# Modeling situations with differential equations > AP Calculus BC · Unit 7: Differential Equations > Source: https://www.owlsprep.com/study/ap-calculus-bc-u7-modeling-situations-with-differential-equations/ This guide covers translating real-world contextual descriptions to differential equations for AP Calculus BC, including standard models like exponential growth, logistic growth, mixing problems, and Newton's Law of Cooling, with worked examples and common pitfalls to avoid. **Prerequisites:** Derivative as a rate of change; Basic algebra for equation setup; Chain rule for composite functions ## Learning objectives - Translate verbal contextual descriptions to differential equations - Recognize and set up standard models including exponential growth/decay, Newton's Law of Cooling, logistic growth, and mixing problems - Avoid common setup errors that lose points on AP FRQs - Confirm sign of rate of change matches context ## Core Skill: Translating Verbal Rate Descriptions The core idea of translating context to differential equations is that any statement about how a quantity changes can be written as $\frac{dy}{dt}$ (or appropriate variables for your problem) equal to the net rate of change of the quantity. A key rule of thumb: if the problem states that the rate of change of $y$ is proportional to some quantity, you add a constant of proportionality $k$ to the right-hand side. **Differential Equation Modeling** — The process of translating a verbal description of a changing quantity into a differential equation, the mandatory first step for all differential equation problems on the AP exam. *Notation:* \frac{dy}{dt} = f(y,t) 1. Name all unknown functions and their independent variables 2. Write the derivative of the unknown function of interest 3. Translate proportionality relationships by adding a constant of proportionality 4. Combine all increasing and decreasing rates to get the final net rate for the right-hand side **Worked example:** The number of customers $C(t)$ at a coffee shop increases at a rate proportional to the difference between the current number of customers and the shop's maximum capacity of 120 customers. Write the differential equation that models $C(t)$. 1. 1. Identify variables: $C(t)$ = number of customers at time $t$, maximum capacity = 120. The difference described in the problem is $120 - C$. 2. 2. The rate of change of $C$ is $\frac{dC}{dt}$, which is proportional to $120 - C$. 3. 3. Since customers are increasing toward capacity, $\frac{dC}{dt}$ is positive when $C < 120$, so the constant of proportionality $k>0$. 4. 4. Final differential equation: 5. $$\frac{dC}{dt} = k(120 - C) \quad k>0$$ > **Exam tip:** Always explicitly name the constant of proportionality in your DE for FRQs; AP exam graders require you to include the constant, even if you do not calculate its value, to earn full setup points. ## Standard Common Contextual Models The AP exam frequently tests a set of standard contextual models that you should recognize immediately, though you should always be able to derive them from the problem description instead of relying solely on memorization. The most common standard models are: - **Exponential growth/decay**: If a quantity changes at a rate proportional to the size of the quantity, $\frac{dy}{dt} = ky$, with $k>0$ for growth, $k<0$ for decay. - **Newton's Law of Cooling/Heating**: The rate of change of an object's temperature is proportional to the difference between the object's temperature $T(t)$ and the constant ambient temperature $T_s$, so $\frac{dT}{dt} = k(T - T_s)$. - **Logistic population growth**: For a population with carrying capacity $K$, the growth rate is proportional to both the population size and remaining capacity: $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ with $r>0$. - **Rectilinear motion**: Acceleration is the rate of change of velocity, so $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$ for motion along a line. **Worked example:** A population of deer in a forest has a carrying capacity of 800 deer, and the initial growth rate constant is 0.3 per year. Write the differential equation for the deer population $P(t)$ at time $t$ years. 1. 1. The description matches the logistic growth model for a population with fixed carrying capacity. 2. 2. The standard logistic differential equation is: 3. $$\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$$ 4. 3. Substitute the given values $r=0.3$ (growth rate constant) and $K=800$ (carrying capacity): 5. $$\frac{dP}{dt} = 0.3P\left(1 - \frac{P}{800}\right)$$ 6. 4. Confirm the sign: when $0 < P < 800$, $\frac{dP}{dt}$ is positive, which matches expected population growth, so this is correct. > **Exam tip:** Always confirm that the sign of $\frac{dP}{dt}$ matches your expectation for the context: for logistic growth, $\frac{dP}{dt}$ must be positive when the population is between 0 and the carrying capacity. ## Net Rate Models for Inflow/Outflow Systems Many AP problems model mixing problems, bank accounts, or pollution systems where a quantity is both added and removed at different rates. The core rule for all these problems is: $$\frac{dQ}{dt} = (\text{rate of } Q \text{ entering the system}) - (\text{rate of } Q \text{ leaving the system})$$ For mixing problems (the most common inflow/outflow problem), the inflow rate of solute is equal to the inflow concentration multiplied by the inflow flow rate. The outflow rate of solute is equal to the current concentration of solute in the well-mixed system multiplied by the outflow rate. Current concentration is $\frac{Q(t)}{V(t)}$, where $V(t)$ is the total volume at time $t$. If inflow and outflow rates are equal, $V(t)$ is constant; if not, $V(t) = V_0 + (r_{in} - r_{out})t$, where $V_0$ is the initial volume. **Worked example:** A 50-liter tank initially contains 10 liters of water with 2 kg of sugar dissolved in it. Water containing 0.5 kg of sugar per liter flows into the tank at 2 liters per minute, and well-mixed sugar water flows out at 1 liter per minute. Write the differential equation for the mass of sugar $Q(t)$ in kg at time $t$ minutes. 1. 1. Calculate the total volume at time $t$: initial volume $V_0=10$ L, net inflow = $2 - 1 = 1$ L/min, so $V(t) = 10 + t$. 2. 2. Calculate inflow rate of sugar: $0.5 \text{ kg/L} \times 2 \text{ L/min} = 1$ kg/min. 3. 3. Calculate outflow rate of sugar: concentration = $\frac{Q(t)}{10 + t}$ kg/L, so outflow rate = $1 \times \frac{Q(t)}{10 + t} = \frac{Q}{10 + t}$ kg/min. 4. 4. Apply the net rate rule to get the final differential equation: 5. $$\frac{dQ}{dt} = 1 - \frac{Q}{10 + t}$$ > **Exam tip:** Always calculate $V(t)$ first in mixing problems; do not assume the volume is equal to the maximum tank capacity unless the problem says the tank is always full. ## AP-Style Concept Check **Check your understanding** Test your understanding of basic setup with this multiple-choice question: 1. The amount of a radioactive substance $A(t)$ decays at a rate proportional to the amount of substance remaining at time $t$. Which of the following differential equations correctly models this situation? - $\frac{dA}{dt} = kA$, $k>0$ - $\frac{dA}{dt} = -kA$, $k>0$ - $\frac{dA}{dt} = kA$, $k=0$ - $\frac{dA}{dt} = k(A - A_0)$, $k>0$ *Answer:* $\frac{dA}{dt} = -kA$, $k>0$ *Why:* The rate of change of $A$ is proportional to $A$, so the form is $\frac{dA}{dt} = kA$. Since decay means $A$ is decreasing, $\frac{dA}{dt}$ is negative for positive $A$, so $\frac{dA}{dt} = -kA$ with $k>0$ is correct. ## Common pitfalls - **Wrong:** Writing $\frac{dT}{dt} = k(T - T_s)$ with $k>0$ for a cooling object and claiming the DE is wrong - Why it fails: Students memorize a specific sign convention for Newton's Law and mark any other form incorrect, or get confused about what sign they need - Correct: Check that $\frac{dT}{dt}$ is negative when $T > T_s$ (cooling) and positive when $T < T_s$ (heating); any DE that satisfies this sign condition is acceptable. - **Wrong:** Omitting the constant of proportionality when the problem says 'proportional', e.g. writing $\frac{dP}{dt} = (K - P)$ instead of $\frac{dP}{dt} = k(K - P)$ - Why it fails: Students think the 'proportional' description is just context and forget that proportional means multiplied by an unknown constant - Correct: Every time you see the word 'proportional' in a problem, add a constant of proportionality to your DE before moving on. - **Wrong:** In mixing problems with different inflow/outflow rates, using the maximum tank volume instead of the current volume $V(t)$ to calculate concentration - Why it fails: Students assume the tank starts full, even when the problem says otherwise - Correct: Always calculate $V(t) = V_0 + (r_{in} - r_{out})t$ explicitly before setting up the outflow rate. - **Wrong:** Writing the logistic DE as $\frac{dP}{dt} = rP\left(\frac{P}{K} - 1\right)$ with $r>0$ for a growing population - Why it fails: Students swap the order of the terms, leading to a negative growth rate when $P < K$ - Correct: Always confirm that $\frac{dP}{dt}$ is positive for $0 < P < K$; if it is negative, flip the order of the terms or change the sign of $r$. - **Wrong:** Writing acceleration as $\frac{dx}{dt}$ when the problem describes the rate of change of velocity - Why it fails: Students confuse position, velocity, and acceleration derivatives - Correct: Remember that velocity is $\frac{dx}{dt}$ and acceleration is $\frac{dv}{dt} = \frac{d^2x}{dt^2}$, so match the left-hand side of your DE to what is changing. ## Cheatsheet | Category | Formula | Notes | | --- | --- | --- | | General Net Rate | $\frac{dy}{dt} = \text{Rate in} - \text{Rate out}$ | Applies to all problems with inflows/outflows of a quantity | | Proportional Rate | $\frac{dy}{dt} = ky$ | Rate of change of $y$ proportional to $y$; $k>0$ = growth, $k<0$ = decay | | Proportional to Difference | $\frac{dy}{dt} = k(y - C)$ | $C$ is constant; sign of $k$ must match direction of change | | Newton's Law of Cooling | $\frac{dT}{dt} = k(T - T_s)$ | $T_s$ = constant ambient temperature; $k<0$ for a cooling object | | Logistic Population Growth | $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ | $K$ = carrying capacity, $r>0$ = growth rate; $\frac{dP}{dt}>0$ for $0