Inverse Proportion to Powers and Roots

When $y$ is inversely proportional to $x^2$ , $x^3$ or $\sqrt{x}$ : forming the equation, finding $k$ and unknown values, working out the effect of a change in $x$ , and deciding from a table which power it is.

Before you start

Lesson 6.5: if $y \propto \frac{1}{x}$ , then $y = \frac{k}{x}$ and $k = xy$ .
Lesson 6.3: if $x$ is multiplied by 2, then $x^2$ is multiplied by 4, $x^3$ by 8 and $\sqrt{x}$ by $\sqrt{2}$ .

Inverse Proportion to the Square

The brightness of the light from a small lamp is inversely proportional to the square of the distance from the lamp. Move twice as far away and the light is not half as bright but a quarter as bright, because the square of the distance is 4 times as large.

Inverse proportion to the square

$y$ is inversely proportional to the square of $x$ , written $y \propto \frac{1}{x^2}$ , if $y = \frac{k}{x^2}$ , where $k$ is a constant.

Then $x^2y$ is the same for every pair of values, and it equals $k$ .

When $x$ is multiplied by 2, $y$ is divided by $2^2 = 4$ ; when $x$ is multiplied by 3, $y$ is divided by $3^2 = 9$ ; when $x$ is halved, $y$ is multiplied by 4.

Example 1: Finding k from one pair

$y$ is inversely proportional to $x^2$ , and $y = 2$ when $x = 4$ . Find
(a) an equation connecting $x$ and $y$ ,
(b) $y$ when $x = 8$ ,
(c) the values of $x$ when $y = 8$ .

Solution

(a) $y = \frac{k}{x^2}$ , so $k = x^2y = 4^2 \times 2 = 16 \times 2 = 32\text{,}$ and $y = \frac{32}{x^2}$ .

(b) $y = \frac{32}{8^2} = \frac{32}{64} = \frac{1}{2}\text{.}$

Check (b) by multipliers: $x$ is doubled from 4 to 8, so $y$ is divided by 4, from 2 to $\frac{1}{2}$ .

(a) $y = \frac{32}{x^2}$ (b) $y = \frac{1}{2}$ (c) $x = 2$ or $x = -2$

Example 2: A table for y = 100/x²

For $y = \frac{100}{x^2}$ , complete a table for $x = 1, 2, 4, 5, 10$ . Show that $x^2y$ is the same for every pair but $xy$ is not.

Solution

Square first, then divide 100 by the square: at $x = 4$ , $y = \frac{100}{16} = 6.25$ .

$x$	$1$	$2$	$4$	$5$	$10$
$y$	$100$	$25$	$6.25$	$4$	$1$
$x^2y$	$100$	$100$	$100$	$100$	$100$
$xy$	$100$	$50$	$25$	$20$	$10$

$x^2y = 100$ in every column, so $y \propto \frac{1}{x^2}$ with $k = 100$ . But $xy$ falls, so $y$ is not inversely proportional to $x$ .

The multipliers: from $x = 1$ to $x = 2$ , $y$ falls from 100 to 25, a quarter; from $x = 1$ to $x = 10$ , $y$ falls to $\frac{1}{100}$ of its value. Compare $y = \frac{100}{x}$ , which at $x = 2$ is still 50: $y = \frac{100}{x^2}$ falls faster at first.

$y = 100, 25, 6.25, 4, 1\text{;}$ $x^2y = 100$ every time, while $xy = 100, 50, 25, 20, 10\text{.}$

Inverse Proportion to a Cube and to a Square Root

Cubes and square roots

$y \propto \frac{1}{x^3}$ means $y = \frac{k}{x^3}$ ; then $x^3y$ is the same for every pair. When $x$ is doubled, $y$ is divided by $2^3 = 8$ .

$y \propto \frac{1}{\sqrt{x}}$ means $y = \frac{k}{\sqrt{x}}$ ; then $\sqrt{x} \times y$ is the same for every pair. When $x$ is multiplied by 4, $y$ is divided by $\sqrt{4} = 2$ .

Example 3: Inverse proportion to the square root

$y$ is inversely proportional to $\sqrt{x}$ , and $y = 6$ when $x = 4$ . Find
(a) an equation connecting $x$ and $y$ ,
(b) $y$ when $x = 16$ ,
(c) $x$ when $y = 4$ .

Solution

(a) $k = \sqrt{x} \times y = \sqrt{4} \times 6 = 2 \times 6 = 12\text{,}$ so $y = \frac{12}{\sqrt{x}}$ .

(b) $y = \frac{12}{\sqrt{16}} = \frac{12}{4} = 3\text{.}$

(a) $y = \frac{12}{\sqrt{x}}$ (b) $y = 3$ (c) $x = 9$

Example 4: Inverse proportion to the cube

$y \propto \frac{1}{x^3}$ , and $y = 2$ when $x = 2$ . Find $y$ when $x = 4$ , and $x$ when $y = 128$ .

Solution

$k = x^3y = 2^3 \times 2 = 16\text{,}$ so $y = \frac{16}{x^3}$ .

When $x = 4$ : $y = \frac{16}{64} = \frac{1}{4}$ . (Doubling $x$ divides $y$ by 8: $2 \div 8 = \frac{1}{4}$ .)

When $y = 128$ : $128x^3 = 16$ , so $x^3 = \frac{16}{128} = \frac{1}{8}$ and $x = \frac{1}{2}$ , since $\left(\frac{1}{2}\right)^3 = \frac{1}{8}\text{.}$

$y = \frac{1}{4}$ when $x = 4$ ; $x = \frac{1}{2}$ when $y = 128$ .

Example 5: Which quantities are inversely proportional?

For each equation, state the two quantities that are inversely proportional to each other.
(a) $y = \frac{2}{x^3}$ (b) $y = \frac{3}{\sqrt{x}}$ (c) $y = \frac{7}{x^2}$

Solution

(a) $x^3y = 2$ , a constant, so $y$ is inversely proportional to $x^3$ . It is not inversely proportional to $x$ : doubling $x$ divides $y$ by 8, not by 2.

(b) $\sqrt{x} \times y = 3$ , so $y$ is inversely proportional to $\sqrt{x}$ .

(a) $y$ and $x^3$ (b) $y$ and $\sqrt{x}$ (c) $y$ and $x^2$

The Effect of a Change in x

In inverse proportion, $y$ is multiplied by the reciprocal of the change in $x^2$ , $x^3$ or $\sqrt{x}$ . The table gives the number $y$ is multiplied by.

$x$ is multiplied by	$2$	$3$	$4$
$y \propto \frac{1}{x}$ : $y$ is multiplied by	$\dfrac{1}{2}$	$\dfrac{1}{3}$	$\dfrac{1}{4}$
$y \propto \frac{1}{x^2}$ : $y$ is multiplied by	$\dfrac{1}{4}$	$\dfrac{1}{9}$	$\dfrac{1}{16}$
$y \propto \frac{1}{x^3}$ : $y$ is multiplied by	$\dfrac{1}{8}$	$\dfrac{1}{27}$	$\dfrac{1}{64}$
$y \propto \frac{1}{\sqrt{x}}$ : $y$ is multiplied by	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{\sqrt{3}}$	$\dfrac{1}{2}$

Example 6: A lamp moved farther away

The brightness of the light from a small lamp is inversely proportional to the square of the distance from the lamp. At $d$ m from the lamp the brightness is $B$ units. When the distance is increased by 400%, the brightness becomes $cB$ units. Find $c$ .

Solution

$B = \frac{k}{d^2}$ , so $k = Bd^2$ .

An increase of 400% makes the distance $d + 4d = 5d$ .

The new brightness is $\frac{k}{(5d)^2} = \frac{Bd^2}{25d^2} = \frac{1}{25}B\text{.}$

$c = \dfrac{1}{25} = 0.04$

Example 7: Percentage changes

$y \propto \frac{1}{x^2}$ . Find the percentage change in $y$ when $x$ is (a) halved, (b) increased by 25%.

Solution

(a) $x$ is multiplied by $\frac{1}{2}$ , so $x^2$ is multiplied by $\frac{1}{4}$ and $y$ by 4. The new $y$ is 400% of the old: an increase of 300%.

(b) $x$ is multiplied by $1.25 = \frac{5}{4}$ , so $x^2$ is multiplied by $\frac{25}{16}$ and $y$ by $\frac{16}{25} = 0.64$ . The new $y$ is 64% of the old: a decrease of 36%.

(a) an increase of 300% (b) a decrease of 36%

Problems, and Deciding the Power

Example 8: Cones of the same volume

For cones of a fixed volume, the height, $h$ cm, is inversely proportional to the square of the base radius, $r$ cm. Cone A has base radius 6 cm and height 5 cm. Cones B and C have the same volume as A. Cone B has base radius 3 cm, and cone C has height 1.25 cm. Find (a) the height of cone B, (b) the base radius of cone C.

Solution

$h = \frac{k}{r^2}$ , so $k = hr^2$ . From cone A: $k = 5 \times 6^2 = 180$ .

(a) $h \times 3^2 = 180$ , so $h = \frac{180}{9} = 20$ .

(b) $1.25 \times r^2 = 180$ , so $r^2 = 144$ and $r = 12$ (a radius is positive, so $-12$ is not kept).

By multipliers: halving the radius from 6 to 3 multiplies the height by 4, from 5 to 20.

(a) 20 cm (b) 12 cm

Testing a table

Work out $xy$ , $x^2y$ , $x^3y$ and $\sqrt{x} \times y$ in every column. The one that is the same in every column gives the power, and its value is $k$ .

Or use multipliers: where $x$ is doubled, $y$ divided by 2 suggests $\frac{1}{x}$ ; by 4, $\frac{1}{x^2}$ ; by 8, $\frac{1}{x^3}$ . Then check every column.

Example 9: Which power?

Is $y$ inversely proportional to $x$ or to $x^2$ ? Find the equation.

$x$	$1$	$2$	$3$	$6$
$y$	$36$	$9$	$4$	$1$

Solution

$x$	$1$	$2$	$3$	$6$
$y$	$36$	$9$	$4$	$1$
$xy$	$36$	$18$	$12$	$6$
$x^2y$	$36$	$36$	$36$	$36$

Only $x^2y$ is the same in every column. The multipliers agree: from $x = 3$ to $x = 6$ , $x$ is doubled and $y$ falls from 4 to 1, a quarter.

$y \propto \frac{1}{x^2}$ , and $y = \frac{36}{x^2}$ .

Example 10 (challenge): Proportion to an expression

$q^2$ is inversely proportional to $p + 3$ , and $q = 5$ when $p = 2$ . Find the values of $q$ when $p = 17$ .

Solution

The quantities in proportion are $q^2$ and $p + 3$ : $q^2 = \frac{k}{p + 3}$ , so $k = q^2(p + 3)$ .

When $p = 2$ , $q = 5$ : $k = 25 \times 5 = 125$ .

When $p = 17$ : $q^2 \times 20 = 125$ , so $q^2 = 6.25$ and $q = 2.5$ or $q = -2.5$ .

$q = 2.5$ or $q = -2.5$

Common Mistakes

Halving $y$ when $x$ is doubled in $y \propto \frac{1}{x^2}$ . Then $y$ is divided by 4.
Squaring the wrong number for $k$ . For $y = 2$ at $x = 4$ : $k = 4^2 \times 2 = 32$ , not $(4 \times 2)^2$ or $4 \times 2$ .
Dropping a solution: $x^2 = 4$ gives $x = 2$ or $x = -2$ , unless $x$ is a length or another quantity that cannot be negative.
Taking "increased by 400%" as "multiplied by 4". It means multiplied by 5.
Saying $y$ is inversely proportional to $x$ when the equation is $y = \frac{7}{x^2}$ . It is inversely proportional to $x^2$ .

Practice Questions

Write $k$ as the product ( $x^2y$ , $x^3y$ or $\sqrt{x} \times y$ ) first, and check each answer with it. The questions get harder as you go.

Question 1

$y \propto \frac{1}{x^2}$ , and $y = 5$ when $x = 2$ . Find $y$ when $x = 4$ , and the values of $x$ when $y = 0.8$ .

ShowHide solution

$k = 2^2 \times 5 = 20$ , so $y = \frac{20}{x^2}$ .

When $x = 4$ : $y = \frac{20}{16} = 1.25$ .

When $y = 0.8$ : $x^2 = \frac{20}{0.8} = 25$ , so $x = 5$ or $x = -5$ .

$y = 1.25$ ; $x = 5$ or $x = -5$

Question 2

$y$ is inversely proportional to $\sqrt{x}$ , and $y = 6$ when $x = 9$ . Find $y$ when $x = 25$ , and $x$ when $y = 9$ .

ShowHide solution

$k = \sqrt{9} \times 6 = 18$ , so $y = \frac{18}{\sqrt{x}}$ .

When $x = 25$ : $y = \frac{18}{5} = 3.6$ .

When $y = 9$ : $\sqrt{x} = \frac{18}{9} = 2$ , so $x = 4$ .

$y = 3.6$ ; $x = 4$

Question 3

$y \propto \frac{1}{x^3}$ , and $y = 5$ when $x = 2$ . Find $y$ when $x = 1$ , and $x$ when $y = 0.32$ .

ShowHide solution

$k = 2^3 \times 5 = 40$ , so $y = \frac{40}{x^3}$ .

When $x = 1$ : $y = 40$ .

When $y = 0.32$ : $x^3 = \frac{40}{0.32} = 125\text{,}$ so $x = 5$ .

$y = 40$ ; $x = 5$

Question 4

For each equation, state which two quantities are inversely proportional, or say that there are none.
(a) $P = \frac{8}{t^2}$ (b) $A = \frac{6}{\sqrt{n}}$ (c) $y = \frac{x^3}{4}$

ShowHide solution

(a) $t^2P = 8$ : $P$ and $t^2$ .

(b) $\sqrt{n} \times A = 6$ : $A$ and $\sqrt{n}$ .

(a) $P$ and $t^2$ (b) $A$ and $\sqrt{n}$ (c) none: $y \propto x^3$

Question 5

$y \propto \frac{1}{\sqrt{x}}$ . By what number is $y$ multiplied when $x$ is (a) multiplied by 9, (b) divided by 4?

ShowHide solution

(a) $\sqrt{x}$ is multiplied by $\sqrt{9} = 3$ , so $y$ is divided by 3: multiplied by $\frac{1}{3}$ .

(b) $\sqrt{x}$ is divided by $\sqrt{4} = 2$ , so $y$ is multiplied by 2.

(a) $\frac{1}{3}$ (b) 2

Question 6

$y \propto \frac{1}{x^2}$ . Find the percentage change in $y$ when $x$ is increased by 100%.

ShowHide solution

An increase of 100% doubles $x$ , so $y$ is multiplied by $\frac{1}{4}$ : 25% of the old value.

That is a decrease of 75%.

$y$ decreases by 75%.

Question 7

$y \propto \frac{1}{x^2}$ . Find the percentage change in $y$ when $x$ is decreased by 20%.

ShowHide solution

$x$ is multiplied by 0.8, so $x^2$ is multiplied by $0.64$ and $y$ by $\frac{1}{0.64} = 1.5625$ .

The new $y$ is 156.25% of the old: an increase of 56.25%.

$y$ increases by 56.25%.

Question 8

The brightness of the light from a small lamp is inversely proportional to the square of the distance from it. At 2 m the brightness is 90 units. (a) Find the brightness at 3 m. (b) At what distance is the brightness 10 units?

ShowHide solution

$B = \frac{k}{d^2}$ : $k = 2^2 \times 90 = 360$ .

(a) $B = \frac{360}{9} = 40$ .

(b) $10d^2 = 360$ , so $d^2 = 36$ and $d = 6$ (a distance is positive).

(a) 40 units (b) 6 m

Question 9

Yi Hao pours the same volume of water into cylinders of different diameters. He says that the height of the water is inversely proportional to the diameter. Do you agree? (The volume of a cylinder is $\pi r^2h$ .) Test his claim with a cylinder of diameter 10 cm in which the water is 8 cm high.

ShowHide solution

With radius $r = \frac{d}{2}$ , the volume is $V = \pi\left(\frac{d}{2}\right)^2h = \frac{\pi d^2h}{4}\text{,}$ so $d^2h = \frac{4V}{\pi}$ , which is the same for every cylinder.

So $h$ is inversely proportional to $d^2$ , not to $d$ .

For the test: $d^2h = 10^2 \times 8 = 800$ . A cylinder of diameter 20 cm holds the water at $h = \frac{800}{400} = 2$ cm, a quarter of 8 cm, not half (4 cm) as Yi Hao would say.

No: the height is inversely proportional to the square of the diameter. Doubling the diameter divides the height by 4.

Question 10

Is $y$ inversely proportional to $x$ or to $\sqrt{x}$ ? Find the equation.

$x$	$1$	$4$	$9$	$16$
$y$	$12$	$6$	$4$	$3$

ShowHide solution

$x$	$1$	$4$	$9$	$16$
$y$	$12$	$6$	$4$	$3$
$xy$	$12$	$24$	$36$	$48$
$\sqrt{x} \times y$	$12$	$12$	$12$	$12$

$\sqrt{x} \times y = 12$ in every column. From $x = 1$ to $x = 4$ , $x$ is multiplied by 4 and $y$ is halved.

$y \propto \frac{1}{\sqrt{x}}$ , and $y = \frac{12}{\sqrt{x}}$ .

Question 11

$y \propto \frac{1}{x^2}$ , and $y = 12$ when $x = 2$ . Daniel writes: "When $x$ is doubled to 4, $y$ is halved to 6." Find his mistake and the correct value of $y$ .

ShowHide solution

Doubling $x$ multiplies $x^2$ by 4, so $y$ is divided by 4, not 2.

$y = 12 \div 4 = 3$ . Check with $k = 2^2 \times 12 = 48$ : $\frac{48}{16} = 3$ .

He treated $y$ as inversely proportional to $x$ . The correct value is $y = 3$ .

Question 12 (challenge)

$y^2$ is inversely proportional to $x + 1$ , and $y = 3$ when $x = 3$ . Find the values of $y$ when $x = 8$ .

ShowHide solution

$y^2 = \frac{k}{x + 1}$ , so $k = y^2(x + 1) = 9 \times 4 = 36\text{.}$

When $x = 8$ : $y^2 = \frac{36}{9} = 4$ , so $y = 2$ or $y = -2$ .

$y = 2$ or $y = -2$

Key Points

$y \propto \frac{1}{x^2}$ means $y = \frac{k}{x^2}$ ( $k = x^2y$ ); $y \propto \frac{1}{x^3}$ means $y = \frac{k}{x^3}$ ; $y \propto \frac{1}{\sqrt{x}}$ means $y = \frac{k}{\sqrt{x}}$ .
Find $k$ from one pair, write the equation, then find $y$ or $x$ ; $x^2 = 4$ gives $x = \pm 2$ when $x$ may be negative.
If $x$ is multiplied by $m$ , then $y$ is divided by $m^2$ , $m^3$ or $\sqrt{m}$ .
Turn a percentage change into a multiplier first: an increase of 400% is × 5.
To decide the power from a table, work out $xy$ , $x^2y$ , $x^3y$ and $\sqrt{x} \times y$ in every column.
Read the equation for what is in proportion: in $y = \frac{7}{x^2}$ , $y$ is inversely proportional to $x^2$ , not to $x$ .