Applications

Example 1

Assume that the number of grams of a radioactive substance is given by [Maple Math] ( [Maple Math] ) where x is in years. Make a graph. Will the graph increase or decrease? Why?

Solution .

We expect the graph to decrease since 0.7, the base of the exponential function, is less than one.

[Maple Plot]

Example 2

Estimate when the number of radioactive grams equals 400.

Solution .

We will first add the line y = 400 to the previous graph.

[Maple Plot]

We see that they intersect with x between 2 and 3 years. If we make a table for g(x) with x between 2 and 3, we can get a better estimate.

x                           f(x)

---------------------------

2.                      490.00

2.100000000 472.8309366

2.200000000 456.2634584

2.300000000 440.2764865

2.400000000 424.8496805

2.500000000 409.9634130

2.600000000 395.5987440

2.700000000 381.7373972

2.800000000 368.3617369

2.900000000 355.4547450

3.000000000 343.000

This table shows that the number of grams is 400 for x between 2.5 and 2.6 years. By making another table with x between 2.5 and 2.6 in steps of 0.01, we can get a closer estimate.

x                          f(x)

---------------------------

2.5                    409.9634130

2.510000000 408.5037808

2.520000000 407.0493456

2.530000000 405.6000886

2.540000000 404.1559916

2.550000000 402.7170362

2.560000000 401.2832040

2.570000000 399.8544769

2.580000000 398.4308365

2.590000000 397.0122649

2.600000000 395.5987440

This table shows that the time is between 2.56 years and 2.57 years. The actual answer is closer to 2.57 than to 2.56

Example 3

In the equation [Maple Math] A is the compound amount, P is the principal, r is the interest rate as a decimal, n is the number of compounding times per year, and t is time in years. If $100 is invested at 5% compounded quarterly, make a table showing the compound amount for the years from 0 through 7.

Solution .

In this problem, r = 0.05, n = 4, and P = 100. Consequently, [Maple Math] Since the base of this exponential function is 1.0125 which is more than 1, A should increase over time. The following table confirms this.

t                           A(t)

------------------------

0                       100.

1.000000000 105.0945337

2.000000000 110.4486101

3.000000000 116.0754518

4.000000000 121.9889548

5.000000000 128.2037232

Example 4

Use a graph to estimate when the investment in example 3 would double (from $100 to $200).

Solution .

We will graph A as well as the line y = 200 and estimate the point of intersection.

[Maple Plot]

From the graph we see that it takes about 14 years to double at 5% compounded quarterly.