# Proportion

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## Proportion

Two variables are proportional if as one variable changes, the other variable changes in a specific way. Variables can either be directly proportional or inversely proportional.

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## Direct Proportion

If two variables are directly proportional, then as one increases, the other increases by the same scale factor (at the same rate). For two variables, say $x$ and $y$, we can write

$y \propto x$

which means “$y$ is directly proportional to $x$” (the $\propto$ symbol means proportional).

This expression is equivalent to writing

$y = \textcolor{orange}{k}x$

where $\textcolor{orange}{k}$ is the constant of proportionality – this tells us how $x$ and $y$ are related to each other.

There are other types of direct proportion, such as $y \propto x^2$ or $y \propto \sqrt{x}$, which can be seen in the table below.

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## Inverse Proportion

If two variables are inversely proportional, then as one increases, the other decreases by the same scale factor (at the same rate). For two variables, say $x$ and $y$, we can write

$y \propto \dfrac{1}{x}$

which means “$y$ is inversely proportional to $x$” or “$y$ is directly proportional to $\dfrac{1}{x}$”.

This expression is equivalent to writing

$y = \dfrac{\textcolor{orange}{k}}{x}$

There are other types of inverse proportion, such as $y \propto \dfrac{1}{x^2}$ or $y \propto \dfrac{1}{\sqrt{x}}$, which can be seen in the table below.

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## Proportionality Graphs

The equations of direct proportion and inverse proportion can be plotted as graphs:

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## Example 1: Direct Proportion

If $y$ is directly proportional to $x^2$ and $y=36$ when $x=3$, find the value of $y$ when $x=5$.

[3 marks]

Step 1: $y \propto x^2$, so we can write this as an equation involving the constant of proportionality: $y = kx^2$

Step 2: We are given that $y = 36$ and $x = 3$. Substitute these into the equation above and solve to find $k$:

\begin{aligned} 36 &= k \times 3^2 \\ 36 &= 9k \\ \textcolor{orange}{k} &\textcolor{orange}{= 4} \end{aligned}

Hence, the equation becomes: $y = \textcolor{orange}{4}x^2$

Step 3: Find the value of $y$ when $x=5$ by substituting in $x=5$ into the equation:

$y = \textcolor{orange}{4} \times 5^2 = 100$

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## Example 2: Inverse Proportion

The density of a solid, $\rho \text{ kg/m}^3$, is modelled as being inversely proportional to the volume of the solid, $V \text{ m}^3$.

a) A solid with density $40 \text{ kg/m}^3$ has a volume of $0.05 \text{ m}^3$. Find the constant of proportionality.

b) Sketch the graph of $\rho$ against $V$.

[4 marks]

a) $\rho \propto \dfrac{1}{V}$ is equivalent to $\rho = \dfrac{k}{V}$

When $\rho = 40$, $V = 0.05$, so

$40 = \dfrac{k}{0.05} \Rightarrow \textcolor{orange}{k} = 40 \times 0.05 = \textcolor{orange}{2}$

b) $\rho = \dfrac{2}{V}$ is of the form $\rho = kV^n$ where $\textcolor{orange}{k = 2}$ and $n = -1$.

The volume cannot be negative, so we only need to sketch the top-right quadrant of the graph.

Note: There will be asymptotes here at $\rho = 0$ and $V = 0$.

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## Proportion Example Questions

Question 1: If $y$ is inversely proportional to $\sqrt{x}$ and $y = 8$ when $x = 9$, find the value of $x$ when $y = 6$.

[3 marks]

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$y \propto \dfrac{1}{\sqrt{x}}$, so we can write this as $y = \dfrac{k}{\sqrt{x}}$

We are given that $y = 8$, when $x = 9$, so substitute these into the equation and solve to find $k$:

\begin{aligned} 8 &= \dfrac{k}{\sqrt{9}} \\[1.2em] 8 &= \dfrac{k}{3} \\[1.2em] k &= 24 \end{aligned}

So, the equation is

$y = \dfrac{24}{\sqrt{x}}$

Then, find the value of $x$ when $y = 6$, by substituting in $y = 6$ into the equation and solving for $x$:

\begin{aligned} 6 &= \dfrac{24}{\sqrt{x}} \\[1.2em] \sqrt{x} &= \dfrac{24}{6} \\[1.2em] \sqrt{x} &= 4 \\[1.2em] x &= 16 \end{aligned}

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Question 2: The kinetic energy, $E \text{ J}$ of an object is directly proportional to the velocity of the object, $v \text{ m/s}$, squared.

a) An object travelling at a velocity of $10 \text{ m/s}$ has a kinetic energy of $3000 \text{ J}$. Find the constant of proportionality.

b) Find the kinetic energy of the object when it is travelling at a velocity of $4 \text{ m/s}$.

[3 marks]

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a) $E \propto v^2$, which can be written as $E=kv^2$

When $v = 10$, $E = 3000$, so

\begin{aligned} 3000 &= k \times 10^2 \\ 3000 &= 100k \\ k &= \dfrac{3000}{100} = 30 \end{aligned}

b) $E = 30v^2$

Substitute in $v=4$ into the equation to find the kinetic energy of the object if it is travelling at a velocity $4 \text{ m/s}$:

$E = 30 \times 4^2 = 30 \times 16 = 480 \text{ J}$

Gold Standard Education

Question 3: The work done by an object, $W \text{ Nm}$, is modelled as being directly proportional to the distance moved by the object, $d \text{ m}$.

a) The work done by an object that is moved by $12 \text{ m}$ is $60 \text{ Nm}$. Find the constant of proportionality.

b) Sketch the graph of $W$ against $d$.

c) Find the work done by the object if it is moved by a distance of $25 \text{ m}$.

[5 marks]

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a) $W \propto d$, which is equivalent to $W = kd$.

When $d = 12$, $W = 60$, therefore

$60 = k \times 12 \Rightarrow k = 60 \div 12 = 5$

b) $W = 5d$ will be a straight line passing through the origin. You will only need to sketch the graph in the top right quadrant since distance cannot be negative.

(The gradient of the line is $5$, but since we are only doing a sketch we do not need to write any values on the axes, so we can just draw any straight line passing through the origin with a positive gradient).

The graph will look like:

c) $W = 5d$

Substitute in $d=25$ into the equation to find the work done by the object if it is moved by $25 \text{ m}$:

$W = 5 \times 25 = 125 \text{ Nm}$