Lecture on Exponential Functions and Basic Modeling

Introduction to Exponential Growth Modeling

• Population Growth Example
  • Population grows exponentially under certain conditions.
  • Example: Population of 1,000 doubles in 2 years.
  • Goal: Find an equation modeling the population size after t years.

Formulating the Model

• Assumed Model Form: ( f(t) = y_0 b^t )
  • ( y_0 ): Initial quantity (population at time 0).
  • ( b ): Base of the exponential function.
• Given Data:
  • ( f(0) = 1000 )
  • ( f(2) = 2000 ) (population doubles in 2 years).

Solving for Unknowns

• Calculate ( y_0 )
  • ( f(0) = 1000 = y_0 \times b^0 )
  • Since ( b^0 = 1 ), ( y_0 = 1000 ).
• Calculate ( b )
  • ( f(2) = 2000 = 1000 \times b^2 )
  • ( b^2 = 2 ), therefore ( b = \sqrt{2} ).
• Final Model Equation: ( f(t) = 1000 (\sqrt{2})^t )

Application Example

• Predict population after 10 years:
  • ( 1000 \times (\sqrt{2})^{10} = 32,000 )

Introduction to Exponential Growth in Finance

• Example: Money invested grows exponentially.
  • Example: $1,000 grows to $1,316 in 7 years.

Modeling Investment Growth

• Formulation
  • ( f(t) = y_0 b^t )
  • ( f(0) = 1000 )
  • ( f(7) = 1316 )
• Solving for ( b )
  • ( 1316 = 1000 b^7 )
  • ( b^7 = \frac{1316}{1000} )
  • ( b \approx 1.04 )
• Final Model Equation: ( f(t) = 1000 (1.04)^t )
• Predict account value after 12 years:
  • ( 1000 \times (1.04)^{12} \approx 1601 )

Introduction to Exponential Decay

• Decay Modeling
  • Similar to growth but with ( 0 < b < 1 ).
  • Example: $1,000 declines to $850 in 2 years.
• Formulation
  • ( f(t) = y_0 b^t )
  • ( f(0) = 1000 )
  • ( f(2) = 850 )
• Solving for ( b )
  • ( b^2 = \frac{850}{1000} = 0.85 )
  • ( b = \sqrt{0.85} )
• Final Model Equation for Decay: ( f(t) = 1000 (\sqrt{0.85})^t )

Conclusion

• Exponential models apply to growth and decay contexts with different base values.
• Future sessions will cover more on this topic.