Direct Variation Revision Notes for VCE SSCE General Mathematics

Direct Variation

Introduction to direct variation

When two variables are related so that one increases at a constant rate as the other increases, we call this a direct variation.

For two variables $x$ and $y$ , if we can write the relationship as $y = kx$ where $k$ is a positive constant number, then we say that " $y$ varies directly as $x$ ".

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Another way to express the same idea is to say " $y$ is directly proportional to $x$ ". Both phrases mean exactly the same thing.

The positive constant $k$ in the equation is called the constant of variation. This number stays the same throughout the relationship and determines how quickly $y$ changes as $x$ changes.

Understanding direct variation notation

We use a special symbol to write direct variation more concisely. The symbol $\propto$ means is proportional to or varies as.

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When $y$ varies directly as $x$ , we write: $y \propto x$

This is read as " $y$ is proportional to $x$ ".

An important characteristic of direct variation is that as $x$ increases, $y$ also increases. Both variables move in the same direction.

The constant of variation

The constant of variation tells us the rate at which $y$ changes compared to $x$ .

For direct variation where $y = kx$ , we can find $k$ using:

$k = \frac{y \text{ value}}{\text{corresponding value of } x}$

Once we know the constant of variation, we can find any $y$ value for a given $x$ value, or find any $x$ value for a given $y$ value.

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For example, if $y = 3x$ , then $y \propto x$ and $3$ is the constant of variation. This means that $y$ is always three times as large as $x$ .

Graphical representation of direct variation

A key property of direct variation is that when we graph $y$ against $x$ , we always get a straight line that passes through the origin (the point $(0, 0)$ ).

Real-world example: distance and time

Consider Emily who drives at a constant speed of $100$ km/h from Appleton to Brownsville, a distance of $600$ km. She records her progress each hour:

Looking at the data, we can see that as time increases, distance also increases. The relationship between time $t$ (in hours) and distance $d$ (in kilometres) is:

$d = 100t$

This is an example of direct variation. The constant of variation is $100$ , representing the constant speed in km/h.

We can say that distance travelled varies directly as time taken, or that $d$ is proportional to $t$ (written as $d \propto t$ ).

The graph shows that $d$ plotted against $t$ forms a straight line passing through the origin, confirming this is direct variation.

Key properties of direct variation

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Key Properties to Remember:

The variable $y$ varies directly as $x$ if $y = kx$ , where $k$ is a positive constant
The constant $k$ is called the constant of variation
In direct variation: $k = \frac{y \text{ value}}{\text{corresponding value of } x}$
We write " $y$ varies directly as $x$ " symbolically as $y \propto x$
If $y \propto x$ , the graph of $y$ against $x$ is a straight line passing through the origin

Determining the constant of variation

If $y$ varies directly as $x$ , we can write $y = kx$ where $k$ is the constant of variation.

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The constant of variation can be found if we know just one value of $x$ and its corresponding value of $y$ .

Example 1: Finding the constant of variation

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Worked Example: Finding the Constant of Variation

Problem: Use the table of values to determine the constant of variation $k$ , and complete the table, given that $y \propto x$ :

Table

Solution:

Step 1: Rewrite the variation expression as an equation with $k$ as the constant of variation.

$y \propto x$

$y = kx$

Step 2: Substitute corresponding values for $x$ and $y$ , then solve for $k$ .

When $x = 3$ , $y = 21$ :

$21 = 3k$

$\therefore k = 7$

Step 3: Substitute $k = 7$ into $y = kx$ .

$y = 7x$

Step 4: Substitute the value for $x$ to find the corresponding $y$ value.

When $x = 5$ :

$y = 7(5)$

$\therefore y = 35$

Step 5: Substitute the value of $y$ to find the corresponding $x$ value.

When $y = 63$ :

$63 = 7(x)$

$\therefore x = 9$

Step 6: Complete the table.

Table

Variation involving powers

Sometimes in direct variation, one of the variables is raised to a power. The concept remains the same, but the relationship involves an exponent.

Example: falling object

Consider a metal ball dropped from the top of a tall building. The distance it falls (in metres) is recorded each second:

As $t$ increases, $d$ also increases. The relationship between time and distance is:

$d = 4.91t^2$

This is another example of direct variation. Here, we say that the distance fallen varies directly as the square of the time taken. In other words, $d$ is proportional to $t^2$ . We write this as $d \propto t^2$ .

Notice that:

The graph of $d$ against $t$ is a parabola (curved)
However, the graph of $d$ against $t^2$ is a straight line passing through the origin

This confirms that $d$ varies directly as $t^2$ .

General rule for powers

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Direct Variation with Powers:

If $y \propto x^n$ , then $y = kx^n$ , where $k$ is a constant of variation
If $y \propto x^n$ , then the graph of $y$ against $x^n$ is a straight line passing through the origin

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Important note: For all examples of direct variation, when one variable increases, the other also increases. However, not all increasing trends are examples of direct variation.

Example 2: Finding the constant of variation involving powers

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Worked Example: Direct Variation with Squares

Problem: Given that $y \propto x^2$ , use the table of values to determine the constant of variation $k$ , and complete the table:

Table

Solution:

Step 1: Rewrite the variation expression as an equation with $k$ as the constant of variation.

$y \propto x^2$

$y = kx^2$

Step 2: Substitute corresponding values for $x$ and $y$ , then solve for $k$ .

When $x = 2$ , $y = 12$ :

$12 = (2^2)k$

$\therefore k = 3$

Step 3: Rewrite the equation for $y$ , substituting $k$ .

$y = 3x^2$

Step 4: Check with other values that the correct value for $k$ has been found.

When $x = 6$ :

$y = 3(6^2) = 108$ ✓

Step 5: Substitute values for $x$ to find corresponding $y$ values.

When $x = 4$ :

$y = 3(4^2)$

$\therefore y = 48$

Step 6: Substitute the value of $y$ to find the corresponding $x$ value.

When $y = 192$ :

$192 = 3(x^2)$

$\therefore x = 8$

Step 7: Complete the table.

Table

Solving practical problems

Direct variation appears in many real-world situations. The key is to identify the relationship, find the constant of variation, and use it to solve for unknown values.

Example 3: Wire resistance problem

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Worked Example: Real-World Application - Wire Resistance

Problem: In an electrical wire, the resistance ( $R$ ohms) varies directly as the length ( $L$ m) of the wire.

a) If a $6$ m wire has a resistance of $5$ ohms, what is the resistance of a $4.5$ m wire?

b) How long is a wire for which the resistance is $3.8$ ohms?

Solution:

Part a:

Step 1: Write down the variation expression.

$R \propto L$

Step 2: Rewrite the expression as an equation with $k$ as the constant of variation.

$R = kL$

Step 3: Find the constant of variation by substituting given values for $R$ and $L$ .

When $L = 6$ , $R = 5$ :

$5 = k(6)$

$\therefore k = \frac{5}{6}$

Step 4: Substitute the value for $k$ and write down the equation.

$R = \frac{5L}{6}$

Step 5: Substitute $L = 4.5$ and solve to find $R$ .

$R = \frac{5 \times 4.5}{6}$

$= 3.75$

Step 6: Write your answer.

A wire of length $4.5$ m has a resistance of 3.75 ohms.

Part b:

Step 1: Write down the equation.

$R = \frac{5L}{6}$

Step 2: Substitute $R = 3.8$ and solve to find $L$ .

$3.8 = \frac{5 \times L}{6}$

$\therefore L = 4.56$

Step 3: Write your answer.

A wire of resistance $3.8$ ohms has a length of 4.56 m.

Remember!

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Key Points to Remember:

Direct variation means two variables are related by $y = kx$ where $k$ is a positive constant
The symbol $\propto$ means is proportional to or varies as
The constant of variation ( $k$ ) is the fixed multiplier that relates the two variables
In direct variation, as one variable increases, the other also increases
The graph of direct variation is always a straight line through the origin $(0, 0)$
Direct variation can involve powers: if $y \propto x^n$ , then $y = kx^n$
To solve problems: find $k$ using known values, then use the equation to find unknown values

Direct Variation (VCE SSCE General Mathematics): Revision Notes