Test 4 Review Material
Test Instructions
Below are the general instructions for all tests
- You will get a chance to retake this. Your highest score counts toward your final grade.
- Follow the guidance for each part and show all work for full credit.
- Non-graphical calculators are allowed.
- One page (two sides) of handwritten (or font size 8) notes are allowed.
- The equation sheet from the textbook is included as part of the test pack (it doesn’t count as note pages)
Test Format
You should be able to do the following for this test
- Interpret what a differential equation means
- Build a differential equation from a verbal model
- Read and interpret slope fields
- Use slope fields to identify long-term behavior
- Solve separable differential equations
- Solve initial value problems
- Determine an unknown rate constant from data
- Identify equilibrium solutions
- Classify equilibria as stable or unstable
- Interpret logistic growth and logistic growth with harvesting
- Explain what the solution means in context
Part A — Short Answer (Concepts)
There may not be a dedicated short-answer section, but there may be interpretation questions attached to calculations.
1. What does a differential equation represent?
Hint
Think about what is being related: a quantity and how fast it changes.
Suggested Answer
A differential equation relates a quantity to its rate of change. It tells us how a system changes over time or with respect to another variable.
2. What is the difference between a function and a differential equation?
Hint
One gives values directly. The other gives a rule for change.
Suggested Answer
A function gives the value of a quantity directly, while a differential equation gives a rule for how that quantity changes.
3. What does it mean if a model says “rate of change is proportional to the amount present”?
Hint
Think exponential growth or decay.
Suggested Answer
It means the derivative is proportional to the current amount, so the model has the form \[ \frac{dP}{dt}=kP. \] This leads to exponential growth if \(k>0\) and exponential decay if \(k<0\).
4. What is a slope field?
Hint
Think of many tiny tangent segments.
Suggested Answer
A slope field is a graphical picture of a differential equation showing the slope of solution curves at many points.
5. What does an equilibrium solution mean?
Hint
What happens when the derivative is zero?
Suggested Answer
An equilibrium solution is a constant solution where the rate of change is zero, so the system stays at the same value over time.
6. How can you tell from a slope field whether a solution is increasing or decreasing?
Hint
Look at the sign of the slopes.
Suggested Answer
If the slope segments tilt upward, the solution is increasing. If they tilt downward, the solution is decreasing.
7. What makes an equilibrium stable?
Hint
What do nearby solutions do?
Suggested Answer
An equilibrium is stable if nearby solutions move toward it over time.
8. What makes an equilibrium unstable?
Hint
Again, think about nearby solutions.
Suggested Answer
An equilibrium is unstable if nearby solutions move away from it over time.
9. What does a logistic differential equation model?
Hint
Think growth with a limiting value.
Suggested Answer
A logistic differential equation models growth that starts approximately exponential, then slows as the population approaches a carrying capacity.
10. What does carrying capacity mean in a logistic model?
Hint
What value does the population level off toward?
Suggested Answer
The carrying capacity is the long-term limiting population value that the system approaches.
11. Why can harvesting create multiple equilibria in a population model?
Hint
Harvesting shifts the growth curve downward.
Suggested Answer
Harvesting subtracts from natural growth, so the population may have new equilibrium levels where growth exactly balances harvesting.
12. What does an initial condition do?
Hint
General solution vs. specific solution.
Suggested Answer
An initial condition determines the constant of integration and selects the specific solution that fits the situation.
13. Why do we separate variables?
Hint
What are we trying to do before integrating?
Suggested Answer
We separate variables so that all terms involving the dependent variable are on one side and all terms involving the independent variable are on the other, allowing us to integrate both sides.
Part B — Building and Interpreting Models
1. Build a differential equation
A population \(P(t)\) grows at a rate proportional to the current population.
Write a differential equation for this model.
Hint
Use a proportionality constant \(k\).
Worked Solution
If growth is proportional to the amount present, then
\[ \frac{dP}{dt}=kP \]
where \(k\) is a constant.
\[ \boxed{\frac{dP}{dt}=kP} \]
2. Build a logistic model
A population \(P(t)\) grows with intrinsic growth rate \(r\) and carrying capacity \(K\).
Write the logistic differential equation.
Hint
Use a growth factor and a limiting factor.
Worked Solution
The logistic equation is
\[ \frac{dP}{dt}=rP\left(1-\frac{P}{K}\right) \]
\[ \boxed{\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)} \]
3. Add harvesting
Suppose constant harvesting \(H\) is added to the logistic model.
Write the new differential equation.
Hint
Subtract harvesting from growth.
Worked Solution
With constant harvesting, the model becomes
\[ \frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)-H \]
\[ \boxed{\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)-H} \]
Part C — Separation of Variables
4. Solve
\[ \frac{dP}{dt}=kP \]
Use separation of variables to solve for \(P(t)\).
Hint
Separate first, then integrate both sides.
Worked Solution
Separate variables:
\[ \frac{1}{P}\,dP = k\,dt \]
Integrate both sides:
\[ \int \frac{1}{P}\,dP = \int k\,dt \]
\[ \ln|P| = kt + C \]
Exponentiate:
\[ P = Ce^{kt} \]
\[ \boxed{P(t)=Ce^{kt}} \]
Part D — Initial Value Problems
Part E — Determining a Rate Constant
9. Find \(k\)
A population follows
\[ P(t)=6e^{kt} \]
and satisfies \(P(4)=48\).
Find \(k\).
Hint
Substitute \(t=4\) and solve for \(k\).
Worked Solution
Plug in \(t=4\):
\[ 48=6e^{4k} \]
\[ 8=e^{4k} \]
Take natural log:
\[ \ln 8 = 4k \]
\[ k=\frac{\ln 8}{4} \]
\[ \boxed{k=\frac{\ln 8}{4}} \]
10. Find \(k\)
A cooling model is
\[ T(t)=Ce^{-kt} \]
If \(T(0)=100\) and \(T(3)=40\), find \(k\).
Hint
First find \(C\), then use the second point.
Worked Solution
From \(T(0)=100\),
\[ 100=Ce^0=C \]
So
\[ T(t)=100e^{-kt} \]
Use \(T(3)=40\):
\[ 40=100e^{-3k} \]
\[ 0.4=e^{-3k} \]
Take natural log:
\[ \ln(0.4)=-3k \]
\[ k=-\frac{\ln(0.4)}{3} \]
\[ \boxed{k=-\frac{\ln(0.4)}{3}} \]
Part F — Slope Fields, Equilibria, and Stability
11. Interpreting signs
Consider
\[ \frac{dP}{dt}=P(4-P) \]
For each interval, determine whether solutions increase or decrease:
- \(P<0\)
- \(0<P<4\)
- \(P>4\)
Hint
Use sign reasoning only.
Worked Solution
The derivative is
\[ P(4-P) \]
- If \(P<0\): negative \(\times\) positive \(=\) negative → decreasing
- If \(0<P<4\): positive \(\times\) positive \(=\) positive → increasing
- If \(P>4\): positive \(\times\) negative \(=\) negative → decreasing
So:
- decreasing for \(P<0\)
- increasing for \(0<P<4\)
- decreasing for \(P>4\)
\[ \boxed{\text{dec},\ \text{inc},\ \text{dec}} \]
12. Find equilibria
For
\[ \frac{dP}{dt}=P(4-P) \]
find the equilibrium solutions.
Hint
Set the derivative equal to zero.
Worked Solution
Equilibria satisfy
\[ P(4-P)=0 \]
So
\[ P=0 \quad \text{or} \quad P=4 \]
\[ \boxed{P=0,\ 4} \]
13. Classify equilibria
For
\[ \frac{dP}{dt}=P(4-P) \]
classify each equilibrium as stable or unstable.
Hint
Look at whether nearby solutions move toward or away.
Worked Solution
From the sign analysis:
- Near \(P=0\), solutions move away from 0 → unstable
- Near \(P=4\), solutions move toward 4 → stable
\[ \boxed{P=0 \text{ unstable},\quad P=4 \text{ stable}} \]
14. Logistic with harvesting
Consider
\[ \frac{dP}{dt}=P(5-P)-4 \]
Find the equilibrium solutions.
Hint
Set the right side equal to zero and solve the quadratic.
Worked Solution
Set equal to zero:
\[ P(5-P)-4=0 \]
\[ 5P-P^2-4=0 \]
\[ P^2-5P+4=0 \]
\[ (P-1)(P-4)=0 \]
So the equilibria are
\[ \boxed{P=1,\ 4} \]
15. Stability with harvesting
For
\[ \frac{dP}{dt}=P(5-P)-4 \]
classify the equilibria \(P=1\) and \(P=4\).
Hint
Test the sign of the derivative on the intervals.
Worked Solution
Factor:
\[ \frac{dP}{dt}=-(P-1)(P-4) \]
Check intervals:
- \(P<1\): derivative negative → decreasing
- \(1<P<4\): derivative positive → increasing
- \(P>4\): derivative negative → decreasing
Thus:
- Near \(P=1\), solutions move away → unstable
- Near \(P=4\), solutions move toward → stable
\[ \boxed{P=1 \text{ unstable},\quad P=4 \text{ stable}} \]
Part G — Interpretation of Solutions
16. Interpret
Suppose a population model has solution
\[ P(t)=\frac{20}{1+9e^{-2t}} \]
What value does the population approach as \(t\to\infty\), and what does that mean?
Hint
Think about what happens to \(e^{-2t}\).
Worked Solution
As \(t\to\infty\),
\[ e^{-2t}\to 0 \]
so
\[ P(t)\to \frac{20}{1+0}=20 \]
Thus the population approaches 20 in the long run, which is the carrying capacity.
\[ \boxed{P(t)\to 20} \]
17. Interpret
A solution curve in a slope field starts below a stable equilibrium. Describe what happens over time.
Hint
Stable means nearby solutions move toward it.
Worked Solution
If the solution starts below a stable equilibrium, it will move toward that equilibrium over time.
18. Interpret
A model has
\[ \frac{dP}{dt}>0 \quad\text{for}\quad 0<P<10 \]
and
\[ \frac{dP}{dt}<0 \quad\text{for}\quad P>10. \]
What does this tell you about \(P=10\)?
Hint
Look at what happens from both sides.
Worked Solution
If solutions below 10 increase and solutions above 10 decrease, then nearby solutions move toward 10.
So \(P=10\) is a stable equilibrium.
\[ \boxed{P=10 \text{ is stable}} \]
Review Sheet: Key Patterns to Recognize
Exponential Growth / Decay
If the equation looks like
\[ \frac{dP}{dt}=kP \]
then:
- \(k>0\) → exponential growth
- \(k<0\) → exponential decay
- solution: \[ P(t)=Ce^{kt} \]
Logistic Growth
If the equation looks like
\[ \frac{dP}{dt}=rP\left(1-\frac{P}{K}\right) \]
then:
- \(P=0\) and \(P=K\) are equilibria
- \(P=K\) is usually stable
- growth is fastest in the middle
- the population levels off at \(K\)
Practice Problems
Solve: \[ \frac{dy}{dx}=4y \]
Solve the IVP: \[ \frac{dy}{dx}=4y, \qquad y(0)=3 \]
Find the equilibria of \[ \frac{dP}{dt}=P(6-P) \]
Classify the equilibria of \[ \frac{dP}{dt}=P(6-P) \]
Find the unknown rate constant if \[ P(t)=2e^{kt}, \qquad P(3)=16 \]
For \[ \frac{dP}{dt}=P(3-P)-2 \] find the equilibrium solutions.
Answer Key
1. \[ y=Ce^{4x} \]
2. \[ y=3e^{4x} \]
3. \[ P=0,\ 6 \]
4. \[ P=0 \text{ unstable},\quad P=6 \text{ stable} \]
5. \[ 16=2e^{3k} \Rightarrow 8=e^{3k} \Rightarrow k=\frac{\ln 8}{3} \]
6. \[ P(3-P)-2=0 \Rightarrow 3P-P^2-2=0 \Rightarrow P^2-3P+2=0 \Rightarrow (P-1)(P-2)=0 \]
\[ P=1,\ 2 \]