In 2.06 we explored inverse variation where as one quantity increases, the other decreases. We also saw that if $x$x and $y$y vary inversely, the equation $y=\frac{k}{x}$y=kx can be used to model the relationship.
So many things that we experience every day involve inverse variation, and can be modelled by the simple reciprocal function $y=\frac{k}{x}$y=kx.
Distance, speed and time
Suppose we want to travel from Sydney to Wagga Wagga. Travelling along the fastest route, the total travel distance $d$d is $480$480 km. Depending on how long we want to take ($t$t), the speed ($s$s) we'll need to travel at can be found using this equation:
$s=\frac{d}{t}$s=dt
Substituting the distance gives us $s=\frac{480}{t}$s=480t.
The less time we want to take, the faster our speed will need to be. So this represents an inversely proportional relationship between speed and time.
- If we want to reach our destination after $10$10 hours of travelling, we would need to travel at $\frac{480}{10}=48$48010=48 km/h.
- If instead we want to reach our destination after $5$5 hours, we would need to travel at $\frac{480}{5}=96$4805=96 km/h.
- In theory we could get from Sydney to Wagga Wagga in $1$1 hour if we travelled at $\frac{480}{1}=480$4801=480 km/h, but this is illegal and unrealistic!
We can use the same approach to find the time for a given speed. Rearranging the equation above gives us $t=\frac{480}{s}$t=480s. If we had to walk the whole way, travelling at a speed of $5$5 km/h, this rearranged equation tells us it would take us $\frac{480}{5}=96$4805=96 hours to reach our destination.
Area, length and width
The area, $A$A, of a rectangle of length $l$l and width $w$w is given by $A=lw$A=lw.
When building a house, a council will often limit how much ground space the house can take up. If an architect needs to design a home on a rectangular ground space measuring $300$300 m2, the possible length and width of the space will vary according to $lw=300$lw=300.
The architect can change the length and width of the rectangular ground space to whatever they want, as long as $lw$lw is always equal to $300$300. If they increase the length, they need to decrease the width. This means that the length and width have an inverse relationship.
Some lengths and widths that satisfy the council requirement of a $300$300 m2 ground space are shown in the table of values below.
| $l$l (m) | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $8$8 | $10$10 | $12$12 | $15$15 | $20$20 |
|---|---|---|---|---|---|---|---|---|---|---|
| $w$w (m) | $300$300 | $150$150 | $100$100 | $75$75 | $60$60 | $37.5$37.5 | $30$30 | $25$25 | $20$20 | $15$15 |
It's not practical to have a house that measures $1$1 metre by $300$300 metres, but notice that as $l$l gets very small, $w$w gets very large, and as $l$l gets very large, $w$w gets very small.
Here is the graph of the hyperbola:
Notice that $l$l and $w$w cannot be equal to $0$0. The line with equation $y=0$y=0 (the $x$x-axis) and the line with equation $x=0$x=0 (the $y$y-axis) are the asymptotes of the hyperbola. If either of these dimensions becomes extremely close to $0$0 (less than a fraction of a mm), the other dimension will shoot up towards $\infty$∞! But in practical situations like this, it isn't possible for either the length or width of a rectangle to be $0$0 or $\infty$∞.
Try exploring this applet, where the top slider changes the area of land available, and the bottom slider changes one dimension of the rectangle. The width changes automatically to ensure that the area of the rectangle stays constant.
Practice questions
Question 1
The density ($D$D) of an object is the mass ($m$m) per unit volume ($V$V). That is, $D=\frac{m}{V}$D=mV.
A metal rod has fixed mass of $12$12 kg and variable volume and density due to the thermal expansion from the sun.
Find the density of the metal rod with volume $0.2$0.2 m3 in kg/m3.
Find the density of the metal rod with volume $0.625$0.625 m3 in kg/m3.
Question 2
Hiring a bus costs $\$950$$950 regardless of the number of passengers.
Write an equation that relates the number of passengers ($n$n) to the unit price per passenger ($p$p).
How much will each passenger pay if there are $38$38 passengers?
Question 3
After a rain storm, a water tank has enough water to cover $300$300 m2 with $10.2$10.2 mm of water, or $375$375 m2 with $8.16$8.16 mm of water.
Write an equation relating the depth of water ($D$D mm) to the area of the land ($A$A m2).
Find the depth of water when the area of land is $425$425 m2.