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Exponential Growth and Decay.

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1- Do the following maths.
2- Hit Submit
3- See Your Results Below.

Review of the Exponential Growth and Decay.

If an amount A changes with a rate r, then the A(t) has the form:

A(t) = A0ert

- If r > 0, it is an exponential growth.
- If r < 0, it is an exponential decay.
- A0 is the initial amount.

Exp Growth

Example problem I: Exponential Growth.

At the start of an experiment, there are 5,000 bacteria, and one hour later, the population has increased to 6,000. How long will it take for the population to reach 8,000?

We are given A0 = 5,000; t = 1; A(t) = 6,000.

We solve for r first :

=> A(t) = A0ert

=> 6000 = 5000 er · 1

=> 6000/5000 = er

=> 1.2 = er

=> ln 1.2 = ln(er)

=> 0.182 = r

Then we solve for t :

=> A(t) = A0e0.182t

=> 8000 = 5000 e0.182t

=> 8000/5000 = e0.182t

=> 1.6 = e0.182t

=> ln 1.6 = ln(e0.182t)

=> 0.470 = 0.182t

=> t = 0.470/0.182

t = 2.6 hrs = 2 hrs 36 mins

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Example problem II: Decay Half Life.

The pesticide DDT decays in exponential way. The half-life of DDT is 20 years. If 100 grams of DDT are present now, how long will it take until only 20 grams of the original sample remains?

We are given A0 = 100; t half life = 20 yrs. At that time A = A0/2

We solve for r first :

=> A(t) = A0ert

=> A0/2 = A0 er · 20

=> 1/2 = e20r

=> ln 0.5 = ln(e20r)

=> -0.693 = 20r

=> r = -0.035

Then we solve for t :

=> A(t) = A0e-0.035t

=> 20 = 100 e-0.0357t

=> 20/100 = e-0.0357t

=> 0.2 = e-0.035t

=> ln 0.2 = ln(e-0.035t)

=> -1.609 = -0.035t

=> t = -1.609/-0.035

t = 46 yrs.

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Do the following problems:


I- At the start of an experiment, there are 4000 bacteria, and 2 hour later, the population has increased to 12000. How long from the begining will it take for the population to reach 36000?

Answer :  

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II- A radiation subtance decays in exponential way. Its half-life is 200 years. If 200 grams of the subtance are present now, how long will it take until only 10% of the original sample remains?

Answer :  


  

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Your Previous Answers and Results :



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