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Interpret and use an exponential model

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One line ignores the constant offset. Which line gives the correct initial value?

Year 11, VCE Units 1 & 2, Chapter 11, 4 difficulty levels, Calculator allowedBuild a worksheet

Exponential models

An exponential model separates a constant offset from the amount that changes. Substitute the time into the exponent, calculate the factor exactly, then apply it to the displacement

\[Q(t)=L+A b^{\frac{t}{s}}\]
  1. Find the initial value At time zero the exponential factor is 1, so add the offset and the initial displacement \(\displaystyle Q(0)=L+A b^{0}=L+A\)
  2. Read the period multiplier The base multiplies the displacement from the offset once every s time units \(\displaystyle Q(t)=L+A b^{\frac{t}{s}}\quad\Rightarrow\quad b\text{ every }s\text{ time units}\)
  3. Convert a multiplier to a rate Subtract 1 from the multiplier and express the result as a percentage \(\displaystyle \text{rate}=100\left(b-1\right)\%\)
  4. Interpret the long-run value For a shifted decay with 0 < b < 1, the value approaches L but does not reach it at a finite time \(\displaystyle 0<b<1\quad\Longrightarrow\quad Q(t)\to L\text{ as }t\to\infty\)

Worked example

A storage tank's volume is modelled in litres, with time measured in hours. Find its volume at 4 hours

\[V(t)=30+160\left(\frac{3}{4}\right)^{\frac{t}{2}}\]
  1. \(\displaystyle V(4)=30+160\left(\frac{3}{4}\right)^{\frac{4}{2}}\)Substitute t = 4 into the model
  2. \(\displaystyle \frac{4}{2}=2\quad\text{and}\quad \left(\frac{3}{4}\right)^{2}=\frac{9}{16}\)Calculate the exponent and then the exact exponential factor
  3. \(\displaystyle V(4)=30+160\left(\frac{9}{16}\right)=30+90=120\)Apply the factor to the displacement and add the offset, giving 120 litres

Common mistakes

Ignoring the constant offset
✕\(\displaystyle Q(t)=20+120\left(\frac{3}{2}\right)^{t/2}\quad\Rightarrow\quad Q(0)=120\)✓\(\displaystyle Q(t)=20+120\left(\frac{3}{2}\right)^{t/2}\quad\Rightarrow\quad Q(0)=20+120=140\)
At t = 0 the exponential factor is 1, so the initial value is L + A rather than A alone
Using a percentage as the multiplier
✕\(\displaystyle 10\%\text{ growth}\quad\Rightarrow\quad b=10\)✓\(\displaystyle 10\%\text{ growth}\quad\Rightarrow\quad b=1+\frac{10}{100}=1.10\)
A percentage growth rate is converted to a multiplier by adding it to 1
Treating half-life as reaching zero
✕\(\displaystyle \text{after one half-life: }Q(t)-L=0\)✓\(\displaystyle \text{after one half-life: }Q(t)-L=\frac{A}{2}\)
A half-life halves the offset-adjusted amount; it does not make that amount zero

Difficulty levels

introThe initial value and multiplier are read before one exact forward prediction
standardA growth or decay rate is interpreted and a rounded prediction or change is calculated
challengeReverse time, shifted decay, half-life, doubling time or long-run behaviour is considered
examA compact model question combines a parameter interpretation, a numeric prediction and a domain or long-run choice

Questions about this topic

Why is the coefficient not always the initial value?+

At time zero, the exponential factor is 1, so the initial value is the offset plus the initial displacement

How is a decay rate converted to a multiplier?+

Subtract the percentage rate from 100 percent and write the remaining decimal as the multiplier. A 15 percent decay gives a multiplier of 0.85

What does a long-run value mean?+

It is the value approached as time increases. A shifted decay approaches its offset but does not attain it at a finite time

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 11: Exponentials and logarithms

In this chapter

Simplify and evaluate powersApply logarithm laws and inverse identitiesConnect exponential and logarithmic formsSolve exponential equations and inequalities exactlyConstruct exponential and logarithmic inverse pairsSolve exponential and logarithmic equationsAnalyse exponential and logarithmic graphsInterpret and use an exponential model