NSW Y11 Maths - Extension 1 Rates of Change Growth And Decay

Resources for Growth And Decay

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Growth And Decay Theory

NSW Syllabus Reference

NSW Syllabus Reference: ME-C1.2: Exponential growth and decay. This will require student to 

  • construct, analyse and manipulate an exponential model of the form \(𝑁(𝑡) = 𝐴𝑒^{𝑘𝑡}\) to solve a practical growth or decay problem in various contexts (for example population growth, radioactive decay or depreciation)
  • establish the modified exponential model, \(\dfrac{𝑑𝑁}{𝑑t}= 𝑘(𝑁 − 𝑃)\), for dealing with problems such as ‘Newton’s Law of Cooling’ or an ecosystem with a natural ‘carrying capacity’
  • solve problems involving situations that can be modelled using the exponential model or the modified exponential model and sketch graphs appropriate to such problems

Ref: https://educationstandards.nsw.edu.au/

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Videos

Videos relating to Growth And Decay.

  • Growth And Decay - Video - Exponential Growth and Decay Calculus, Relative Growth Rate, Differential Equations, Word Problems

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  • Growth And Decay - Video - Newton's Law of Cooling

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Theory

The exponential function has many practical applications.

In growth: population growth and bacteria growth.

In decay: decay of a radioactive substance and rate of cooling of a body.

The exponential function \(N = A{e^{kt}}\) is known as the growth function for \(k > 0\) and the decay function for \(k < 0.\)

For \(N = A{e^{kt}}\) then \(\dfrac{{dN}}{{dt}} = kA{e^{kt}} \to \dfrac{{dN}}{{dt}} = kN\)

Hence the growth function \(N = A{e^{kt}}\) satisfies the equation \(\dfrac{{dN}}{{dt}} = kN\).

The equation \(\dfrac{{dN}}{{dt}} = kN\) states that the rate of change of \(N\) is proportional to \(N\). The constant \(k\) is the growth constant and the constant \(A\) is the initial amount.

In practical applications more complicated exponential functions are used to describe the process of growth and decay.

A population \(N\) has a rate of change proportional to the difference between \(N\) and a constant \(P\) can be expressed as \(\dfrac{{dN}}{{dt}} = k\left( {N - P} \right)\). Integrating this differential equation we have \(N = P + A{e^{kt}}\) where \(A\) is an arbitrary constant.