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5.8 Exponential Growth and Decay Mon Dec 7 • Do Now • In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k = 0.41. Assume that 1000 bacteria are present at time = 0 • 1) Find the formula for the # of bacteria P(t) at time t • 2) How large is the population after 5 hours? • 3) When will the population reach 10,000?

5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

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Exponential growth and decay When P0 is the initial size at t = 0 If k > 0, then P(t) grows exponentially If k < 0 then P(t) decreases exponentially K is either known as the growth or decay constant. We’ve done these things in Pre-calc

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Page 1: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

5.8 Exponential Growth and DecayMon Dec 7

• Do Now• In the laboratory, the number of Escherichia

coli bacteria grows exponentially with growth constant k = 0.41. Assume that 1000 bacteria are present at time = 0

• 1) Find the formula for the # of bacteria P(t) at time t

• 2) How large is the population after 5 hours?• 3) When will the population reach 10,000?

Page 2: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

HW Review p.339

Page 3: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Exponential growth and decay

•When P0 is the initial size at t = 0•If k > 0, then P(t) grows exponentially•If k < 0 then P(t) decreases exponentially•K is either known as the growth or decay constant.

•We’ve done these things in Pre-calc

Page 4: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Y’ = ky

• If y(t) is a differentiable function satisfying the differential equation

then , where P0 is the initial value P0 = y(0)

Page 5: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Notes

• This theorem tells us that a process obeys an exponential law precisely when its rate of change is proportional to the amount present.

• A population grows exponentially because each present organism contributes to growth

Page 6: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Ex

• Find all solutions of y’ = 3y. Which solution satisfies y(0) = 9?

Page 7: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Ex

• Pharmacologists have shown that penicillin leaves a person’s bloodstream at a rate proportional to the amount present.

• A) Express this as a differential equation• B) Find the decay constant if 50 mg of penicillin

remains in the bloodstream 7 hours after an injection of 450 mg

• C) Under the hypothesis of (B), at what time was 200 mg present?

Page 8: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Doubling Time / Half-life

• If , then

is the doubling time if k >0

is the half-life if k < 0

Page 9: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Compound Interest

• If P0 dollars are deposited into an account earning interest at an annual rate r, compounded M times yearly, then the value of the account after t years is

• If compounded continuously, is

Page 10: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Present Value• The concept of present value (PV) is used to

compare payments made at different times

• The PV of P dollars received at time t is

• Having money now means you can get interest from it right away.

• Getting money later means you will lose out on any interest you could have gotten

Page 11: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Ex

• Is it better to receive $2000 today or $2200 in 2 years? Consider a 3% and 7% compounded interest rate

Page 12: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Ex• You must decide whether to upgrade your

company’s computer system. The upgrade costs $400,000 and will save $150,000 per year for the next 3 years. Is this a good investment if r = 7%?

Page 13: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Income stream

• An income stream is a sequence of periodic payments that continue over an interval of T years

• This is like winning the lottery and taking several payments over the years vs taking a lump sum

Page 14: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

PV of an income stream

• If the interest rate is r, the present value of an income stream paying out R(t) dollars per year continuously for T years is

Page 15: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Ex

• An investment pays out 800,000 pesos per year, continuously for 5 years. Find the PV of the investment for r = 0.04

Page 16: 5.8 Exponential Growth and Decay Mon Dec 7 Do Now In the laboratory, the number of Escherichia coli bacteria grows exponentially with growth constant k

Closure

• For the investment that paid out 800,000 pesos per year for 5 years, find the PV of the investment if r = 0.06

• HW: p.350 #3 7 13 14 15 39 44 45