class 051109
Exponential functions happen when the rate of change of a quantitiy y with respect to another quantity t is directly proportional to y.
For example, if you have $1000 at 5% interest then in a year you will gain $50, so the average rate of change is $50 / yr. If you have $2000 at 5%, then you gain $100 in a year so the ave rate of change is $100 / yr.
Note that in the above examples the growth rate is .05, the growth factor is 1.05, the same for both situations. The exponential functions would be different; one would be $1000 * 1.05^t and the other would be $2000 * 1.05^t. The time period t is 1 year in each example.
The average rate of change of y with respect to t was $50 / yr in the first example and $100 / yr in the second. The ave rate of change of y with respect to t is not to be confused with growth rate or growth factor.
What would be the average rate of change of y with respect to t for each of the two functions
y = $1000 * 1.05^t and
y = $2000 * 1.05^t
over the 2-year period of the first two years?
The ave rate of change of y with respect to t is
ave rate = change in y / change in t.
We are finding ave rate from t = 0 to t = 2.
For the first function we have
y(0) = $1000 (1.05)^0 = $1000 and
y(2) = $1000 ( 1.05)^2 = $1102.50
so that
ave rate = ($1102.50 - $1000) / (2 yr - 0 yr) = $102.50 / (2 yr) = $51.25 / yr.
For the first two years our principle goes up by $51.25 / yr, on the average. In the first year it goes up only $50/yr, but in the second it goes up be $52.50 / yr.
On a graph of y vs. t the $102.50 increase over the 2-year period is represented by the rise of the graph between the t = 0 and t = 2 points, and the 2-year period itself by the run. So the average rate of change is represented by slope = rise / run.
The calculations for the second function are identical to those for the first, except that all our y values are twice as great and the rate of change is twice as great.
Using DERIVE we can author the function
p(t):=1000*1.05^t
and then since p(2) will be our t = 2 principle and p(0) our t = 0 principle we evaluate the expression
(p(2) - p(0)) / ( 2 - 0)
to get the same result.
Note that the function
aveRate(t):=(p(t) - p(0)) / ( t - 0)
will give us the average rate between clock time 0 and clock time t. So for example
aveRate(1) will evaluate to 50,
aveRate(2) will evaluate to 51.5 and
aveRate(3) will evaluate to 52.54, approx..
1. (a^b)^c, a^b * a^c, a^(b + c), (a+b)^c and a^(bc) are not all equal quantities, but some of these quantities are equal to one another. Explain in terms of common sense which must be equal to which.
2. Evaluate the exponential function P(t) = 1000 * 1.05^t for a series of 4 different values of t, where each t value is greater than the one before it by the number of letters in your last name. Write the results as a sequence.
Figure out the successive ratios of that sequence. What pattern do you observe?
3. The function y = 100 * 2^(.03 t) can also be written in the form y = 100 * b^t. What is the value of b for this function?
4. Which of the following recurrence relations result in an exponential functions?