6.4 Exponential Growth and Decay
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Transcript of 6.4 Exponential Growth and Decay

6.4 Exponential Growth and DecayGreg Kelly, Hanford High School, Richland, WashingtonGlacier National Park, MontanaPhoto by Vickie Kelly, 2004

The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.)So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interestbearing account.If the rate of change is proportional to the amount present, the change can be modeled by:

Rate of change is proportional to the amount present.Divide both sides by y.Integrate both sides.

Integrate both sides.Exponentiate both sides.When multiplying like bases, add exponents. So added exponents can be written as multiplication.

Exponentiate both sides.When multiplying like bases, add exponents. So added exponents can be written as multiplication.

Exponential Change:If the constant k is positive then the equation represents growth. If k is negative then the equation represents decay.

Continuously Compounded InterestIf money is invested in a fixedinterest account where the interest is added to the account k times per year, the amount present after t years is:If the money is added back more frequently, you will make a little more money.The best you can do is if the interest is added continuously.

Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine.We could calculate:but we wont learn how to find this limit until chapter 8.Since the interest is proportional to the amount present, the equation becomes:
(The TI89 can do it now if you would like to try it.)

Radioactive DecayThe equation for the amount of a radioactive element left after time t is:This allows the decay constant, k, to be positive.The halflife is the time required for half the material to decay.

Halflife

Newtons Law of CoolingEspresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air.(It is assumed that the air temperature is constant.)If we solve the differential equation:we get:p
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