Download 3.5b Notes - Jessamine County Schools

3.5 – Exponential and Logarithmic Growth Models

Law of uninhibited growth or decay, where
represents the original amount (when t = 0) and t represents
time.
(where growth happens when k > 0 and decay happens when k < 0)

Uninhibited Growth of Cells
- Gives the number N of cells in the culture after a time t has passed, where
is the initial number
of cells and k is a positive constant that represents the growth rate of the cells
Ex) A colony of bacteria grows according to the law of uninhibited growth
measured in grams and t is measured in days.
(a)
(b)
(c)
(d)
(e)
, where N is
Determine the initial amount of the bacteria
What is the growth rate of the bacteria?
What is the population after 5 days?
How long will it take for the population to reach 140 grams?
What is the doubling time of the population?
(a) The initial amount of bacteria,
, is obtained when t = 0.
(Can also be found knowing that 100 represents the initial quantity in the equation)
(b) The k value represents the growth rate of the bacteria. Since k = 0.045, the growth rate is 4.5%
(c) The population after 5 days is
(d) For the population to reach 140 grams, N(t) = 140. Set up the equation and solve.
Set up equation
Divide both sides by 100
Take ln of both sides
Power rule and fact that ln e = 1
Divide by 0.045
(e) The population doubles when N(t) = 200 grams, so we solve this equation.
Set up equation
Divide by 100
Power Rule and fact ln e = 1
Divide by 0.045

Radioactive Decay
- The amount A of a radioactive material present at a a time t, where
is the original amount of
the radioactive material and k is a negative number that represents the rate of decay.
, where k < 0
Note: all radioactive substances have a specific half-life, which is the time required for half of the
radioactive substance to decay.
Ex) Traces of burned wood along with ancient stone tools in an archeological dig in Chile were found to
contain approximately 1.67% of the original amount of carbon 14.
(a) If the half-life of carbon 14 is 5600 years, approximately when was the tree cut and burned?
(a) Using the decay formula, the amount of carbon 14 present at time t can be determined by the
fact that the half-life of carbon 14 is 5600 years.
Substitute into formula
Divide by
Power Rule and fact ln e = 1
Gives us the rate of decay
This means the actual decay equation is
Since there is 1.67% of the original amount of carbon 14 left, we know that we have
0.0167 . Using this fact, set up an equation and solve.
Set up the new equation with calculated k value
Divide both sides by
Rewrite as logarithm

Newton’s Law of Cooling
- States that the temperature of a heated object decreases over time toward the temperature of
the surrounding medium.
- The temperature u of a heated object at a given time t, where T is the constant temperature of
the surrounding medium and is the initial temperature of the heated object and k is a
negative constant.
, where k < 0
Ex) An object is heated to
temperature is
(degrees Celsius) and is then allowed to cool in a room whose air
.
(a) If the temperature of the object is
(a) Using the formula with T = 30 and
after 5 minutes, when will it’s temperature be
, the temperature of the object at time t (in
minutes) is
In order to find k, we use the fact that u = 80 and t = 5
Substitute into formula
Subtract 30
Divide by 70
Power Rule and fact ln e = 1
Gives us the rate of cooling
This means the actual cooling equation is
We want to find t when u =
. Using this fact, set up an equation and solve.
Set up the new equation with calculated k value
Subtract 30
Divide by 70
Power Rule and fact ln e = 1
?