AP®MathsCollege BoardCalculus ABRevision NotesUnit 7: Differential EquationsFirst-Order Differential EquationsExponential Models

Exponential Models (College Board AP® Calculus AB)

Revision Note

Roger B

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Roger B

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Maths

Differential equations for exponential models

What type of differential equation corresponds to an exponential model?

  • There are many situations where assuming that the rate of change of a quantity is proportional to the size of the quantity provides a good model

    • For example a population of bacteria

      • The more bacteria there are, the more new bacteria will be being produced

    • Or a radioactive sample

      • The more radioactive atoms there are, the greater the number of atoms that will be undergoing decay

  • This is known as the exponential growth and decay model

  • The mathematical way of expressing this for a quantity y is

    • fraction numerator d y over denominator d t end fraction equals k y

      • fraction numerator d y over denominator d t end fraction is the rate of change of y

      • k is the constant of proportionality

    • If the quantity y is increasing

      • then k is positive

    • If the quantity y is decreasing

      • then k is negative

      • Or alternatively, assume that k is positive and write fraction numerator d y over denominator d t end fraction equals negative k y

Solutions to exponential growth & decay models

How do I find the solutions for exponential growth and decay models?

  • The solution to the exponential growth and decay model fraction numerator d y over denominator d t end fraction equals k y, with the initial condition y equals y subscript 0 when t equals 0, is

    • y equals y subscript 0 e to the power of k t end exponent

  • It is a good idea to remember this result

    • but it can also be derived using separation of variables

Solving the exponential growth and decay model using separation of variables

  • Start with fraction numerator d y over denominator d t end fraction equals k y

    • Separate the variables

      • 1 over y fraction numerator d y over denominator d t end fraction equals k

    • Integrate both sides with respect to t

      • integral 1 over y space d y equals integral k space d t

    • Integrate, including a constant of integration

      • ln open vertical bar y close vertical bar equals k t plus C

    • If y represents a population then y can never be negative, so we can ignore the modulus sign

      • ln y equals k t plus C

    • y equals y subscript 0 when t equals 0, so

      • ln open parentheses y subscript 0 close parentheses equals k open parentheses 0 close parentheses plus C space space rightwards double arrow space space C equals ln open parentheses y subscript 0 close parentheses

    • Rearrange

      • table row cell ln y end cell equals cell k t plus ln open parentheses y subscript 0 close parentheses end cell row cell e to the power of ln y end exponent end cell equals cell e to the power of k t plus ln open parentheses y subscript 0 close parentheses end exponent end cell row y equals cell e to the power of k t end exponent times e to the power of ln open parentheses y subscript 0 close parentheses end exponent end cell row y equals cell y subscript 0 e to the power of k t end exponent end cell end table

Exam Tip

If an exam question is based on a real world example, be sure that your answers are given in the context of the question.

Worked Example

At any point in time, the rate of growth of a colony of bacteria is proportional to the current population size, P.

(a) Write a differential equation to model the size of the population of bacteria.

Answer:

This is a description of an exponential growth model.

fraction numerator d P over denominator d t end fraction equals k t

 

At time t equals 0 hours, the population size is 5000.

(b) Write down the particular solution of the differential equation from part (a).

Answer:

For fraction numerator d y over denominator d t end fraction equals k y, with the initial condition y equals y subscript 0 when t equals 0, the particular solution isy equals y subscript 0 e to the power of k t end exponent

P equals 5000 e to the power of k t end exponent

 

After 1 hour, the population has grown to 7000.

(c) Determine how long it will take from time t equals 0, according to the model, for the population of bacteria to grow to 100 000.

Answer:

The question doesn't say what units of time to use, but looking at the information given it will be easiest to use hours

Substitute the values for t equals 1 hour into the particular solution, and solve for k

table row 7000 equals cell 5000 e to the power of k open parentheses 1 close parentheses end exponent end cell row cell e to the power of k end cell equals cell 7000 over 5000 equals 7 over 5 equals 1.4 end cell row k equals cell ln open parentheses 1.4 close parentheses end cell end table

Using that value of k, substitute P equals 100000 into the particular solution and solve for t

table row 100000 equals cell 5000 e to the power of ln open parentheses 1.4 close parentheses t end exponent end cell row cell e to the power of ln open parentheses 1.4 close parentheses t end exponent end cell equals cell 100000 over 5000 equals 20 end cell row cell ln open parentheses 1.4 close parentheses t end cell equals cell ln open parentheses 20 close parentheses end cell row t equals cell fraction numerator ln open parentheses 20 close parentheses over denominator ln open parentheses 1.4 close parentheses end fraction equals 8.903356... end cell end table

8.903 hours (to 3 decimal places)

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    Roger B

    Author: Roger B

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.