Deriving and Applying a Model for Exponential Growth and Decay
In AP Calculus AB and BC, exponential growth and decay models are fundamental for understanding how quantities change over time. These models describe situations where the rate of change of a quantity is proportional to the quantity itself. This concept is essential in fields such as biology, economics, and physics. By deriving the exponential function from a simple differential equation, students can apply it to various real-world scenarios, making it a crucial topic for mastering both AP Calculus exams.
Learning Objectives
When studying “Deriving and Applying a Model for Exponential Growth and Decay” for the AP Calculus AB and BC exams, you should focus on understanding how to derive the exponential growth and decay models from first-order differential equations. Learn to interpret the significance of the growth/decay constant π, and apply these models to solve real-world problems involving population growth, radioactive decay, and continuously compounded interest. Master solving problems related to half-life, doubling time, and continuous growth/decay scenarios using these models.
Understanding Exponential Growth and Decay
![Understanding Exponential Growth and Decay](https://images.examples.com/wp-content/uploads/2024/08/Understanding-Exponential-Growth-and-Decay.png)
In real-world scenarios, many processes, such as population growth, radioactive decay, and continuously compounded interest, can be modeled by exponential functions.
- Exponential Growth occurs when a quantity increases over time.
- Exponential Decay occurs when a quantity decreases over time.
The general form of an exponential function is:
\(y(t) = y_0 \cdot e^{kt} \)
Where:
- y(t) is the quantity at time t.
- y0β is the initial quantity at t = 0.
- k is the growth (k > 0) or decay (k < 0) rate.
- e is the base of the natural logarithm, approximately equal to 2.718.
Deriving the Exponential Growth and Decay Model
![Deriving the Exponential Growth and Decay Model](https://images.examples.com/wp-content/uploads/2024/08/Deriving-the-Exponential-Growth-and-Decay-Model.png)
The derivation of the exponential model starts with the assumption that the rate of change of a quantity y(t) is proportional to the quantity itself. Mathematically, this is expressed as :
\(\frac{dy}{dt} = ky(t) \)
This is a first-order differential equation. The solution to this differential equation gives us the model for exponential growth or decay.
Step 1: Separate Variables
- To solve this differential equation, we first need to separate the variables y and t. This involves manipulating the equation so that all terms involving y are on one side and all terms involving t are on the other side:
- This separation allows us to integrate both sides with respect to their respective variables, which leads us to the next step.
\(\frac{1}{y} \, dy = k \, dt \)
Step 2: Integrate Both Sides
- Now that we have separated the variables, the next step is to integrate both sides to find the general solution to the differential equation.
\(\int \frac{1}{y} \, dy = \int k \, dt \)
\(\ln |y| = kt + C \)
This equation represents the logarithm of the absolute value of y as a linear function of time t.
Step 3: Solve for y(t)
- To find the explicit solution for y(t), we need to solve the equation obtained in Step 2 for y.
- We start by exponentiating both sides of the equation to eliminate the natural logarithm. Exponentiating both sides gives
\(|y| = e^{kt+C} \)
Since ekt+C can be rewritten as ektβ eC , and because eC is itself a constant, we can write:
\(y(t) = Ae^{kt} \)
Apply the initial condition y(0) = y0β
\(y_0 = Ae^{k \cdot 0} = A \)
Therefore, the solution is
\(y(t) = y_0 e^{kt} \)
This is the exponential growth and decay model.
Applying the Exponential Growth and Decay Model
![Applying the Exponential Growth and Decay Model](https://images.examples.com/wp-content/uploads/2024/08/Applying-the-Exponential-Growth-and-Decay-Model.png)
1. Exponential Growth
- When k > 0, the function \( y(t) = y_0 e^{kt} \) represents exponential growth.
- Common applications include population growth, interest accumulation, and spread of diseases.
Example: Suppose a population of 100 bacteria grows at a rate of 5% per hour. The growth model is: \(y(t) = 100 \cdot e^{0.05t} \)
To find the population after 10 hours: \(y(10) = 100 \cdot e^{0.05 \cdot 10} \approx 164.87 \)
2. Exponential Decay
- When k < 0, the function \( y(t) = y_0 e^{kt} \) represents exponential decay.
- Common applications include radioactive decay, depreciation of assets, and cooling of objects.
Example: A radioactive substance decays at a rate of 3% per hour. If the initial amount is 200 grams, the decay model is: \(y(t) = 200 \cdot e^{-0.03t} \)
To find the amount remaining after 5 hours: \(y(5) = 200 \cdot e^{-0.03 \cdot 5} \approx 171.77 \text{ grams} \)
Key Concepts for AP Calculus AB and BC
1. Rate of Change Interpretation
- The differential equation \( \frac{dy}{dt} = ky \) shows that the rate of change is proportional to the quantity. This is crucial for understanding how rapidly quantities grow or decay.
2. Half-Life and Doubling Time
- Half-Life (for decay) is the time it takes for a quantity to reduce to half its initial value. It can be derived using the equation \( t_{1/2} = \frac{\ln 2}{|k|} \)β.
- Doubling Time (for growth) is the time it takes for a quantity to double. It can be derived using \( t_{2x} = \frac{\ln 2}{k} \)β.
Example (Doubling Time): If a population grows at a rate of 7% per year, the doubling time is: \(t_{2x} = \frac{\ln 2}{0.07} \approx 9.9 \text{ years} \)
3. Solving Real-World Problems
- On the AP Calculus exam, you might be given a scenario and asked to set up and solve an exponential model. Be sure to understand how to extract key information like the initial value, growth/decay rate, and time intervals from word problems.
Examples
Example 1: Bacterial Population Growth
Suppose a certain type of bacteria doubles in number every 3 hours. If you start with 500 bacteria, the population can be modeled by the exponential growth function y(t) = 500 β ekt. To find the growth rate k, use the fact that the population doubles when t = 3, so 1000 = 500β e3k. Solving this gives \( k = \frac{\ln 2}{3} \approx 0.231 \). Therefore, the population at any time t is y(t) = 500β e0.231t.
Example 2: Radioactive Decay
A radioactive substance has a half-life of 10 years. If the initial amount is 120 grams, the decay of the substance can be modeled by y(t) = 120β ekt. The half-life condition gives 60 = 120β e10k. Solving this, \( k = \frac{\ln 0.5}{10} \approx -0.0693 \). The amount of the substance after t years is then given by y(t) = 120β eβ0.0693t.
Example 3: Continuous Compound Interest
An initial investment of $5,000 is placed in an account that earns 4% interest per year, compounded continuously. The amount of money in the account after t years is given by the model A(t) = 5000β e0.04t. To find the balance after 8 years, substitute t = 8 into the equation, yielding A(8) β 5000β e0.32 β 6835.38. This shows that the investment will grow to approximately $6,835.38 after 8 years.
Example 4: Carbon-14 Dating
The amount of Carbon-14 in a fossil decays exponentially over time, with a half-life of 5,730 years. If a fossil originally had 10 grams of Carbon-14 and now has only 2 grams, the time elapsed can be calculated using the decay model y(t) = 10β ekt. Knowing \( k = \frac{\ln 0.5}{5730} \approx -0.00012097 \), we can solve 2 = 10β eβ0.00012097t to find t β 13344 years. This suggests the fossil is approximately 13,344 years old.
Example 5: Pharmacokinetics
A drug is administered intravenously, and its concentration in the bloodstream decreases exponentially as the body metabolizes it. Suppose the drug has an initial concentration of 200 mg/L and a half-life of 4 hours. The decay of the drug’s concentration over time can be modeled by \( C(t) = 200 \cdot e^{kt} \), where \( k = \frac{\ln 0.5}{4} \approx -0.1733 \). The concentration of the drug after 12 hours is C(12) = 200β eβ0.1733β 12 β 25 mg/L, indicating that the concentration has dropped significantly over this period.
Multiple Choice Questions
Question 1
The population of a certain species of fish in a lake is modeled by the function P(t) = 200β e0.04t, where P(t) represents the population at time t in years. What is the approximate population after 5 years?
A) 200
B) 244
C) 245
D) 250
Answer: C) 245
Explanation:
To find the population after 5 years, substitute t = 5 into the given exponential model:
P(5) = 200β e0.04Γ5
Calculate the exponent:
P(5) = 200β e0.2
Using e0.2 β 1.221, we get:
P(5) β 200β 1.221 = 244.2
Rounding to the nearest whole number, the population is approximately 245.
Question 2
A substance decays according to the function N(t) = N0ββ eβ0.1t, where N(t) is the amount remaining at time t and N0β is the initial amount. If N0β = 100 grams, how much of the substance remains after 10 hours?
A) 36.8 grams
B) 50 grams
C) 60.6 grams
D) 90.5 grams
Answer: A) 36.8 grams
Explanation:
To find the amount remaining after 10 hours, substitute t = 10 into the exponential decay model:
N(10) = 100β eβ0.1 Γ 10
Simplify the exponent:
N(10) = 100β eβ1
Using eβ1 β 0.368, we get:
N(10) β 100β 0.368 = 36.8Β grams
So, after 10 hours, approximately 36.8 grams of the substance remains.
Question 3
If a population doubles every 5 years, what is the value of the constant k in the exponential growth model P(t) = P0ββ ekt?
A) \( \frac{\ln 2}{5} \)
B) \( \ln 2 \)
C) \( 5 \ln 2 \)
D) \( \frac{1}{5} \)
Answer: A) \( \frac{\ln 2}{5} \)
Explanation:
The doubling time of a population is the time Tdβ it takes for the population to double, which satisfies:
\(P(T_d) = 2P_0 = P_0 \cdot e^{kT_d} \)
Dividing both sides by P0β:
\(2 = e^{kT_d} \)
Taking the natural logarithm of both sides:
\(\ln 2 = kT_d \)
Solving for k:
\(k = \frac{\ln 2}{T_d} \)
Given that the doubling time Tdβ is 5 years:
\(k = \frac{\ln 2}{5} \)
Therefore, the value of k is \( \frac{\ln 2}{5} \)β.
These questions and detailed explanations are designed to test and reinforce your understanding of exponential growth and decay, a crucial topic for AP Calculus.