Exponential Functions

5.6 Exponential Functions as Mathematical Models

Exponential Growth

• The exponential function describes a quantity Q(t) that is initially present in the amount of Q(0) = Q0.

• The growth of the quantity is characterized by the equation:

  $Q(t) = Q_0 e^{kt}$

  where:

  • Q(t) = quantity at time t

  • k = constant of proportionality (growth rate)

• The derivative expressing the rate of growth at time t is given by:

  $Q'(t) = rac{d}{dt}(Q<em>0 e^{kt}) = Q</em>0 k e^{kt}$

• This indicates that the rate of change of the quantity (Q'(t)) is directly proportional to the current amount of the quantity present at time t.

• A function exhibiting this kind of growth is said to exhibit unrestricted exponential growth.

Example 1: Bacterial Growth

• Initial quantity: 10,000 bacteria

• Quantity after 2 hours: 60,000 bacteria

  • Part a: Calculate the number of bacteria at the end of 4 hours:

  1. Determine k from the growth equation:
     $60,000 = 10,000 e^{2k}$
     $6 = e^{2k}$
     $2k = ext{ln}(6)$

     $k = rac{ ext{ln}(6)}{2}$

  2. Using k to find the population at t = 4:
     $Q(4) = 10,000 e^{4k} = 10,000 e^{4( rac{ ext{ln}(6)}{2})} = 10,000 imes 6^4$
     $Q(4) = 10,000 imes 1296 = 12,960,000$

  • Part b: Determine the rate of growth after 4 hours using Q'(t):
    $Q'(4) = 10,000 k e^{4k}$

Exponential Decay

• A quantity exhibits exponential decay if it decreases at a rate proportional to its size, represented as:

  $Q(t) = Q_0 e^{-kt}$

  where k is a positive constant that represents the decay rate.

• The derivative of this function is given by:

  $Q'(t) = rac{d}{dt}(Q<em>0 e^{-kt}) = -Q</em>0 k e^{-kt}$

Half-Life

• The half-life of a radioactive substance is defined as the time required for a quantity to reduce to half its initial amount, represented mathematically as:

  $Q(t<em>{1/2}) = rac{Q</em>0}{2}$

Example 2: Carbon-14 Decay

• Given that the half-life of Carbon-14 is 5730 years, determine the decay constant:

  • By using the half-life formula and solving for k in the equation:

  $rac{Q<em>0}{2} = Q</em>0 e^{-k imes 5730}$

• This leads to:
  $rac{1}{2} = e^{-k imes 5730}$

  Taking the natural logarithm, we find:
  $-k imes 5730 = ext{ln}(0.5)$

  Therefore:
  $k = - rac{ ext{ln}(0.5)}{5730}$

Example 3: Age of Wood Deposits

• Wood deposits contain 20% of the original Carbon-14. To find out how long ago the tree died:

  • Again, use the half-life related calculations and set up the equation:

  $0.2Q<em>0 = Q</em>0 e^{-kt}$

  • By solving this, deduce the time since the tree died.

Supplemental Examples

Example 4: Global Population Modeling

• World population in 1990: 5.3 billion. Assuming a growth rate of 2%/year, find:

  • Part a: The function Q(t) expresses the world population (in billions) as:

  $Q(t) = 5.3 e^{0.02t}$
  where t is measured in years from 1990.

  • Part b: Estimate the world population in 2020:

  $Q(30) = 5.3 e^{0.02 imes 30}$

  • Part c: Based on the function, determine the rate of growth in 2020:

  $Q'(t) = Q_0 (r) e^{rt}$
  where r = 0.02.

  • Part d: To calculate the time required for the population to triple:

  Find t0 such that:
  $3 imes 5.3 = 5.3 e^{0.02t_0}$

  • Part e: If growth rate is reduced to 1.8%/year, apply the earlier findings to determine population size.

Example 5: Bank Failures Post-Financial Crisis

• Notable peak in bank failures in 2010 at 157.

• The function modeling failures from 2010-2012 is:

  $f(t) = 157 e^{-0.55t}$
  where t is years after 2010.

  • Part a: Determine the rate of decrease in failures in 2011 by computing:

  $f'(1) = -157 imes 0.55 e^{-0.55 imes 1}$

  • Part b: Predict the number of bank failures by 2013 using the same model.