataset. Just by looking at the dataset, guess the doubling period and write a guess of the appropriate exponential function. Evaluate that exponential function (using a calculator or computer) and compare the values with the data.
Table 3.4.3. Data Set 2 for Exponential Growth
t 0 1 2 3 4 5 6 7 8 9 10
p 3.0 4.2 6.3 10.5 14.7 22.8 33.3 50.1 73.8 102.3 144.6
Solution.
It looks like this data doubles is something a little less than two years. I will guess a doubling period of \(\frac{9}{5}\text{.}\) The starting value is 3. From the notes, I can use the provided structure for this exponential function: if \(a\) is the starting value and \(d\) is the doubling period, the function should be \(a2^{\frac{t}{d}}\text{.}\) Using the values from this data, here is the exponential function.
\begin{equation*} p(t) = (3) 2^{\frac{5t}{9}} \end{equation*}
If I wanted, I could change base to write this as an exponential function with base \(e\text{.}\) (This is useful to do in many cases, since the base \(e\) is the most common and most useful base for exponential functions in calculus.)
\begin{equation*} p(t) = (3)e^{\frac{t 5\ln 2}{9}} \end{equation*}
The following table shows the values of this function to compare with the data set.
Table 3.4.4. Data Set 2 for Exponential Growth - Values
t 0 1 2 3 4 5 6 7 8 9 10
p 3.0 4.2 6.3 10.5 14.7 22.8 33.3 50.1 73.8 102.3 144.6
\((3)2^{\frac{5t}{9}}\) 3.00 4.40 6.48 9.52 14.00 20.58 30.24 44.44 65.32 96.00 141.09
The function produces numbers which are reasonably close to the original, so I can conclude that the guess was a decent one.

Activity 3.4.3.

Here is a dataset. Just by looking at the dataset, guess the doubling period and write a guess of the appropriate exponential function. Evaluate that exponential function (using a calculator or computer) and compare the values with the data.
Table 3.4.5. Data Set 3 for Exponential Growth
t 0 1 2 3 4 5 6 7 8 9 10
p 5 10 17 32 62 118 212 383 698 1258 2362
Solution.
It looks like this data doubles every year, perhaps a little bit more than a year in some cases. I guess a doubling period of \(\frac{11}{10}\text{.}\) The starting value is 5. From the notes, I can use the provided structure for this exponential function: if \(a\) is the starting value and \(d\) is the doubling period, the function should be \(a2^{\frac{t}{d}}\text{.}\) Using the values from this data, here is the exponential function.
\begin{equation*} p(t) = (5) 2^{\frac{10t}{11}} \end{equation*}
If I wanted, I could change base to write this as an exponential function with base \(e\text{.}\) (This is useful to do in many cases, since the base \(e\) is the most common and most useful base for exponential functions in calculus.)
\begin{equation*} p(t) = (5)e^{\frac{t 10\ln 2}{11}} \end{equation*}
The following table shows the values of this function to compare with the data set.
Table 3.4.6. Data Set 3 for Exponential Growth - Values
t 0 1 2 3 4 5 6 7 8 9 10
p 5 10 17 32 62 118 212 383 698 1258 2362
\(p(t) = (5)e^{\frac{t 10\ln 2}{11}}\) 5 9 18 33 62 117 219 412 773 1452 2727
This guess was perhaps a little weaker. The function I produced is similar to the data, but seems to outpace it but a bit. A slightly smaller doubling period might have been more appropriate.

Activity 3.4.4.

Draw a phase line for this differential equation.
\begin{equation*} \frac{dp}{dt} = p^2 - 9p + 18 \end{equation*}
Determine the steady states and trajectories, then classify the steady states. You may assume this is a population model, so that \(p \geq 0\) and 0 is always a steady state.
Solution.
Figure 3.4.7. Phase Line for \(p^2 - 9p + 18\)
To calculate the phase line, I need the zeros of the right side of the equation. This is a quadratic and it factors as \((p-6)(p-3)\text{.}\) Therefore, there are steady states at \(p=3\) and \(p=6\text{.}\) I label these points on the phase line. To determine the trajectories, I check the value of the right side of the equation between the steady states.
  • At \(p=1\text{,}\) the right side evaluates to 10, which is positive. The trajectory between 0 and 3 is upward.
  • At \(p=4\text{,}\) the right side evaluates to -2, which is negative. The trajectory between 3 and 6 is downward.
  • At \(p=7\text{,}\) the right side evaluates to 4, which is positive. The trajectory above 6 is upward.
This lets me classify the steady states.
  • The (imposed) steady state at 0 only has an trajectory pointing away, so this is an unstable steady state.
  • The steady state at 3 has trajectories pointing toward it, so it is a stable steady state.
  • The steady state at 6 has trajectories pointing away from it, so it is an unstable steady state.

Activity 3.4.5.

Draw a phase line for the differential equation
\begin{equation*} \frac{dp}{dt} = p^3 - 6p^2 + 11p - 6 \end{equation*}
Determine the steady states and trajectories, then classify the steady states. You may assume this is a population model, so that \(p \geq 0\) and 0 is always a steady state.
Solution.
Figure 3.4.8. Phase Line for \(p^3 - 6p^2 + 11p - 6\)
To calculate the phase line, I need the zeros of the right side of the equation. This is a cubic: I can try to look for the roots by inspection or ask a computer for the roots. In any case, I find that \(p=1\text{,}\) \(p=2\) and \(p=3\) are the roots. I label these points on the phase line. To determine the trajectories, I check the value of the right side of the equation between the steady states.
  • At \(p=\frac{1}{2}\text{,}\) the right side evaluates to \(\frac{-15}{8}\text{,}\) which is negative. The trajectory between 0 and 1 is downward.
  • At \(p=\frac{3}{2}\text{,}\) the right side evaluates to \(\frac{3}{8}\text{,}\) which is positive. The trajectory between 1 and 2 is upward.
  • At \(p=\frac{5}{2}\text{,}\) the right side evaluates to \(\frac{-3}{8}\text{,}\) which is negative. The trajectory between 2 and 3 is downward.
  • At \(p=4\text{,}\) the right side evaluates to 6, which is positive. The trajectory above 4 is upward.
This lets me classify the steady states.
  • The (imposed) steady state at 0 only has an trajectory pointing towards, so this is a stable steady state.
  • The steady state at 1 has trajectories pointing away from it, so it is an unstable steady state.
  • The steady state at 2 has trajectories pointing toward it, so it is a stable steady state.
  • The steady state at 3 has trajectories pointing away from it, so it is an unstable steady state.

Activity 3.4.6.

Draw a phase line for the differential equation
\begin{equation*} \frac{dp}{dt} = p^3 - 10p^2 + 32 p - 32 \end{equation*}
Determine the steady states and trajectories, then classify the steady states. You may assume this is a population model, so that \(p \geq 0\) and 0 is always a steady state.
Solution.
Figure 3.4.9. Phase Line for \(p^3 - 10p^2 + 32p - 32\)
To calculate the phase line, I need the zeros of the right side of the equation. This is a cubic; I can find the factors by inspection or by asking a computer. The roots are \(p=2\) and \(p=4\text{.}\) I label these points on the phase line. To determine the trajectories, I check the value of the right side of the equation between the steady states.
  • At \(p=1\text{,}\) the right side evaluates to -9, which is negative. The trajectory between 0 and 2 is downward.
  • At \(p=3\text{,}\) the right side evaluates to 1, which is positive. The trajectory between 2 and 4 is upward.
  • At \(p=5\text{,}\) the right side evaluates to 5, which is positive. The trajectory above 4 is upward.
This lets me classify the steady states.
  • The (imposed) steady state at 0 only has an trajectory pointing towards, so this is a stable steady state.
  • The steady state at 2 has trajectories pointing away from it, so it is an unstable steady state.
  • The steady state at 4 has one trajectory pointing towards and one pointing away. This is neither stable or unstable; we can call it partially stable. If I want to be more specific, I can say that it is stable from below and unstable from above.

Subsection 3.4.2 Conceptual Review Questions

  • What is percentage growth? Why does it lead to a graph that looks exponential?
  • What is a differential equation? What is a solution to one?
  • What is a stable point for an autonomous differential equation?
  • What does a phase line represent? How does it carry meaning about the solution of the differential equation?
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