AP Calculus AB/BC · Unit 7 · Topic 7.1
Modeling Situations with Differential Equations
Write an equation that relates a quantity's rate to current variables and parameters.
1. Topic Focus
Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.
This topic: Write an equation that relates a quantity's rate to current variables and parameters.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
A population growing proportionally to its size satisfies dP/dt=kP.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
A differential equation turns a description of change into a mathematical rate law. If \(Q(t)\) is the quantity being modeled, a first-order model has the form
The left side is a rate, so every term on the right must have the same rate units. The initial condition \(Q(t_0)=Q_0\) is separate from the differential equation: the equation describes a family of possible behaviors, while the initial value identifies the behavior that fits the situation.
A reliable modeling workflow
- Name the variables and units. State what \(t\) measures and what the dependent variable represents.
- Identify parameters. Record which constants are positive and attach units when useful.
- Build the net rate. Use “rate in minus rate out” or “production minus loss.”
- Translate each verbal relationship. For example, “proportional to the amount” becomes \(kQ\).
- Add initial data. Write \(Q(t_0)=Q_0\) if a starting value is given.
- Audit the model. Check units, signs, equilibria, physical restrictions, and the interval on which the assumptions are reasonable.
Common verbal cues
| Description | Rate model | Interpretation |
|---|---|---|
| increases at a constant rate \(r\) | \(Q'=r\) | the same amount is added per unit time |
| decreases at a constant rate \(r\) | \(Q'=-r\) | \(r>0\) is the magnitude of the loss |
| changes proportionally to its amount | \(Q'=kQ\) | growth if \(k>0\), decay if \(k<0\) |
| moves toward a level \(M\) | \(Q'=k(M-Q)\) | the sign automatically points toward \(M\) |
| input and output occur together | \(Q'=R_{\mathrm{in}}-R_{\mathrm{out}}\) | convert both terms to quantity per time |
| growth is limited by capacity \(K\) | \(P'=rP(1-P/K)\) | growth slows as \(P\) approaches \(K\) |
Units, signs, and equilibria
In \(P'=kP\), if \(t\) is measured in years, then \(k\) has units \(\text{year}^{-1}\). In \(T'=-k(T-A)\), the same reciprocal-time units for \(k\) make both sides temperature per time. An equilibrium is a constant value \(Q=Q_*\) for which the modeled rate is zero. Values on either side of an equilibrium reveal whether solutions move toward it or away from it.
An equation such as \(Q'=F(Q)\) is autonomous: its rate depends on the current state but not explicitly on time. A model such as \(Q'=4+\sin t-0.1Q\) is nonautonomous because its input varies with time.
Model assumptions matter
A model is an approximation, not the situation itself. A mixing model may assume a perfectly stirred tank and constant volume; a cooling model may assume constant ambient temperature; a population model may ignore migration or changing resources. State restrictions such as \(Q\ge 0\), and stop using a model when its assumptions produce physically impossible predictions.
6. Detailed Worked Example and Error Check
Example 1: Population with constant harvesting
A population \(P(t)\) has a per-capita birth rate of \(0.18\) per year and a per-capita death rate of \(0.07\) per year. A fixed \(120\) individuals are removed each year, and \(P(0)=2500\).
Every term has units individuals per year. The equilibrium is \(P=120/0.11\approx1091\). Above it the model predicts growth; below it the fixed harvest exceeds natural net growth. Because the formula could eventually predict a negative population, its physical use must stop at \(P=0\).
Example 2: Newton's law of cooling
A hot object at \(90^\circ\mathrm C\) is placed in a room held at \(22^\circ\mathrm C\). Its temperature changes at a rate proportional to the difference from the room temperature.
When \(T>22\), the derivative is negative; if \(T<22\), it is positive. Thus the sign of the model always moves the temperature toward \(22\), which is its equilibrium.
Example 3: A well-mixed tank
A \(200\)-liter tank contains \(S(t)\) grams of salt. Brine enters and leaves at \(3\) liters per minute, so volume remains constant. The incoming concentration is \(2\) grams per liter.
The outflow concentration is \(S/200\), not simply \(S\). Setting \(S'=0\) gives the equilibrium \(S=400\) grams, the amount corresponding to the incoming concentration throughout the tank.
Example 4: Medication infusion
A medicine enters a patient's bloodstream at \(5\) milligrams per hour and is eliminated at a rate proportional to the amount \(M(t)\), with proportionality constant \(0.12\) per hour.
The equilibrium amount is \(5/0.12\approx41.7\) milligrams. If \(M\) is below that level, the net rate is positive; above it, elimination is faster than infusion.
Example 5: Time-dependent input
Water enters a reservoir at \(8+2\sin(\pi t/12)\) cubic meters per hour and leaves at \(5\) cubic meters per hour. If \(V(0)=300\), then
This model is nonautonomous because the inflow depends explicitly on time. The derivative, rather than \(V\), equals the net flow.
Example 6: Limited population growth
A population has intrinsic growth constant \(0.4\) per year and carrying capacity \(500\). A standard logistic model is
The factors identify equilibria \(P=0\) and \(P=500\). For \(0<P<500\), both factors are positive, so the population grows; above \(500\), the model predicts a decrease.
Example 7: Motion with linear resistance
Take downward velocity \(v(t)\) as positive. Gravity contributes acceleration \(g\), while resistance of magnitude \(kv\) acts upward when the object is falling.
The equilibrium velocity \(v=g/k\) is the terminal velocity. Choosing the positive direction first prevents a sign guess from replacing physical reasoning.
Common modeling errors
- Equating an amount to a rate, such as \(S=6-3S/200\).
- Using inflow concentration as an inflow rate without multiplying by volume per time.
- Writing \(T'=-k|T-A|\), which incorrectly predicts cooling even when \(T<A\).
- Omitting the initial condition or hiding a parameter's required sign.
- Solving an equation before checking whether it represents the verbal situation.
7. AP Reasoning Routine
Translate the context into a rate equation, verify candidate solutions by substitution, carry constants through integration, and apply initial conditions last.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Write a differential equation and any stated initial condition for each situation. Define positive constants when needed, but do not solve the equations.
(a) A lake loses \(15\) cubic meters of water per day.
(b) A culture begins with \(600\) cells and grows at a rate proportional to its current population.
(c) An object at \(20^\circ\mathrm C\) is heated in an oven held at \(180^\circ\mathrm C\), with rate proportional to the temperature difference.
(d) A medication enters at \(8\) milligrams per hour and \(20\%\) of the current amount is eliminated per hour. Find the equilibrium and describe the sign of the rate on each side.
(e) A well-mixed \(100\)-liter tank receives and drains \(4\) liters per minute. Incoming brine contains \(0.5\) gram per liter. Let \(S(t)\) be salt in grams.
(f) A population has carrying capacity \(1200\) and intrinsic growth constant \(0.3\) per year.
(g) A falling object's downward velocity is positive. Gravity contributes \(g\), and air resistance produces acceleration of magnitude \(kv\) opposite the motion. The object is released from rest.
(h) If \(N'=kN\) and \(t\) is measured in days, what units must \(k\) have?
(i) Explain why \(T'=-k|T-20|\), \(k>0\), is not an appropriate model for an object warming from \(5^\circ\mathrm C\) toward a \(20^\circ\mathrm C\) room.
(j) A tank receives liquid at \(4+t\) liters per minute and drains at a rate equal to \(Q/50\) liters per minute, where \(Q(t)\) is its volume and \(Q(0)=100\).
Check the solution
(a) If \(V\) is volume, \(V'=-15\).
(b) \(P'=kP\), \(P(0)=600\), with \(k>0\).
(c) \(T'=k(180-T)\), \(T(0)=20\), with \(k>0\). The derivative is positive below the oven temperature.
(d) \(M'=8-0.20M\). The equilibrium is \(M=40\) milligrams; \(M'>0\) below \(40\) and \(M'<0\) above \(40\).
(e) Inflow is \((4)(0.5)=2\) grams per minute and outflow is \(4(S/100)=S/25\), so \(S'=2-S/25\).
(f) \(P'=0.3P(1-P/1200)\).
(g) \(v'=g-kv\), \(v(0)=0\), with \(k>0\).
(h) Since \(N'\) has units amount per day, \(k\) must have units \(\text{day}^{-1}\).
(i) At \(T=5\), the proposed right side is negative, so it predicts further cooling. A model that points toward room temperature is \(T'=k(20-T)\).
(j) Net rate equals inflow minus outflow: \(Q'=4+t-Q/50\), \(Q(0)=100\). This is nonautonomous because the inflow varies with \(t\).