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Grade 12

Modelling non-linear Relationships

Lesson

In many real life experiences we find quantities varying against each other - one quantity, say $y$y, varying in a predictable way simply because another related quantity, say $x$x, is varying. This might be the height of a cricket ball varying because of a changing angle of projection. It might be the the distance to the horizon varying as the observer changes her distance from the earth. It might be the area of a rectangle varying as its dimensions change.

Whenever this occurs, not only might it be possible to formulate why the variation happens, but also the way it happens. 

Linear variation is variation in $y$y that occurs as a fixed multiple $m$m of the variation in $x$x. It gets its name from the type of graph that is produced when $y$y is plotted against $x$x - a straight line of slope $m$m.

But there are many other types of variation, and these are often characterised as non-linear. We will explore two instances of non-linear variation here.

Example 1: The Area of an enclosure. 

A property owner, Anne, decides to set up a rectangular grassed enclosure for her horse. She has 200 metres of fencing, and in an effort to increase the grazing area, has used a straight section of a creek as one of the sides (see diagram below).

She originally thought that no matter what the dimensions of the rectangular enclosure she erects, the feeding area for the horse wouldn't change. However, after doing some mathematics, she realised that the area varies as the dimensions vary. 

Her mathematical logic was clever. Setting the two shorter sides as the variable length $x$x metres, Anne determines that the longer side must be $200-2x$2002x. The area then can be expressed as a function of the shorter sides, given as $A=l\times b=x\left(200-2x\right)$A=l×b=x(2002x).

Expanding this last expression we have $A=200x-2x^2$A=200x2x2. Anne could try various values of $x$x to see how the area would vary, but instead decides to graph the area as a function of $x$x. It's an inverted parabola with roots of $0$0 and $100$100.

She immediately recognises that the area can be maximised if $x=50$x=50 is chosen. This means, by substitution, that $A=200\left(50\right)-2\left(50\right)^2=5000$A=200(50)2(50)2=5000 square metres of grazing is possible. 

 

Example 2: Kepler's Third Law

Johannes Kepler (1571 - 1630) discovered an important relationship between a planet's mean distance from the Sun and its period (the time taken to get around the Sun).

If we call the distance the Earth is from the Sun $1$1 astronomical unit ($1$1AU for short) then we can write down each planet's distance in AUs and each planet's period (in "Earth" years). The tables shows this data broken up into the inner and outer planets of our solar system.

Table 1: Inner Planets Mercury Venus Earth Mars
Distance (R) 0.387 0.723 1 1.523
Period (T) 0.241 0.615 1 1.881

 

Table 2: Outer Planets Jupiter Saturn Uranus Neptune
Distance (R) 5.203 9.537 19.191 30.069
Period (T) 11.9 29.5 84.0 164.8

Kepler was able to prove, given certain assumptions, that the relationship between the period $T$T in Earth years and the distance $R$R was approximately given by:

$T=R\sqrt{R}$T=RR

Looking at the inner planets first, Kepler's formula determines the periods as: 

$T_{Me}$TMe $=$= $0.387\times\sqrt{0.387}=0.241$0.387×0.387=0.241
$T_{Ve}$TVe $=$= $0.723\times\sqrt{0.723}=0.615$0.723×0.723=0.615
$T_{Ea}$TEa $=$= $1\times\sqrt{1}=1$1×1=1
$T_{Ma}$TMa $=$= $1.523\times\sqrt{1.523}=1.880$1.523×1.523=1.880

We could do the same for the outer planets. For example, Kepler's Law for the planet Neptune shows that $T_{Ne}=30.069\times\sqrt{30.069}=164.884$TNe=30.069×30.069=164.884, which is remarkably close for such a far off planet. Perhaps you could check the other 3 planets yourself.

We can sketch the function $T=R\sqrt{R}$T=RR across the restricted domain $0\le R\le35$0R35 to understand the non-linearity of the relationship. Below is such a sketch showing the approximate relative positions of Jupiter, Saturn, Uranus and Neptune. What a special natural relationship this is.

 

Further Examples

Question 1

A rectangle is to be constructed with $80$80 metres of wire. The rectangle will have an area of $A=40x-x^2$A=40xx2, where $x$x is the length of one side of the rectangle.

  1. Using the equation, state the area of the rectangle if one side is $12$12 metres long.

  2. The graph below displays all the possible areas that can be obtained using this amount of wire. From the graph, determine the nearest value for the longer side of a rectangle that has an area of $256$256 square metres.

    Loading Graph...

    A Cartesian plane featuring an horizontal axis labeled "x," ranging from 0 to $40$40, marked at intervals of $4$4. The vertical axis labeled "A," extends from 0 to $400$400, marked at intervals of $100$100 with minor tick marks every $20$20 units. A parabola is graphed on this plane, with its vertex positioned at ($20$20, $400$400) opening downward and passes through the origin and ($40$40, 0). Its coordinates of the vertex and y-intercept are not explicitly labeled or given.

    $33$33 m

    A

    $9$9 m

    B

    $8$8 m

    C

    $32$32 m

    D
  3. Using the graph, what is the greatest possible area of a rectangle that has a perimeter of $80$80 m?

  4. Using the graph, state the dimensions of the rectangle with the maximum area.

    Length $=$=$\editable{}$ m

    Width $=$=$\editable{}$ m

Question 2

The formula for the surface area of a sphere is $S=4\pi r^2$S=4πr2, where $r$r is the radius in centimetres.

  1. Fill in the following table of values for $S=4\pi r^2$S=4πr2, writing your answers correct to two decimal places where necessary.

    $r$r $1$1 $2$2 $3$3 $4$4 $5$5 $6$6 $7$7 $8$8
    $S$S $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. Choose the correct curve for $S=4\pi r^2$S=4πr2

    Loading Graph...

    A

    Loading Graph...

    B

    Loading Graph...

    C

    Loading Graph...

    D
  3. From your graph, determine the closest value for the surface area of a sphere with radius $5.5$5.5 cm.

    $S=410$S=410

    A

    $S=360$S=360

    B

    $S=380$S=380

    C

    $S=350$S=350

    D
  4. From your graph, determine the closest value for the radius of a sphere with a surface area of $201$201 $cm^2$cm2.

    $r=4$r=4

    A

    $r=2$r=2

    B

    $r=5$r=5

    C

    $r=6$r=6

    D

Question 3

$\$700$$700 is invested at $4$4% p.a. compound interest. The value of the money ($A$A) after $n$n years is given by $A=P\left(1+\frac{r}{100}\right)^n$A=P(1+r100)n, where $P$P is the principal invested at a rate of $r$r per cent. The relationship has been graphed.

Loading Graph...
A graph of $A=700\left(1+\frac{4}{100}\right)^n$A=700(1+4100)n on a cartesian plane. The horizontal axis is labeled "n (years)" with markings from 0 to 15, indicating the time in years. The vertical axis is labeled "A" and has values marked from 0 to $1750$1750 in increments of $350$350.

 

  1. Using the equation $A=P\left(1+\frac{r}{100}\right)^n$A=P(1+r100)n, determine how much money would be in the account after $7$7 years, to the nearest dollar.

    Remember that rounding should only be performed at the final step.

  2. According to the graph, approximately how many years does it take for the balance to reach $\$1120.72$$1120.72?

    $14$14

    A

    $17$17

    B

    $12$12

    C

    $10$10

    D
  3. How much interest is earned in the $7$7th year? Give your answer to 2 decimal places.

Outcomes

12C.A.3.4

Solve multi-step problems requiring formulas arising from real-world applications (e.g., determining the cost of two coats of paint for a large cylindrical tank)

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