Exponential Models with Differential Equations - AP Calculus BC Study Guide

The Core Idea: Exponential Models with Differential Equations

This topic explores a fundamental relationship in calculus and the natural world: situations where the rate of change of a quantity is directly proportional to the amount of the quantity itself. This relationship is modeled by the differential equation $dy/dt=ky$. The solution to this equation is not linear or polynomial, but rather an exponential function, which accurately describes many real-life phenomena.

The core task is to recognize this proportional relationship, translate it into the specific differential equation $dy/dt=ky$, and then use given conditions to find the particular exponential function that models the situation. This framework is the foundation for modeling processes such as population growth, where a larger population leads to a faster rate of growth, and radioactive decay, where the rate of decay slows as the amount of substance decreases.

Key Formulas

The entire topic is built upon a single differential equation and its corresponding general solution.

The Differential Equation:

This equation states that the rate of change of a quantity $y$ with respect to time $t$ is directly proportional to $y$.

$$\frac{dy}{dt}=ky$$

• $y$ is the quantity at time $t$.

• $k$ is the constant of proportionality.

The General Solution:

The family of functions that satisfies the differential equation above is given by:

$$y(t)=C{e}^{kt}$$

• $C$ is the initial value of the quantity, i.e., the value of $y$ when $t=0$.

• $k$ is the same constant of proportionality from the differential equation, representing the relative growth or decay rate.

Understanding the Proportionality Constant and Initial Value

The power of the exponential model $y=C{e}^{kt}$ lies in the specific meanings of the constants $C$ and $k$. Mastering their interpretation is essential.

The Initial Value, $C$

The constant $C$ is always the value of the function at $t=0$. This can be shown by substituting $t=0$ into the general solution:

$$y(0)=C{e}^{k(0)}=C{e}^0=C(1)=C$$

Therefore, in any problem involving this model, if you are given the initial amount of a substance or an initial population, you are being given the value of $C$.

The Constant of Proportionality, $k$

The constant $k$ is the relative rate of growth or decay. It dictates how quickly the quantity $y$ changes in proportion to its current size.

• If $k>0$: The model represents exponential growth. The rate of change is positive, so the quantity $y$ increases over time. This is common in population models.

• If $k<0$: The model represents exponential decay. The rate of change is negative, so the quantity $y$ decreases over time. This is characteristic of radioactive decay.

The value of $k$ is determined by using a data point other than the initial condition. For example, if you know the population at a later time $t_1$, you can plug $t_1$ and $y(t_1)$ into the equation $y(t)=C{e}^{kt}$ (after finding $C$) and solve for $k$.

Core Concepts & Rules

• A statement that "the rate of change of a quantity is proportional to the quantity" mathematically translates to the differential equation $\frac{dy}{dt}=ky$.

• The general solution to the differential equation $\frac{dy}{dt}=ky$ is always the exponential function $y=C{e}^{kt}$. You should know this relationship directly without needing to solve the differential equation by separation of variables.

• The constant $C$ in the solution represents the initial value of the quantity $y$, which is the value at $t=0$.

• The constant $k$ is the constant of proportionality. It is also referred to as the relative growth rate (if positive) or the relative decay rate (if negative).

• This model is specifically used for real-world applications such as population growth and radioactive decay.

Step-by-Step Example 1: Basic Application (Population Growth)

Problem: The number of cells in a culture grows at a rate proportional to the number of cells present. At time $t=0$, there are 1,000 cells. After 3 hours, there are 8,000 cells. Find an equation for the number of cells, $P(t)$, at any time $t\ge 0$.

Step 1: Set up the Differential Equation

The problem states that the growth rate is proportional to the number of cells. This translates directly to:

$$\frac{dP}{dt}=kP$$

Step 2: Write the General Solution

Based on the Essential Knowledge for this topic, the solution to this differential equation is:

$$P(t)=C{e}^{kt}$$

Step 3: Use the Initial Condition to Find $C$

We are given that the initial population is 1,000 cells. This means $P(0)=1000$. The constant $C$ is the initial value.

$$C=1000$$

The model is now a particular solution in terms of $k$:

$$P(t)=1000e^{kt}$$

Step 4: Use the Second Data Point to Find $k$

We are given that after 3 hours, there are 8,000 cells. This means $P(3)=8000$. We substitute $t=3$ and $P(3)=8000$ into our model and solve for $k$.

$$8000=1000e^{k(3)}$$

Divide both sides by 1000:

$$8=e^{3k}$$

Take the natural logarithm of both sides to isolate the exponent:

$$\ln (8)=\ln (e^{3k})$$

$$\ln (8)=3k$$

Solve for $k$:

$$k=\frac{\ln (8)}{3}$$

Step 5: Write the Final Particular Solution

Substitute the value of $k$ back into the model from Step 3.

$$P(t)=1000e^{\left(\frac{\ln (8)}{3}\right)t}$$

This is the final equation for the number of cells at time $t$.

Step-by-Step Example 2: Exam-Style Application (Radioactive Decay)

Problem: A radioactive substance decays at a rate proportional to the amount present. A sample of the substance has an initial mass of 200 grams. After 50 years, the mass is 160 grams.

(a) Find an expression for the mass of the substance at time $t$.

(b) What is the half-life of the substance?

(a) Find an expression for the mass of the substance at time $t$

Step 1: Define Variables and Set Up the Model

Let $M(t)$ be the mass in grams at time $t$ in years. The problem states $\frac{dM}{dt}=kM$. The general solution is:

$$M(t)=C{e}^{kt}$$

Step 2: Find the Initial Value $C$

The initial mass is 200 grams, so $M(0)=200$. Therefore, $C=200$.

$$M(t)=200e^{kt}$$

Step 3: Find the Proportionality Constant $k$

After 50 years, the mass is 160 grams, so $M(50)=160$.

$$160=200e^{k(50)}$$

$$\frac{160}{200}=e^{50k}$$

$$0.8=e^{50k}$$

$$\ln (0.8)=50k$$

$$k=\frac{\ln (0.8)}{50}$$

Note that since $0.8<1$, $\ln (0.8)$ is negative, which correctly models decay.

Step 4: Write the Final Expression

The expression for the mass at time $t$ is:

$$M(t)=200e^{\left(\frac{\ln (0.8)}{50}\right)t}$$

(b) What is the half-life of the substance?

Step 5: Set up the Half-Life Equation

The half-life is the time $t$ required for the mass to decay to half of its initial amount. The initial amount was 200 grams, so half is 100 grams. We need to find $t$ such that $M(t)=100$.

$$100=200e^{\left(\frac{\ln (0.8)}{50}\right)t}$$

Step 6: Solve for $t$

Divide by 200:

$$0.5=e^{\left(\frac{\ln (0.8)}{50}\right)t}$$

Take the natural logarithm of both sides:

$$\ln (0.5)=\left(\frac{\ln (0.8)}{50}\right)t$$

Isolate $t$:

$$t=\frac{50\ln (0.5)}{\ln (0.8)}$$

The half-life of the substance is $\frac{50\ln (0.5)}{\ln (0.8)}$ years.

Using Your Calculator

The methods for solving problems in this topic are primarily analytical, involving algebra and the properties of logarithms. A calculator is not used to find the solution directly but is essential for two key tasks:

1. Numerical Approximation: In a calculator-active section of the exam, you may be asked for a decimal answer. For Example 2, you would use your calculator to compute the final value for the half-life.

   • To calculate $t=\frac{50\ln (0.5)}{\ln (0.8)}$, you would enter $(50\ast ln(0.5))/ln(0.8)$, which gives approximately $155.3$ years.

2. Checking Your Work: You can verify your final equation by graphing it or evaluating it at the given points.

   • For Example 1, you could graph $Y_1=1000e^{\left(\frac{\ln (8)}{3}\right)X}$. Then, check the table of values or trace the graph to confirm that at $X=0$, $Y=1000$ and at $X=3$, $Y=8000$. This confirms your particular solution is correct.

There are no specific calculus functions on the calculator (like numerical integration or differentiation) that are used to solve these problems. The process is algebraic, and the calculator serves as a tool for computation and verification.

AP Exam Quick Hit

Common Question Types

• Finding the Particular Solution: You will be given a word problem stating that a quantity's rate of change is proportional to its size, along with an initial condition and one other data point. You will be asked to find the specific function $y(t)$ that models the situation. (This is identical to Example 1).

• Solving for Time or Amount: Given the differential equation (e.g., $\frac{dy}{dt}=-0.02y$) and an initial value (e.g., $y(0)=500$), you might be asked to find the time $t$ when $y$ reaches a certain value (like finding a half-life) or to find the value of $y$ at a specific time.

• Interpreting Constants: A question may provide the final model, such as $P(t)=150e^{0.04t}$, and ask for the initial population and the relative growth rate. You must identify that $C=150$ is the initial population and $k=0.04$ is the relative growth rate.

Common Mistakes

• Not Memorizing the Solution: Students attempt to solve $\frac{dy}{dt}=ky$ using separation of variables during the exam. While correct, this is slow and can lead to errors with integration or constants. You are expected to know that the solution is $y=C{e}^{kt}$.

• Logarithm Errors: When solving for $k$ from an equation like $A=B{e}^{kt}$, students make algebraic mistakes. A common error is writing $\ln (A)=\ln (B)+\ln (e^{kt})$ instead of first dividing by $B$ to get $\frac{A}{B}=e^{kt}$ and then taking the logarithm: $\ln (\frac{A}{B})=kt$.

• Confusing Initial Value: Using a non-initial data point to determine $C$. The constant $C$ is only the value of $y$ when $t=0$. Any other data point must be used to find $k$.

• Sign Errors in $k$: Forgetting that $k$ must be negative for decay problems. If you are modeling radioactive decay and your calculated $k$ is positive, you have made an error. This often happens by incorrectly evaluating a logarithm, such as thinking $\ln (0.5)$ is positive when it is negative.