EXPONENTIAL AND LOGARITHMIC FUNCTIONS

PROBLEM 6.

A satellite has a radioisotope power supply. The power output in watts is given by the equation

P = 50e^(-t/250)

where t is the time in days and e is the base of natural logarithms.

a. How much power will be available at the end of one year?

Solution: Applying the given equation, we have

P = 50e^(-365/250)

= 5Oe^(-1.46)

= 50 x 0.232

= 11 .6

Thus approximately 11.6 watts will be available at the end of one year.

b. What is the half-life of the power supply? In other words, how long will it take for the power to drop to half its original strength?

Solution:
To find the half-life, we solve the equation

25 = 50e^(-t/250)

for t and obtain

- t/250 = ln O.5

= -0.693

t = 250 x 0.693

= 173.

Thus the half-life of the power supply is approximately 173 days. (Note that In x is a shorter expression for log[base e] x.)

c. The equipment aboard the satellite requires 10 watts of power to operate properly. What is the operational life of the satellite?

Solution: Solving the equation

10 = 50e^(-t/250)

for t gives

-t/250 = ln(10/50)

= In 0.2

= - 1 .609

t = 250 x 1.609

= 402.

Hence the operational life of the satellite is 402 days.

NEXT PAGE

PREVIOUS PAGE

RETURN TO SPACE MATHEMATICS HOME PAGE


INDEX

SPACE EDUCATORS' HANDBOOK HOME PAGE

SCIENCE FICTION SPACE TECHNOLOGY HOME PAGE

SPACE MOVIES CINEMA HOME PAGE

SPACE COMICS

SPACE CALENDAR