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Section 15.3- APPLICATIONS:
A radioactive element decays in such a way that the amount present each year is 0.95 of the amount present the previous year. At the start of 1990, there were 50 mg. of the element present.
a) Produce a table of values showing how much of the element is present at the start of each year from 1990 to 1998.
b) The rule for this situation is given by N = k x at, where N mg is the amount of element present t years after the start of 1990. Find the value of a and k.
c) sketch the graph of N = k x at.
d) How long will it be before there is 25 mg of this element remaining?
SOLUTION:
a) Initially there are 50 mg of the element present. At the start of the following year there will be 95% of that left over, i.e., there will be 0.95 x 50 = 47.5 mg left over. At the start of the year there will be 95% of 47.5 mg left over. i.e., there will be 0.95 x 47.5 = 45.125 mg, and so on. Continuing in this manner we can complete the table of values
|
t |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
N |
50 |
47.5 |
41.13 |
42.87 |
40.73 |
38.69 |
36.75 |
34.92 |
33.17 |
b) The initial amount is 50 mg, therefore, k = 50. The quantity is decreasing by 95% each time, therefore, a = 0.95.
d) Using the TI-83 we are solving the equation 25 = 50 x (0.95)t or (as is required for the TI-83), we need to solve the equation 50 x 0.95t - 25 = 0
Put into calculator as shown: Solve(50(.95)^x-25,x,10)
Solution received= 13.51340733
31.51 years for this amount of radioactive element to halve.
EXAMPLE for the Natural Exponential Function:
500 dollars id deposited into a bank account that pays 5.40 % interest per annum.
a) How much money will there be in the account at the end of six years if
i) The interest is compounded quarterly?
ii) The interest is compounded continually?
b) Under continuous compounding, how long will it be before the money in the account is doubled?
c) Show the behavior of continuous compounding using a graph.
SOLUTION: Let the amount of money in the account after t years be $A(t), t>0.
a) i) Because interest is compounded quarterly we need to first determine the interest per quarter i.e., 5.4/4 = 1.35. Next, over a period of six years the interest would have been compounded 6 x 4 = 24 times. Therefore, the amount in the account will be given by A = 500 (1+(1.35/100))24 = 689.826...
That is, after six years there will be $689.83 in the account
ii) For interest compounded continuously over the six year period we have that the compounding term is given by lim (1+ (0.054/n))n = e 0.054 so that the amount in the account is given by A= 500 (e 0.054 )6= 691.323...
That is, after six years there will be $691.32 in the account
b) Initially there is 500 dollars in the account, meaning that we want to find the time taken for the amount to reach one thousand dollars (assuming continuous compounding). That is, we want the value of t such that 1000 = 500 e 0.054t . Using the TI-83, we have that t = 12.84.
That is, it will take approximately 12.84 years
