Exponential and Logarithmic Models Summary
Exponential and Logarithmic Models
Four common types of mathematical models involving exponential and logarithmic functions:
Exponential Decay: A = A_0e^{kt}, where k < 0 is the decay rate.
Newton's Law of Cooling: T(t) = Ae^{kt} + Ts, where Ts is the surrounding temperature.
Logistic Growth: f(x) = \frac{c}{1 + Ae^{-Bx}}.
Exponential Growth
Model: A(t) = A_0e^{kt}, where:
A(t) is the population at time t.
A_0 is the initial population size.
k is the growth rate.
t is time.
Example: Fruit Flies
Initial setup to find k includes two points:
After 2 days, 100 flies: (2, 100)
After 4 days, 300 flies: (4, 300)
*Solving for k: k = \frac{ln(3)}{2}Finding A0: A0 = \frac{100}{3} = 33.33
Resulting function: A(t) = 33.33e^{\frac{ln(3)}{2}t}
*Solving for A(5) yields approximately 520 flies.A(5) = 33.33e^{\frac{ln(3)}{2}5} \approx 520
Example: Bacteria doubling every two hours
Equation: 200 = 100e^{k \cdot 2}
Solve for k: k = \frac{ln(2)}{2}. Function is A(t) = A_0e^{\frac{ln(2)}{2}t}.
Doubling Time
Formula: A(t) = A_0 \cdot 2^{\frac{t}{a}}, where a is the doubling time.
Example: Bacterial culture doubling every 15 minutes
Initial: A_0 = 2000, doubling time a = 15 minutes.
Amount of bacteria in 1 hour (60 minutes): A(60) = 2000 \cdot 2^{\frac{60}{15}} = 32,000
Amount after 6 hours (360 minutes): A(360) = 2000 \cdot 2^{\frac{360}{15}} = 3.53 \times 10^{75}
Exponential Decay
Example: Radioactive dye
Initial dose: 10 mg. After 5 minutes, 6 mg remain. Detector sounds alarm if more than 2 mg are present.
Finding k: 6 = 10e^{k \cdot 5}, so k = \frac{ln(0.6)}{5}.
Time until 2mg remains: 2 = 10e^{\frac{ln(0.6)}{5}t}. Solving for t gives t \approx 15.75 minutes.
Half-Life
Time for a quantity to reduce to half its original amount.
Formula: k = \frac{ln(0.5)}{t} = \frac{-ln(2)}{t}
Example: Plutonium-239
Half-life of 24,100 years. Amount remaining as a function of time:
A(t) = A0e^{\frac{-ln(2)}{24100}t} = A0 \cdot (\frac{1}{2})^{\frac{t}{24100}}
Alternate Half-Life Formula:A(t) = A_0(\frac{1}{2})^{\frac{t}{a}}, where a is the half-life.
Radiocarbon Dating
*Used to estimate the age of organic matter.
Carbon-14 half-life: 5730 years.
Formula for Carbon-14 remaining: A = A_0e^{\frac{ln(0.5)}{5730}t}
Where,A is the remaining amount of Carbon-14.
A_0 is the intial amount of carbon-14 when the plant/animal began decaying.
Solving for age: t = \frac{ln(r)}{-0.000121}, where r = \frac{A}{A_0}. R is the percentage of carbon-14 in object to the percentage of carbon-14 in the air.
Example: Fossil with 35% carbon-14 compared to a living sample
Age: t = \frac{ln(0.35)}{-0.000121} \approx 8,676 years.
Newton's Law of Cooling
Formula: T(t) = Ae^{kt} + Ts, where: * T(t) is the temperature at time t. * Ts is the surrounding air temperature.
* A is the difference between the initial temperature of the object and T_s.
* k is a constant.
Example: Coffee cooling
Coffee starts at 180°F, room temperature is 76°F. After 5 minutes, coffee is 168°F.
A = 180 - 76 = 104. Initial function: T(t) = 104e^{kt} + 76.
Solving for k using the information provided
Substitute values into the equation168 = 104e^{k \cdot 5} + 76 so k = \frac{ln(\frac{92}{104})}{5}.
Function: T(t) = 104e^{\frac{ln(\frac{92}{104})}{5}t} + 76.
Finding time for coffee to reach 155°F:
Set T(t)=155.155 = 104e^{\frac{ln(\frac{92}{104})}{5}t} + 76. Solving for t gives t \approx 11.21 minutes.
Logistic Growth
Curve shape: Rapid growth that tapers off to an upper limit. Formula: f(x) = \frac{c}{1 + Ae^{-Bx}}, where:
\frac{c}{1 + A} is the initial value.
c is the carrying capacity or limiting value.
B is a constant determined by the growth rate.
Example: Endangered species
100 animals released, carrying capacity of 1000, model: P(t) = \frac{1000}{1 + 9e^{-0.1656t}}
Population after 5 months: P(5) = \frac{1000}{1 + 9e^{-0.1656 \cdot 5}} \approx 203 animals.
Number of months until the population is 500:
Set P(t)=500 and Input500 = \frac{1000}{1 + 9e^{-0.1656t}}. Solving for t gives t \approx 13 months.