IBMathsDPApplications & Interpretation (AI)HLRevision Notes2. FunctionsModelling with FunctionsExponential Models

Exponential Models (DP IB Applications & Interpretation (AI)): Revision Note

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on

DP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): SL

Exponential models

What are the parameters of an exponential model?

  • An exponential model is of the form

    •  space f left parenthesis x right parenthesis equals k a to the power of x plus c or space f left parenthesis x right parenthesis equals k a to the power of negative x end exponent plus c for space a greater than 0

    • space f open parentheses x close parentheses equals k straight e to the power of r x end exponent plus c

      • Where e is the mathematical constant 2.718…

    • The value of c represents the boundary for the function

      • It can never be this value

    • The value of a or r describes the rate of growth or decay

      • The bigger the value of a or the absolute value of r the faster the function increases/decreases

What can be modelled as an exponential model?

  • Exponential growth or decay

    • Exponential growth is represented by

      • a to the power of x where a greater than 1

      • a to the power of negative x end exponent where 0 less than a less than 1

      • straight e to the power of r x end exponent where r greater than 0

    • Exponential decay is represented by

      • a to the power of x where 0 less than a less than 1

      • a to the power of negative x end exponent where a greater than 1 

      • straight e to the power of r x end exponent where r less than 0

  • They can be used when there is a constant percentage increase or decrease

    • Such as functions generated by geometric sequences

  • For example, suppose V open parentheses t close parentheses equals 24000 open parentheses 0.95 close parentheses to the power of t plus 6000 is the value of a vehicle in dollars t years after it was purchased

    • The value of the car decreases by 5% each year

    • The initial value of the car is $24000+$6000 = $30000

    • The boundary is $6000

      • The car will never reach this value

  • Examples include:

    • V(t) is the value of car after t years

    • S(t) is the amount in a savings account after t years

    • B(t) is the amount of bacteria on a surface after t seconds

    • T(t) is the temperature of a kettle t minutes after being boiled

Examiner Tips and Tricks

These models are different to quadratic and cubic models, the constant term is not the initial value.

  • The initial value of f open parentheses x close parentheses equals 100 straight e to the power of 2 x end exponent is 100

  • The initial value of f open parentheses x close parentheses equals 100 straight e to the power of 2 x end exponent plus 50 is 100 plus 50 equals 150

What are possible limitations of an exponential model?

  • An exponential growth model does not have a maximum

    • In real-life this might not be the case

      • The function might reach a maximum and stay at this value

  • Exponential models are monotonic

    • In real-life this might not be the case

      • The function might fluctuate

How can I find the half-life using an exponential model?

  • You may need to find the half-life of a substance

    • This is the time taken for the mass of a substance to halve

  • Given an exponential model space f open parentheses t close parentheses equals k a to the power of negative t end exponent or space f open parentheses t close parentheses equals k straight e to the power of negative r t end exponent the half-life is the value of t such that:

    • space f left parenthesis t right parenthesis equals k over 2

    • You can solve for t using your GDC

  • For space f open parentheses t close parentheses equals k a to the power of negative t end exponent the half-life is given by t equals fraction numerator ln 2 over denominator ln a end fraction

    • k over 2 equals k a to the power of negative t end exponent

    • a to the power of t equals 2

    • t space ln a equals ln 2

  • For space f open parentheses t close parentheses equals k straight e to the power of negative r t end exponent the half-life is given by t equals fraction numerator ln 2 over denominator r end fraction

    • k over 2 equals k straight e to the power of negative r t end exponent

    • straight e to the power of r t end exponent equals 2

    • r t equals ln 2

  • For example, supposeM open parentheses t close parentheses equals 100 straight e to the power of negative 3 t end exponent is the mass of a substance after t hours

    • Solve 100 straight e to the power of negative 3 t end exponent equals 50 to find the half-life

      • straight e to the power of negative 3 t end exponent equals 1 half

      • straight e to the power of 3 t end exponent equals 2

      • 3 t equals ln 2

      • t equals 1 third ln 2 equals 0.2310...

Worked Example

The value of a car, V (NZD), can be modelled by the function

space V open parentheses t close parentheses equals 25125 cross times 0.8 to the power of t plus 8500 comma blank t greater or equal than 0

where t is the age of the car in years.

a) State the initial value of the car.

2-3-3-ib-ai-sl-exponential-models-a-we-solution

b) Find the age of the car when its value is 17500 NZD.

2-3-3-ib-ai-sl-exponential-models-b-we-solution

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    Author
    Reviewer
    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

    Lucy Kirkham

    Reviewer: Lucy Kirkham

    Expertise: Head of Content Creation

    Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.