PRECALCULUS I

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# PRECALCULUS I - PowerPoint PPT Presentation

PRECALCULUS I. EXPONENTIAL &amp; LOG MODELS. Dr. Claude S. Moore Danville Community College. FIVE COMMON TYPES OF MATHEMATICAL MODELS. 1. Exponential Growth 2. Exponential Decay 3. Gaussian Model 4. Logistics Growth 5. Logarithmic Model. 1. EXPONENTIAL GROWTH.

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PRECALCULUS I

EXPONENTIAL & LOG MODELS

Dr. Claude S. MooreDanville Community College

FIVE COMMON TYPES OF MATHEMATICAL MODELS

1. Exponential Growth

2. Exponential Decay

3. Gaussian Model

4. Logistics Growth

5. Logarithmic Model

1. EXPONENTIAL GROWTH

Find the annual rate (%) for a \$10,000

investment to double in 5 years with

continuous compounding.

A = P ert with P = 10000, A = 20000, t = 5

20000 = 10000er(5) or 2 = e5r

ln 2 = ln e5r gives ln 2 = 5r(ln e) = 5r

r = (ln 2)/5 = 0.1386 or r is 13.9%.

2. EXPONENTIAL DECAY

The half life of carbon 14 is 5730 years.

Find the equation y = a e bx if a = 3 grams.

0.5(3) = 3 eb(5730) or 0.5 = e5730b

ln 0.5 = ln e5730b gives

ln 0.5 = 5730b(ln e) = 5730b

b = (ln 0.5)/5730 = -0.12097

Thus the equation is y = 3 e -0.12097x .

EXPONENTIAL EQUATION

3. Write the exponential equation of the line that passes through (0,5) and (4,1).

The equation is of the form y = a e bx .

(0,5) yields 5 = a eb(0) or 5 = a e0 or a = 5.

(4,1) yields 1 = 5 eb(4) or 0.2 = eb(4)

ln 0.2 = ln e4b gives ln 0.2 = 4b(ln e) = 4b

b = (ln 0.2)/4 = -0.402359

Thus the equation is y = 5 e -0.402359x .

BACTERIA GROWTH

4. The number of bacteria N is given by the model N = 250 e kt with t in hours.

If N = 280 when t = 10, estimate time for bacteria to double.

The point (10,280) yields 280 = 250 eb(10)

1.12 = e10b or ln 1.12 = ln e10b

ln 1.12 = 10b(ln e) = 10b

b = (ln 1.12)/10 = 0.0113329

Thus the equation is y = 250 e 0.0113329t .

TIME OF DEATH

5. The time, t, elapsed since death and the body temperature, T, at room temperature of 70 degrees is given by

If the body temperature at 9:00 a.m. was 85.7 degrees, estimate time of death.

TIME OF DEATH concluded

5. If the body temperature at 9:00 a.m. was 85.7 degrees, estimate time of death.

t = -2.5 ln 0.54895 = 1.499 or t = 1.5 hrs

So time of death was 1.5 hrs before 9 a.m Thus the time of death was 7:30 a.m.