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Exponential Growth and Decay
AP Calculus 5-8 Exponential Growth and Decay
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π π‘ = π 0 π ππ‘ π¦= π 0 π ππ‘ Find π¦β² π¦ β² =π π 0 π ππ‘ π¦ β² =ππ¦
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π 0 = Initial Population π= Growth Constant π‘= Time
π(π‘)= π 0 π ππ‘ π 0 = Initial Population π= Growth Constant π‘= Time
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In a laboratory, a number of Escherichia coli (E
In a laboratory, a number of Escherichia coli (E. coli) bacteria grows exponentially with a growth constant of π= βππ’ππ β1 . Assume that 1000 bacteria are present at time π‘=0. Find the formula for the number of bacterial π(π‘) at time π‘. How large is the population after 5 hours? When will the population reach 10,000?
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π π‘ =1000 π 0.41π‘ π 5 =1000 π 0.41(5) π 5 =7768 ππππ‘ππππ 10,000=1000 π 0.41π‘ 10= π 0.41π‘ ln 10 =0.41π‘ π‘=5.62 βππ’ππ
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π¦ π‘ =πΆ π 3π‘ , where πΆ is the initial value πΆ=π¦ 0 .
Find all solutions to π¦ β² =3π¦. Which solution satisfies π¦ 0 =9? π¦ π‘ =πΆ π 3π‘ , where πΆ is the initial value πΆ=π¦ 0 . π¦ π‘ =9 π 3π‘
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Doubling Time/Half-life
ln 2 π If π is positive you are finding doubling time If π is negative you are finding half-life
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The Nazi Zombie Virus (NZV) is spreading quickly around the world
The Nazi Zombie Virus (NZV) is spreading quickly around the world. The virus is doubling time of the virus is days. What is the function for the virus? If the virus started in Norway, how long before everyone is infected in Norway? Norwayβs Population is 5,100,000 people How long before the world is infected? 7 billion people
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7,000,000,000= π 0.06π‘ ln 7,000,000, =π‘ π‘= πππ¦π
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ln 2 π =11.55245301 π=0.06 π π‘ =1 π 0.06π‘ 5,100,000= π 0.06π‘ ln 5,100,000 0.06 =π‘ π‘=257.41 πππ¦π
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Dr. Strangelove has found an antidote that will cure all of those that are infected and prevent the virus from spreading. He finds that the decay rate of the virus is 0.09. Write the function for the virus What is the half-life of the virus? If after 300 days the antidote is starting to be distributed, how long before the virus is infected by only 10 people?
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π π‘ = π 0 π β0.09π‘ ln = πππ¦π
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π 0.06(300) =65,659,969 65,659,969 π β0.09π‘ =10 ln 10 65,659,969 β0.09 =π‘ π‘= πππ¦π
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Problems 5-8 page 350 #1-21
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