Coaching Centre
Reciprocal models and inverse variation
Inverse variation y = k/x, the shape of the hyperbola and its asymptotes, finding k from one pair of values, using reciprocal models for shared work, speed and time, and their limitations.
More painters finish a house sooner. A faster speed gets you there in less time. A pizza shared among more people gives fewer slices each. In each case, one quantity going up makes the other go down, and their product stays the same. That is inverse variation, and its graph is a curve called a hyperbola.
Inverse variation
$y = \dfrac{k}{x}$, or equally $xy = k$
$k$ is the constant of variation: the product of any matching pair of values.
Why is the product constant? A job of $36$ "painter-days" can be done by $1$ painter in $36$ days, $2$ in $18$, $3$ in $12$, $4$ in $9$. Each time, painters × days $= 36$: the total amount of work does not change, it is just shared out.
| Painters $n$ | $1$ | $2$ | $3$ | $4$ | $6$ | $9$ |
|---|---|---|---|---|---|---|
| Days $d = \dfrac{36}{n}$ | $36$ | $18$ | $12$ | $9$ | $6$ | $4$ |
The hyperbola
- For $k \gt 0$ and $x \gt 0$ the curve falls steeply at first, then levels out.
- It never touches either axis. As $x$ grows, $y$ gets closer to $0$; as $x$ gets close to $0$, $y$ grows without limit. Both axes are asymptotes.
- There is no point at $x = 0$, because you can't divide by $0$.
- For negative $x$ (not usually meaningful in practical problems) there is a matching branch in the third quadrant.
On Painting, move the number of painters from $3$ to $6$ and watch the days. Check that the shaded rectangle always has area $36$. Then pick Road trip and find the time at $80$ km/h. Finally go back to Painting and push the painters up to $12$: read the note and decide whether $3$ days is believable.
Building a model from two values
Four people take $15$ hours to pick a crop. Assuming inverse variation, write a model and find how long $6$ people would take.
A $600$ km trip takes $t$ hours at an average speed of $s$ km/h. Write a model and find the time at $80$ km/h and the speed needed to do it in $6$ hours.
The number of tiles $T$ needed to cover a floor varies inversely with the area $A$ cm² of each tile. With $900$ cm² tiles, $160$ tiles are needed. How many $1200$ cm² tiles are needed, and what is the floor area?
Limitations
- Too many workers. Twelve painters in one small house get in each other's way, so the time does not keep halving.
- Whole numbers. You can't hire $2.5$ painters.
- Real limits on $x$. A car can't average $300$ km/h legally, and near $x = 0$ the model gives impossibly large answers.
- Average values. A speed model uses the average speed, which hides stops and traffic.
Your turn
$y$ varies inversely with $x$, and $y = 8$ when $x = 5$. Find $y$ when $x = 4$, and $x$ when $y = 2$.
Show the working
$k = 5 \times 8 = 40$, so $y = \dfrac{40}{x}$. When $x = 4$, $y = 10$. When $y = 2$, $x = 40 \div 2 = 20$.
Four people take $15$ hours to do a job. What is $k$ in $h = \dfrac{k}{n}$?
$k = 4 \times 15 = 60$.
Using $h = \dfrac{60}{n}$, how many hours do $6$ people take?
$60 \div 6 = 10$ hours.
Which table shows inverse variation?
In the third table $xy = 12$ every time. The first is linear ($xy$ is $10, 16, 18$), the second is direct variation and the fourth is $y = x^2$.
A $600$ km trip at an average of $80$ km/h takes how many hours?
$t = 600 \div 80 = 7.5$ hours.
What average speed, in km/h, does the $600$ km trip in $6$ hours?
$s = 600 \div 6 = 100$ km/h.
Why does the graph of $y = \dfrac{12}{x}$ have no point at $x = 0$?
$12 \div 0$ has no value, so the curve never reaches the $y$-axis: it is an asymptote.
If $y = \dfrac{k}{x}$ and $x$ is tripled, $y$ is:
$\dfrac{k}{3x} = \dfrac{1}{3} \times \dfrac{k}{x}$, so $y$ becomes a third of its value.
$y$ varies inversely with $x$, and $y = 8$ when $x = 5$. Find $x$ when $y = 2$.
$k = 5 \times 8 = 40$. $x = 40 \div 2 = 20$.
Exam style: The number of tiles $T$ varies inversely with the tile area $A$. With $900$ cm² tiles, $160$ are needed. How many $1200$ cm² tiles are needed?
$k = 900 \times 160 = 144{,}000$. $T = 144{,}000 \div 1200 = 120$.
Exam style: A model says $d = \dfrac{36}{n}$ days for $n$ painters. Which is the best criticism of using it for $n = 36$?
$36 \div 36 = 1$ is a whole number and the maths is fine. The limitation is practical: the model assumes every extra worker adds the same amount of work, which fails when they crowd each other.