International Baccalaureate · MYP 4

Direct and Inverse Proportion

⌚ ~154 min ▢ 17 blocks ✓ 73 exercises ➣ Prerequisites : Coordinate Geometry and Straight Lines, Quadratic Functions, Ratio, Rates and Speed, Exponents and Standard Form, Similarity and Congruence

I Direct Proportion

I.1 Directly Proportional Quantities

Discover

Two situations. Table A gives the price $y$, in dollars, of $x$ kilograms of apples. Table B gives the time $y$, in hours, needed to drive $120$ km at a constant speed of $x$ km/h. Table A:

$x$	$1$	$2$	$4$	$8$
$y$	$3$	$6$	$12$	$24$

Table B:

$x$	$30$	$40$	$60$	$120$
$y$	$4$	$3$	$2$	$1$

In each table, what happens to $y$ when $x$ is doubled?
In each table, find a calculation on $x$ and $y$ that gives the same result in every column.

Answer

In table A, when $x$ goes from $1$ to $2$, from $2$ to $4$ and from $4$ to $8$, $y$ goes from $3$ to $6$, from $6$ to $12$ and from $12$ to $24$: doubling $x$ doubles $y$. In table B, when $x$ goes from $30$ to $60$ and from $60$ to $120$, $y$ goes from $4$ to $2$ and from $2$ to $1$: doubling $x$ halves $y$.
In table A the quotient $\dfrac{y}{x}$ is always $3$: $\dfrac{3}{1} = \dfrac{6}{2} = \dfrac{12}{4} = \dfrac{24}{8} = 3$, the price of one kilogram. In table B the product $xy$ is always $120$: $30 \times 4 = 40 \times 3 = 60 \times 2 = 120 \times 1 = 120$, the length of the trip.

Two ways for two quantities to be linked: in table A, $y$ is directly proportional to $x$ (a constant quotient); in table B, $y$ is inversely proportional to $x$ (a constant product). This chapter studies both, then quantities linked through a power of $x$.

Definition — Direct Proportion

A quantity $y$ is directly proportional to a quantity $x$ (one also says that $y$ varies directly as $x$) when there is a nonzero number $k$ such that $$y = kx$$ for every pair of matching values. The number $k$ is the constant of proportionality. We write $y \propto x$, read “$y$ is proportional to $x$”. When $x \neq 0$, $k = \dfrac{y}{x}$: the quotient of matching values is the same for every pair.

Proposition — Scaling in a Direct Proportion

If $y \propto x$, then multiplying $x$ by a number $m$ multiplies $y$ by the same number $m$. For instance, doubling $x$ doubles $y$, and dividing $x$ by $3$ divides $y$ by $3$.

Proof

Let $y = kx$. When $x$ is replaced by $mx$, the quantity $y$ takes the value $$\begin{aligned}[t] k \times (mx) &= (k \times m) \times x && \text{(associativity)}\\ &= (m \times k) \times x && \text{(commutativity)}\\ &= m \times (kx) && \text{(associativity)}\\ &= m \times y. \end{aligned}$$ So $y$ is multiplied by $m$.

Example

The mass $M$, in grams, of a copper wire is directly proportional to its length $L$, in metres. A piece $5$ m long has a mass of $180$ g.

Find the constant of proportionality and write $M$ in terms of $L$.
Find the mass of $12$ m of this wire.
What length of wire has a mass of $450$ g?
The length of a piece is multiplied by $4$. What happens to its mass?

Answer

Since $M \propto L$, $M = kL$. The pair $L = 5$, $M = 180$ gives $$\begin{aligned}[t] 180 &= k \times 5 && \text{(the known pair)}\\ k &= \frac{180}{5} && \text{(divide both sides by } 5\text{)}\\ &= 36. \end{aligned}$$ So $M = 36L$: the wire has a mass of $36$ g per metre.
$M = 36 \times 12 = 432$ g.
We look for $L$ such that $36L = 450$: $$L = \frac{450}{36} = 12.5 \text{ m}.$$ Check: $36 \times 12.5 = 450$.
By the Proposition “Scaling in a Direct Proportion”, the mass is multiplied by $4$. For instance, $5$ m weigh $180$ g and $20$ m weigh $36 \times 20 = 720 = 4 \times 180$ g.

Skills to practice

Recognising Direct Proportion
Finding and Using the Constant of Proportionality
Justifying Direct Proportion

I.2 Proportionality on a Graph

Method — From a Table to a Graph

Each column of the table becomes a point: the first quantity on the horizontal axis, the second on the vertical axis.
Example. The proportional table $\textcolor{colordef}{1}, \textcolor{colordef}{2}, \textcolor{colordef}{3} \to \textcolor{colorprop}{2}, \textcolor{colorprop}{4}, \textcolor{colorprop}{6}$ gives the points below.

Proposition — Graphical Characterisation of Proportionality

A situation is proportional exactly when the points of its graph are aligned with the origin.

In the middle, the points are aligned but the line does not pass through the origin; on the right, the curve through the points passes through the origin but the points are not aligned. Neither situation is proportional: both conditions are needed.

Example

This graph shows the price of cherries depending on the mass. Is the price proportional to the mass? Read the price of $\textcolor{colordef}{3}$ kg.

Answer

The points are aligned with the origin, so the price is proportional to the mass.
Following the dotted lines from $\textcolor{colordef}{3}$ kg up to the graph and across to the vertical axis: $\textcolor{colordef}{3}$ kg cost $\textcolor{colorprop}{9}$ dollars.

Proposition — The Graph of Proportionality and the Intercept Theorem

The points of a proportional situation are aligned with the origin $O$. Let $A$ and $B$ be two of them, and let $H$ and $K$ be the points of the horizontal axis below them.

The lines $\Line{HA}$ and $\Line{KB}$ are both vertical, hence parallel; the points $O$, $H$, $K$ are aligned and the points $O$, $A$, $B$ are aligned. The intercept theorem applies in the triangles $OHA$ and $OKB$: $$\dfrac{OH}{OK} = \dfrac{HA}{KB} = \dfrac{OA}{OB}.$$ The lengths $OH$ and $HA$ are the coordinates of $A$, the lengths $OK$ and $KB$ those of $B$. The first equality therefore reads $$\dfrac{x_A}{x_B} = \dfrac{y_A}{y_B}, \qquad \text{that is} \qquad \dfrac{y_A}{x_A} = \dfrac{y_B}{x_B}.$$ The quotient of the two coordinates is the same at every point of the graph other than $O$: this common value is the coefficient of proportionality.

Example

A shop sells cherries at a fixed price per kilogram. On the graph, the point $A$ says that $2$ kg cost $5$ dollars, and the point $B$ is the point with $x$-coordinate $6$. Write the equality that the intercept theorem gives in the triangles $OHA$ and $OKB$, then deduce the price of $6$ kg.

Answer

Here $OH = 2$, $HA = 5$ and $OK = 6$; the price of $6$ kg is the length $KB$. $$\begin{aligned}[t] \dfrac{OH}{OK} &= \dfrac{HA}{KB} && \text{(intercept theorem)} \\ \dfrac{2}{6} &= \dfrac{5}{KB} \\ 2 \times KB &= 6 \times 5 && \text{(cross products)} \\ KB &= \dfrac{30}{2} = 15 \end{aligned}$$ So $6$ kg cost $15$ dollars. The two quotients of coordinates agree: $$\dfrac{5}{2} = 2.5 \qquad \text{and} \qquad \dfrac{15}{6} = 2.5,$$ and $2.5$ dollars per kilogram is the coefficient of proportionality.

Skills to practice

Recognising Proportionality on a Graph
Justifying Proportionality with the Intercept Theorem

I.3 The Direct-Proportion Function

A proportional situation is often given as a table: each value in the second row is obtained from the first by multiplying by the same number. Nothing forces us to stop at a table — that same rule can be read as a machine that takes any number and multiplies it by the coefficient. Written as a function, a proportional situation becomes the simplest function of all.

Report an error

Definition — Direct-Proportion Function

Let $a$ be a nonzero number. The function that maps every number $x$ to the number $ax$ is called the direct-proportion function of coefficient $a$. It is written $$f : x \longmapsto ax, \qquad\text{that is}\qquad f(x) = ax.$$ The number $a$ is the coefficient of the function: it is exactly the proportionality coefficient of the situation being modelled.

Proposition — Graph of a Direct-Proportion Function

The graph of the direct-proportion function $f(x) = ax$ is a \textcolor{colorprop}{straight line passing through the origin}. Conversely, any non-vertical straight line passing through the origin, other than the $x$-axis, is the graph of a direct-proportion function — a vertical line is the graph of no function at all, since it would give one input several outputs, and the $x$-axis is the graph of $x \longmapsto 0$, whose coefficient $0$ is excluded.

Reading the coefficient off the graph is immediate: $a$ is the value of $f(1)$, that is the $y$-coordinate of the point of the line with $x$-coordinate $1$.

Method — Modelling a Situation with a Direct-Proportion Function

Name the two quantities and choose which one is the input $x$: the other one is then $f(x)$.
Check that the situation is proportional: every pair of matching values with a nonzero input must give the same quotient output/input — and an input of $0$ must give an output of $0$.
Compute the coefficient $a$ by dividing one output by the matching input — any pair with a nonzero input gives the same $a$.
Write the function $f(x) = ax$, then use it: an input gives its output by multiplication and, since $a$ is not zero, an output gives its input by division.

Example

A tap fills an empty tank at a steady rate. After $4$ minutes, the tank holds $18$ litres.

Justify that the volume is proportional to the time, and model the situation by a direct-proportion function $V$.
How many litres are in the tank after $10$ minutes?
After how long does the tank hold $27$ litres?

Answer

The rate is steady, so twice the time gives twice the volume, and no time at all gives no water: the situation is proportional. The coefficient is obtained from the pair of matching values $(4, 18)$: $$a = \dfrac{18}{4} = 4.5,$$ so the tap delivers $4.5$ litres per minute and $$V(x) = 4.5\,x,$$ where $x$ is the time in minutes and $V(x)$ the volume in litres.
After $10$ minutes, $$V(10) = 4.5 \times 10 = 45 \text{ litres}.$$
We look for the input whose output is $27$, so we divide by the coefficient: $$x = \dfrac{27}{4.5} = 6 \text{ minutes}.$$ Check: $4.5 \times 6 = 27$.

Skills to practice

Recognising a Direct-Proportion Function
Computing with a Direct-Proportion Function
Solving Problems with a Direct-Proportion Function

II Inverse Proportion

II.1 Inversely Proportional Quantities

Definition — Inverse Proportion

A quantity $y$ is inversely proportional to a quantity $x$ (one also says that $y$ varies inversely as $x$) when there is a nonzero number $k$ such that $$y = \frac{k}{x}, \qquad\text{that is}\qquad xy = k,$$ for every pair of matching values, with $x \neq 0$. The number $k$ is the constant of proportionality: the product of matching values is the same for every pair. We write $y \propto \dfrac{1}{x}$.

Proposition — Scaling in an Inverse Proportion

If $y \propto \dfrac{1}{x}$, then multiplying $x$ by a nonzero number $m$ divides $y$ by $m$. For instance, doubling $x$ halves $y$, and dividing $x$ by $3$ multiplies $y$ by $3$.

Proof

Let $y = \dfrac{k}{x}$. When $x$ is replaced by $mx$, with $m \neq 0$, the quantity $y$ takes the value $$\begin{aligned}[t] \frac{k}{mx} &= \frac{1}{m} \times \frac{k}{x} && \text{(product of fractions)}\\ &= \frac{y}{m}. \end{aligned}$$ So $y$ is divided by $m$. Dividing $x$ by $3$ is multiplying it by $m = \dfrac{1}{3}$, so $y$ is divided by $\dfrac{1}{3}$, that is multiplied by $3$.

Proposition — Graph of an Inverse Proportion

Let $k > 0$. For $x > 0$, the graph of $y = \dfrac{k}{x}$ is a curve that goes down as $x$ increases and that comes as close as we like to each axis without ever meeting it: it is not a straight line. Each point $(x, y)$ of the curve is a corner of a rectangle drawn with the two axes, whose area is $xy = k$ — the same for every point. Here is the graph of $y = \dfrac{12}{x}$, with the rectangles of the points $(2, 6)$ and $(6, 2)$: their areas are $2 \times 6 = 12$ and $6 \times 2 = 12$.

Proof

Let $0 < x_1 < x_2$. The same positive number $k$ is divided by a larger positive number, so $\dfrac{k}{x_2} < \dfrac{k}{x_1}$: the curve goes down. The value $y = \dfrac{k}{x}$ is never $0$, because $k \neq 0$, so the curve never meets the $x$-axis; and $x = 0$ is excluded, so it never meets the $y$-axis. As $x$ grows, $\dfrac{k}{x}$ becomes as small as we like, and as $x$ gets close to $0$, it becomes as large as we like. The curve is not a straight line: a straight line that goes down meets the $x$-axis, which the curve never does. Finally, the rectangle with corners $(0, 0)$ and $(x, y)$ has sides $x$ and $y$, so its area is $xy = k$.

Decreasing Is Not Enough

Two quantities can vary in opposite directions without being inversely proportional. A $50$ L tank empties at $5$ L per minute: after $x$ minutes it holds $y = 50 - 5x$ litres. The volume $y$ goes down when $x$ goes up, but the product $xy$ is not constant: $2 \times 40 = 80$ and $4 \times 30 = 120$. Doubling $x$ from $2$ to $4$ does not halve $y$: it goes from $40$ to $30$, not to $20$. Only a constant product $xy$ shows an inverse proportion.

Report an error

Method — Solving an Inverse Proportion Problem

Check that the product $xy$ is the same for the known pairs, or read in the statement that $y$ is inversely proportional to $x$.
Compute the constant $k = xy$ from one known pair.
Write the formula $y = \dfrac{k}{x}$.
Use it: a value of $x$ gives $y = \dfrac{k}{x}$, and a value of $y$ gives $x = \dfrac{k}{y}$. Check the result with the constant product.

Example

Working at the same rate, $6$ workers build a wall in $10$ days. The time $T$, in days, is inversely proportional to the number $n$ of workers.

Find the constant of proportionality and write $T$ in terms of $n$.
How long do $4$ workers take? And $15$ workers?
How many workers are needed to build the wall in $5$ days?

Answer

The constant is the product of the known pair: $k = nT = 6 \times 10 = 60$ (the wall takes $60$ days of work for one worker). So $$T = \frac{60}{n}.$$
For $4$ workers, $T = \dfrac{60}{4} = 15$ days; for $15$ workers, $T = \dfrac{60}{15} = 4$ days.
We look for $n$ such that $\dfrac{60}{n} = 5$: $$n = \frac{60}{5} = 12 \text{ workers}.$$ Check: $12 \times 5 = 60$. As expected, $12$ workers, twice as many as $6$, take $5$ days, half of $10$ days.

The model holds only while every worker keeps the same rate: with too many workers on one wall, they would get in each other's way.

Example

$y$ is inversely proportional to $x$, and $y = 8$ when $x = 3$. Complete the table, then plot the points and join them with a smooth curve.

$x$	$1$	$2$	$3$	$4$	$6$	$12$
$y$	$?$	$?$	$8$	$?$	$?$	$?$

Answer

The constant is $k = xy = 3 \times 8 = 24$, so $y = \dfrac{24}{x}$:

$x$	$1$	$2$	$3$	$4$	$6$	$12$
$y$	$24$	$12$	$8$	$6$	$4$	$2$

In every column the product is $24$. The points lie on a curve, not on a straight line:

From Inverse Proportion to the Reciprocal Function

The reciprocal function $x \mapsto \dfrac{1}{x}$ is the inverse proportion $y = \dfrac{k}{x}$ with $k = 1$, now taken for every non-zero number $x$, negative numbers included.

Report an error

Definition — Reciprocal Function

The reciprocal function is defined on $\mathbb{R} \setminus \{0\}$ (all real numbers except $0$) by $f(x) = \frac{1}{x}$.

Domain: $\mathbb{R} \setminus \{0\}$, also written $(-\infty, 0) \cup (0, +\infty)$.
Shape: A curve called a hyperbola.

Skills to practice

Recognising Inverse Proportion
Solving Inverse Proportion Problems
Finding Images and Inputs
Justifying Inverse Proportion

III Proportion with a Power

III.1 Direct and Inverse Proportion with a Power

Definition — Proportion with a Power

Let $n$ be a positive whole number and $k$ a nonzero number.

$y$ is directly proportional to $x^n$ when $y = kx^n$ for every pair of matching values. We write $y \propto x^n$. For $n = 2$, $y$ is proportional to the square of $x$; for $n = 3$, to the cube of $x$.
$y$ is inversely proportional to $x^n$ when $y = \dfrac{k}{x^n}$ for every pair of matching values, with $x \neq 0$. We write $y \propto \dfrac{1}{x^n}$. For $n = 2$, this is the inverse-square law.

For $n = 1$, these are direct proportion and inverse proportion.

Proposition — Scaling with a Power

Let $m$ be a nonzero number.

If $y \propto x^n$, multiplying $x$ by $m$ multiplies $y$ by $m^n$.
If $y \propto \dfrac{1}{x^n}$, multiplying $x$ by $m$ divides $y$ by $m^n$.

For instance, doubling $x$ multiplies $y$ by $2^2 = 4$ when $y \propto x^2$, by $2^3 = 8$ when $y \propto x^3$, and divides $y$ by $4$ when $y \propto \dfrac{1}{x^2}$.

Proof

If $y = kx^n$, replacing $x$ by $mx$ gives $$\begin{aligned}[t] k(mx)^n &= k \times m^n x^n && \text{(Exponent Law 4)}\\ &= m^n \times kx^n && \text{(commutativity and associativity)}\\ &= m^n \times y. \end{aligned}$$ If $y = \dfrac{k}{x^n}$, replacing $x$ by $mx$ gives $$\begin{aligned}[t] \frac{k}{(mx)^n} &= \frac{k}{m^n x^n} && \text{(Exponent Law 4)}\\ &= \frac{1}{m^n} \times \frac{k}{x^n} && \text{(product of fractions)}\\ &= \frac{y}{m^n}. \end{aligned}$$ So $y$ is multiplied by $m^n$ in the first case and divided by $m^n$ in the second.

Telling the Laws Apart

Doubling $x$ is a quick way to guess the law that links $y$ to $x$, or to rule one out:

law	$y \propto x$	$y \propto x^2$	$y \propto x^3$	$y \propto \frac{1}{x}$	$y \propto \frac{1}{x^2}$
$x$ doubled: $y$ is	$\times 2$	$\times 4$	$\times 8$	$\div 2$	$\div 4$

In particular, an inverse-square law is not an inverse proportion: when the distance to a lamp doubles, the light received is divided by $4$, not by $2$. Likewise, a quantity proportional to the square of $x$ is not proportional to $x$.

Report an error

Example

On a dry road, the braking distance $d$, in metres, of a car is proportional to the square of its speed $v$, in km/h. At $40$ km/h, the braking distance is $8$ m.

Find the constant $k$ and write $d$ in terms of $v$.
Find the braking distance at $80$ km/h and at $100$ km/h.
At what speed is the braking distance $18$ m?

Answer

Since $d \propto v^2$, $d = kv^2$. The pair $v = 40$, $d = 8$ gives $$\begin{aligned}[t] 8 &= k \times 40^2 && \text{(the known pair)}\\ 8 &= 1600k\\ k &= \frac{8}{1600} && \text{(divide both sides by } 1600\text{)}\\ &= 0.005. \end{aligned}$$ So $d = 0.005v^2$.
At $80$ km/h, $d = 0.005 \times 80^2 = 0.005 \times 6400 = 32$ m: the speed is doubled, so the distance is multiplied by $4$ ($4 \times 8 = 32$). At $100$ km/h, $d = 0.005 \times 100^2 = 0.005 \times 10\,000 = 50$ m.
We look for $v > 0$ such that $0.005v^2 = 18$: $$\begin{aligned}[t] v^2 &= \frac{18}{0.005} && \text{(divide both sides by } 0.005\text{)}\\ &= 3600, \end{aligned}$$ so $v = 60$ km/h, the negative solution $-60$ being rejected, since a speed here is positive. Check: $0.005 \times 3600 = 18$.

Method — Recognising the Law from a Table

For each column of the table (with $x \neq 0$):

compute $\dfrac{y}{x}$ and $xy$, then, if neither is constant, $\dfrac{y}{x^2}$ and $x^2y$, then $\dfrac{y}{x^3}$ and $x^3y$;
keep the calculation that gives the same number in every column: if $\dfrac{y}{x^n}$ is constant, $y \propto x^n$; if $x^ny$ is constant, $y \propto \dfrac{1}{x^n}$; that number is the constant $k$;
if no calculation is constant, the table shows none of the laws tested here ($n = 1$, $2$ or $3$) — one column that breaks the pattern is enough.

Two columns in which $x$ is doubled give a first guess (Remark “Telling the Laws Apart”), which the calculation then confirms.

Example

Each table comes from an experiment. For each one, find which law of this section links $y$ to $x$, if any, and the constant $k$.

The mass $y$, in grams, of a metal cube of edge $x$ cm.

$x$ $1$ $2$ $3$ $4$

$y$ $5$ $40$ $135$ $320$
The illuminance $y$, in lux, at $x$ metres from a small lamp.

$x$ $1$ $2$ $4$ $5$

$y$ $100$ $25$ $6.25$ $4$
The height $y$, in cm, of a plant $x$ weeks after it was measured for the first time.

$x$ $1$ $2$ $3$ $4$

$y$ $6$ $8$ $10$ $12$

Answer

From $x = 1$ to $x = 2$, $y$ is multiplied by $8$: a first guess is $y \propto x^3$. The calculation confirms it: $\dfrac{y}{x^3}$ is $\dfrac{5}{1} = 5$, $\dfrac{40}{8} = 5$, $\dfrac{135}{27} = 5$ and $\dfrac{320}{64} = 5$. So $y \propto x^3$, with $k = 5$: $y = 5x^3$.
From $x = 1$ to $x = 2$, $y$ is divided by $4$: a first guess is $y \propto \dfrac{1}{x^2}$. The product $xy$ is not constant ($1 \times 100 = 100$, $2 \times 25 = 50$), but $x^2y$ is: $1 \times 100 = 100$, $4 \times 25 = 100$, $16 \times 6.25 = 100$ and $25 \times 4 = 100$. So $y \propto \dfrac{1}{x^2}$ with $k = 100$: $y = \dfrac{100}{x^2}$, an inverse-square law.
$\dfrac{y}{x}$ is $6$ then $4$; $xy$ is $6$ then $16$; $\dfrac{y}{x^2}$ is $6$ then $2$; $x^2y$ is $6$ then $32$; $\dfrac{y}{x^3}$ is $6$ then $1$; $x^3y$ is $6$ then $64$: from the second column on, no calculation stays the same. The height goes up by $2$ cm each week, $y = 2x + 4$: the table shows none of these laws (the points lie on a straight line that does not pass through the origin).

Skills to practice

Identifying the Effect of a Power
Using Proportion with a Power
Choosing a Law from Data

IV Going further

Skills to practice

Solving Problems with Proportion
Investigating Patterns in Proportion
Quiz

\(x\)	\(1\)	\(2\)	\(4\)	\(8\)
\(y\)	\(3\)	\(6\)	\(12\)	\(24\)

\(x\)	\(30\)	\(40\)	\(60\)	\(120\)
\(y\)	\(4\)	\(3\)	\(2\)	\(1\)

\(x\)	\(1\)	\(2\)	\(3\)	\(4\)	\(6\)	\(12\)
\(y\)	\(?\)	\(?\)	\(8\)	\(?\)	\(?\)	\(?\)

\(x\)	\(1\)	\(2\)	\(3\)	\(4\)	\(6\)	\(12\)
\(y\)	\(24\)	\(12\)	\(8\)	\(6\)	\(4\)	\(2\)

law	\(y \propto x\)	\(y \propto x^2\)	\(y \propto x^3\)	\(y \propto \frac{1}{x}\)	\(y \propto \frac{1}{x^2}\)
\(x\) doubled: \(y\) is	\(\times 2\)	\(\times 4\)	\(\times 8\)	\(\div 2\)	\(\div 4\)