Inverse Variation

A journey of $120$ km is driven at various speeds. Work out the time taken at $40$ km/h, at $60$ km/h and at $120$ km/h. Now multiply each speed by its time. What do you notice, and why must it happen for this particular journey?

Two quantities are in inverse variation when their product is constant: doubling one halves the other. Written as an equation, that is $y = \dfrac{k}{x}$, and the fixed number $k$ is again the constant of variation.

Two quantities are in inverse variation when multiplying one by a number divides the other by that same number. Double one and the other halves; triple one and the other becomes a third.

Equivalently, and more usefully for checking: their product never changes.

$$xy = k \quad \text{for every pair}$$

In the table above, $2 \times 30$, $4 \times 15$, $5 \times 12$ and $10 \times 6$ all equal $60$.

Typical examples all share a common structure: some fixed total is being shared out or covered.

the time for a journey against the speed, for a fixed distance;
the time to finish a job against the number of workers, for a fixed amount of work;
the number of items each person receives against the number of people, for a fixed supply;
the pressure of a gas against its volume at fixed temperature.

The constant is that fixed total: the distance, the total work, the total supply. Identifying what the product represents is often the quickest way to see that a situation is inverse variation at all.

Three phrases, all meaning the same thing:

"$y$ is inversely proportional to $x$"
"$y$ is proportional to the reciprocal of $x$"
"$y$ varies inversely as $x$"

The middle one explains the other two. Saying $y$ is proportional to $\tfrac{1}{x}$ means $y = k \times \tfrac{1}{x}$, which is the direct-variation pattern from Lesson 1 applied not to $x$ but to its reciprocal.

$$y \propto \frac{1}{x} \quad \Longleftrightarrow \quad y = \frac{k}{x}$$

Multiplying both sides by $x$ gives the form that is easiest to check against a table:

$$xy = k$$

Both forms are useful. Use $y = \tfrac{k}{x}$ when you want to calculate a value of $y$, and $xy = k$ when you want to test whether a table shows inverse variation at all.

The method is the same three steps as last lesson, with a different equation.

Suppose $y$ varies inversely as $x$, and $y = 8$ when $x = 3$.

Step 1. Write the equation with the unknown constant.

$$y = \frac{k}{x}$$

Step 2. Substitute the known pair and solve for $k$.

$$8 = \frac{k}{3} \quad \Longrightarrow \quad k = 24$$

Multiplying the pair together directly gives the same thing, $3 \times 8 = 24$, and is faster.

Step 3. Write the completed equation and use it.

$$y = \frac{24}{x}$$

If $x = 6$, then $y = 4$. If $y = 2$, then $2 = \tfrac{24}{x}$, so $x = 12$.

Notice the doubling behaviour in those answers. From $(3,8)$ to $(6,4)$: $x$ doubled and $y$ halved. From $(3,8)$ to $(12,2)$: $x$ was multiplied by $4$ and $y$ divided by $4$.

Not everything that decreases is inverse variation. Compare three ways $y$ might fall as $x$ rises:

$$y = 20 - x, \qquad y = \frac{20}{x}, \qquad y = 20 - x^2$$

Only the middle one is inverse variation. Test each at $x = 2$ and $x = 4$:

$y = 20 - x$ gives $18$ then $16$: the product changes from $36$ to $64$, and doubling $x$ has not halved $y$ at all.

$y = \tfrac{20}{x}$ gives $10$ then $5$: the product is $20$ both times, and doubling $x$ halved $y$ exactly.

$y = 20 - x^2$ gives $16$ then $4$: products $32$ and $16$, so not constant.

The first of those is a straight line with negative gradient, which is a perfectly ordinary linear relationship and has nothing to do with variation.

Two further structural signals. In inverse variation neither quantity can be zero: $x = 0$ makes $\tfrac{k}{x}$ undefined, and $y = 0$ would need $k = 0$, which would make $y$ zero everywhere. And $y$ never becomes negative if $k$ is positive and $x$ is positive, however large $x$ grows: it approaches zero without reaching it.

Five items on inverse variation. Work each one, then reveal the answer.

1. 1 $y$ varies inversely as $x$, and $y = 9$ when $x = 4$. Find $k$.

   Show answer
   Multiply the pair: $k = xy$.$k = 36$

2. 2 $y = \dfrac{50}{x}$. Find $y$ when $x = 10$.

   Show answer
   Substitute directly.$y = 5$

3. 3 Is $y = 12 - x$ inverse variation?

   Show answer
   At $x=2$, $xy = 20$; at $x=3$, $xy = 27$.No

4. 4 If $y \propto \dfrac{1}{x}$ and $x$ is multiplied by $5$, what happens to $y$?

   Show answer
   The product must stay fixed.It is divided by $5$

5. 5 Why can $x = 0$ never occur in $y = \dfrac{k}{x}$?

   Show answer
   Consider what the expression would require.Division by zero is undefined