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# Hprec5.3

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• 14533.06
• F(x)=2.5(1.0185)^x
• 351/Ex7: 7600=(1+r)^5, solve for 1+r then r. Then f(x)=1000(1+r)^24 (r=.5002)
• 11.88 mg
• .93P=P(.5)^(x/5730 ;xMax=1000-Solve graphically. Ans:599 years prior to 1988 or about 1389. Use 2ndCalc 2
• ### Hprec5.3

1. 1. 5-3: Applications of Exponential Functions © 2007 Roy L. Gover (www.mrgover.com) Learning Goals: •Create and use exponential models
2. 2. Definition Compound Interest Model If P dollars is invested at interest rate r (expressed as a decimal) per time period t, then A is the amount after t periods. (1 )t A P r= + An exponential grow function
3. 3. Try This If you invest \$9000 at 4% interest compounded annually, use the compound interest model to find how much you have after the end of 12 years. \$14409.29
4. 4. ExampleIf you invest \$9000 at 4% interest compounded monthly, use the compound interest model to find how much you have after 12 years.
5. 5. Try This If you invest \$9000 at 4% interest compounded weekly, use the compound interest model to find how much you have after the end of 12 years. There are 52 weeks in a year. \$14541.99
6. 6. Definition The Continuous Compounding Model If P dollars are invested at an annual interest rate of r compounded continuously then A is the amount after t years. rt A Pe=
7. 7. Try This If you invest \$9000 at 4% interest compounded continuously, use the continuous compounding model to find how much you have after the end of 12 years. \$14544.67
8. 8. Important Idea Compounding Amount* Annual \$14409.29 Monthly \$14533.06 Weekly \$14541.99 Continuous \$14544.67 *\$9000 @ 4% for 12 yrs.
9. 9. Example The world population in 1950 was about 2.5 billion people and has been increasing at approximately 1.85% per year. Write the equation that gives the world population in year x when x corresponds to 1950.
10. 10. Important Idea (1 )t A P r= + ( ) (1 )x f x P r= + Exponentia l Grow Model: Compound Interest Model: The models are the same.
11. 11. Example At the beginning of an experiment, a culture contains 1000 bacteria. Five hours later there are 7600 bacteria. Assuming the bacteria are growing exponentially, how many bacteria will there be after 24 hours?
12. 12. Try This A newly formed lake is stocked with 900 fish. After 6 months, biologists estimate there are 1710 fish. Assuming exponential growth, how many fish will there be after 24 months? 11729
13. 13. Definition Exponential Decay is of the form where( ) (1 )x f x P r= − ( )f x is the amount at time x, P is the beginning amount (at t=0) and r is the decay rate.
14. 14. Example Each day 15% of the chlorine in a swimming pool evaporates. Use the Exponential Decay Model to predict how many days are required for 60% of the chlorine to evaporate.
15. 15. Definition The half life of a radioactive substance is the time required for a given quantity of the substance to decay to one-half its original mass.
16. 16. Definition The amount of a radioactive substance that remains is given by ( ) (.5) x hf x P= where is the remaining amount, P is the initial amount and h is the half life of the substance. ( )f x
17. 17. Example An isotope of strontium-90 has a half-life of 25 years. How much of an 18mg. sample be left after 15 years?
18. 18. Try This Carbon 14 is a radio-active substance with a half life of 5730 years. How much of a 5 mg. sample will be left after 300 years? 4.82 mg.
19. 19. Example In 1988, the Vatican authorized the British Museum to date a cloth relic know as the Shroud of Turin. This cloth which first surfaced in the 14th century, contains the negative image of a human body that was widely believed to be that of Jesus.
20. 20. The report of the museum showed that the fibers of the cloth contained about 93% of their original carbon-14. The half-life of carbon-14 is 5730 years. Estimate the age of the shroud. Adapted from Calculus, Howard Anton, 6th ed
21. 21. Try This If a turtle shell is found and tested to have 45% of its original carbon-14, how old is the shell? The half- life of carbon-14 is 5730 years. About 6599 years
22. 22. Lesson Close Exponential models are widely used to solve compound interest, population growth and resource depletion problems.
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