Direct Variation

A taxi charges $\$4$ to get in plus $\$2$ per kilometre. A second taxi charges $\$2$ per kilometre with no flag fall. For each taxi, work out the cost for $2$ km and for $4$ km. In which taxi does doubling the distance double the cost? Say why the other one does not.

Two quantities are in direct variation when their ratio is constant: doubling one doubles the other, tripling one triples the other. Written as an equation, that is $y = kx$, where the fixed number $k$ is the constant of variation.

Two quantities are in direct variation when multiplying one by a number multiplies the other by the same number. Double one and the other doubles; halve one and the other halves.

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

$$\frac{y}{x} = k \quad \text{for every pair}$$

In the table above, $\tfrac{5}{2}$, $\tfrac{10}{4}$, $\tfrac{15}{6}$ and $\tfrac{20}{8}$ all equal $2.5$. That fixed value is the constant of variation.

Typical examples, all of which you have met before under other names:

the total cost of identical items against the number bought;
the distance travelled at a constant speed against the time taken;
the circumference of a circle against its diameter, where the constant is $\pi$;
the mass of a uniform material against its volume, where the constant is its density.

In every case the constant is a rate: dollars per item, kilometres per hour, kilograms per cubic metre. Naming the units of $k$ is a good check that you have set the relationship up the right way round.

Three phrases appear in questions and mean exactly the same thing:

"$y$ is directly proportional to $x$"
"$y$ is proportional to $x$"
"$y$ varies directly as $x$"

All three are written with the proportionality symbol:

$$y \propto x$$

That statement is not yet something you can calculate with, because it says the ratio is fixed without saying what it is fixed at. Introducing a letter for the fixed value turns it into an equation:

$$y \propto x \quad \Longleftrightarrow \quad y = kx \ \text{ for some constant } k$$

So the move from $\propto$ to $=$ always introduces $k$, and finding $k$ is always the first calculation. The symbol states the shape of the relationship; the equation states the relationship itself.

One known pair of values is enough to find $k$, and after that everything else is substitution.

Suppose $y \propto x$ and $y = 18$ when $x = 4$.

Step 1. Write the equation with the unknown constant.

$$y = kx$$

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

$$18 = k(4) \quad \Longrightarrow \quad k = 4.5$$

Step 3. Write the completed equation.

$$y = 4.5x$$

Now any further question is answered by substituting. If $x = 10$, then $y = 45$. If $y = 27$, then $27 = 4.5x$ gives $x = 6$.

That three-step shape, write, find $k$, then use it, is the method for every variation problem in this focus area, including the inverse ones in the next lesson.

The most common error is treating any straight-line relationship as direct variation. It is not.

Compare two taxi fares for a journey of $d$ kilometres:

$$C = 2d \qquad \text{and} \qquad C = 2d + 4$$

Both are straight lines with the same steepness. Only the first is direct variation.

Test by doubling. In the first, $d = 2$ gives $C = 4$ and $d = 4$ gives $C = 8$: doubled. In the second, $d = 2$ gives $C = 8$ and $d = 4$ gives $C = 12$, which is not double. The ratio $\tfrac{C}{d}$ is $4$ at the first point and $3$ at the second, so it is not constant.

The structural reason is the flag fall. Direct variation forces $y = 0$ when $x = 0$, since $k \times 0 = 0$. A relationship with a starting amount does not pass through the origin, so it cannot be direct variation however straight its graph is.

Straight line is not enough; it must also go through the origin. Lesson 3 makes that visible on a graph.

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

1. 1 $y \propto x$ and $y = 24$ when $x = 6$. Find $k$.

   Show answer
   Divide: $k = \tfrac{24}{6}$.$k = 4$

2. 2 Is $y = 7x$ direct variation?

   Show answer
   It has the form $y = kx$ with no constant added.Yes, $k = 7$

3. 3 Is $y = 7x - 1$ direct variation?

   Show answer
   At $x = 1$ the ratio is $6$; at $x = 2$ it is $6.5$.No

4. 4 $P \propto t$ with $k = 15$. Find $P$ when $t = 8$.

   Show answer
   Substitute into $P = 15t$.$P = 120$

5. 5 If $y \propto x$ and $x$ is tripled, what happens to $y$?

   Show answer
   Multiplying one multiplies the other by the same factor.It is tripled