 # 6.4 Exponential Growth and Decay Greg Kelly, Hanford High School, Richland, Washington Glacier National Park, Montana Photo by Vickie Kelly, 2004.

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6.4 Exponential Growth and Decay Greg Kelly, Hanford High School, Richland, Washington Glacier National Park, Montana Photo 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 interest-bearing 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.

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. Since is a constant, let.

At,. This is the solution to our original initial value problem.

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

Continuously Compounded Interest If money is invested in a fixed-interest 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 have not learned how to find this type of limit. Since the interest is proportional to the amount present, the equation becomes: Continuously Compounded Interest: You may also use: which is the same thing.

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

Half-life Half-life:

Newton’s Law of Cooling Espresso 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: Newton’s Law of Cooling where is the temperature of the surrounding medium, which is a constant.

We have used the exponential growth equation to represent population growth. The exponential growth equation occurs when the rate of growth is proportional to the amount present. If we use P to represent the population, the differential equation becomes: The constant k is called the relative growth rate.

The population growth model becomes: However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal. There is a maximum population, or carrying capacity, M. A more realistic model is the logistic growth model where growth rate is proportional to both the amount present ( P ) and the carrying capacity that remains: ( M-P )

The equation then becomes: Logistics Differential Equation We can solve this differential equation to find the logistics growth model.

Logistics Growth Model

Example: Logistic Growth Model Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

At time zero, the population is 10.

After 10 years, the population is 23.

Years Bears We can graph this equation and use “trace” to find the solutions. y=50 at 22 years y=75 at 33 years y=100 at 75 years 

Assignment Page 338 # 1-21 odd Page 347 # 3, 14, 17, 18

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