Direct Proportion to Powers and Roots

When $y$ is directly 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 6.2: if $y \propto x$ , then $y = kx$ ; one pair of values gives $k$ .
Powers and roots: $5^2 = 25$ , $2^3 = 8$ , $\sqrt{36} = 6$ , and $(-4)^2 = 16$ , so $x^2 = 16$ has two solutions, $x = 4$ and $x = -4$ .

Proportion to the Square of a Quantity

The area of a circle is $A = \pi r^2$ . If the radius is doubled, the area is not doubled: it becomes $\pi(2r)^2 = 4\pi r^2$ , four times as large. The area is not directly proportional to $r$ , but it is directly proportional to $r^2$ , with $k = \pi$ .

Direct proportion to the square

$y$ is directly proportional to the square of $x$ , written $y \propto x^2$ , if $y = kx^2$ , where $k$ is a constant.

Then $\frac{y}{x^2}$ is the same for every pair of values (with $x \neq 0$ ), and it equals $k$ .

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

Example 1: A table for y = 3x²

For $y = 3x^2$ , complete a table of values for $x = 1, 2, 3, 4$ . Show that $\frac{y}{x^2}$ is the same for every pair but $\frac{y}{x}$ is not.

Solution

Square first, then multiply by 3: at $x = 2$ , $y = 3 \times 4 = 12$ ; at $x = 4$ , $y = 3 \times 16 = 48$ .

$x$	$1$	$2$	$3$	$4$
$x^2$	$1$	$4$	$9$	$16$
$y$	$3$	$12$	$27$	$48$
$\dfrac{y}{x^2}$	$3$	$3$	$3$	$3$
$\dfrac{y}{x}$	$3$	$6$	$9$	$12$

$\frac{y}{x^2} = 3$ in every column, so $y \propto x^2$ with $k = 3$ . But $\frac{y}{x}$ grows, so $y$ is not directly proportional to $x$ .

Look at the multipliers: from $x = 1$ to $x = 2$ , $x$ is doubled and $y$ goes from 3 to 12, four times as large; from $x = 1$ to $x = 3$ , $y$ goes from 3 to 27, nine times as large.

$y = 3, 12, 27, 48$ ; $\frac{y}{x^2} = 3$ for every pair, while $\frac{y}{x} = 3, 6, 9, 12$ .

The graph of $y = 3x^2$ is the curve of chapter 5. Drawn for $x \ge 0$ , it starts at the origin and gets steeper as $x$ increases. It passes through the origin, but it is not a straight line, so $y$ is not directly proportional to $x$ ; it is directly proportional to $x^2$ .

Example 2: Finding k from one pair

$y$ is directly proportional to $x^2$ , and $y = 12$ when $x = 2$ . Find
(a) an equation connecting $x$ and $y$ ,
(b) $y$ when $x = 5$ ,
(c) the values of $x$ when $y = 48$ .

Solution

(a) $y = kx^2$ . Put in $x = 2$ , $y = 12$ : $12 = k \times 2^2 = 4k$ , so $k = 3$ and $y = 3x^2$ .

(b) $y = 3 \times 5^2 = 3 \times 25 = 75\text{.}$

If $x$ is a length or another quantity that cannot be negative, only $x = 4$ is kept.

(a) $y = 3x^2$ (b) $y = 75$ (c) $x = 4$ or $x = -4$

Proportion to a Cube and to a Square Root

Cubes and square roots

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

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

In every case, find $k$ from one pair, write the equation, then find the unknown.

Example 3: Proportion to the cube

$y \propto x^3$ , and $y = 54$ when $x = 3$ . Find $y$ when $x = 2$ , and $x$ when $y = 250$ .

Solution

$y = kx^3$ : $54 = k \times 3^3 = 27k$ , so $k = 2$ and $y = 2x^3$ .

When $x = 2$ : $y = 2 \times 2^3 = 2 \times 8 = 16\text{.}$

When $y = 250$ : $2x^3 = 250$ , so $x^3 = 125$ and $x = 5$ , since $5^3 = 125$ . (Only $x = 5$ : unlike $x^2 = 25$ , the equation $x^3 = 125$ has one solution, because $(-5)^3 = -125$ , not 125.)

$y = 16$ when $x = 2$ ; $x = 5$ when $y = 250$ .

Example 4: Proportion to the square root

$y$ is directly proportional to the square root of $x$ , and $y = 10$ when $x = 4$ . Find $y$ when $x = 25$ , and $x$ when $y = 15$ .

Solution

$y = k\sqrt{x}$ : $10 = k\sqrt{4} = 2k$ , so $k = 5$ and $y = 5\sqrt{x}$ .

When $x = 25$ : $y = 5\sqrt{25} = 5 \times 5 = 25\text{.}$

When $y = 15$ : $5\sqrt{x} = 15$ , so $\sqrt{x} = 3$ and $x = 3^2 = 9$ .

Check: $\frac{15}{\sqrt{9}} = 5 = k$ .

$y = 25$ when $x = 25$ ; $x = 9$ when $y = 15$ .

Example 5: Pizzas

The price of a pizza is directly proportional to the square of its diameter. A pizza 20 cm across costs $8. Find the price of a pizza 30 cm across.

Solution

Let the price be $ $P$ for a diameter of $d$ cm: $P = kd^2$ .

$8 = k \times 20^2 = 400k$ , so $k = \frac{8}{400} = 0.02$ and $P = 0.02d^2$ .

When $d = 30$ : $P = 0.02 \times 900 = 18$ .

By multipliers: the diameter is multiplied by $\frac{30}{20} = 1.5$ , so the price is multiplied by $1.5^2 = 2.25$ : 2.25 × $8 = $18.

The 30 cm pizza costs $18.

The Effect of a Change in x

If $x$ is multiplied by a number, $x^2$ is multiplied by its square, $x^3$ by its cube and $\sqrt{x}$ by its square root. So $y$ changes by that multiplier, whatever $k$ is.

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

Example 6: A change written with a letter

(a) $y \propto x^2$ , and $y = a$ for a certain value of $x$ . Find $y$ , in terms of $a$ , when this value of $x$ is doubled.
(b) $y \propto \sqrt{x}$ , and $y = b$ for a certain value of $x$ . Find $y$ when this value of $x$ is multiplied by 9.

Solution

(a) Let the value be $x_0$ , so $a = kx_0^2$ . When $x = 2x_0$ : $y = k(2x_0)^2 = 4kx_0^2 = 4a\text{.}$

(b) Let the value be $x_0$ , so $b = k\sqrt{x_0}$ . When $x = 9x_0$ : $y = k\sqrt{9x_0} = 3k\sqrt{x_0} = 3b\text{.}$

(a) $y = 4a$ (b) $y = 3b$

A percentage change in $x$ is a multiplier too. "Increased by 200%" means the new value is 100% + 200% = 300% of the old, so $x$ is multiplied by 3, not by 2. "Decreased by 50%" means multiplied by 0.5; "increased by 10%" means multiplied by 1.1.

Example 7: Braking distance

The braking distance of a car is directly proportional to the square of its speed. The speed is increased by 200%. Find the percentage increase in the braking distance.

Solution

Let the speed be $v$ m/s and the braking distance $d$ m, so $d = kv^2$ .

The new speed is $v + 2v = 3v$ , so the new distance is $k(3v)^2 = 9kv^2 = 9d$ .

The increase is $9d - d = 8d$ , which is $\frac{8d}{d} \times 100\% = 800\%$ of the old distance.

The braking distance increases by 800%: it becomes 9 times as long.

Example 8: Percentage decreases and increases

$y \propto x^2$ . Find the percentage change in $y$ when $x$ is (a) decreased by 50%, (b) increased by 10%.

Solution

(a) $x$ is multiplied by 0.5, so $y$ is multiplied by $0.5^2 = 0.25$ . The new $y$ is 25% of the old: a decrease of 100% − 25% = 75%.

(b) $x$ is multiplied by 1.1, so $y$ is multiplied by $1.1^2 = 1.21$ . The new $y$ is 121% of the old: an increase of 21%.

(a) $y$ decreases by 75%. (b) $y$ increases by 21%.

Example 9: Finding k from a difference

$y$ is directly proportional to $x^2$ . The value of $y$ when $x = 3$ is 32 more than the value of $y$ when $x = 1$ . Find $y$ when $x = -2$ .

Solution

$y = kx^2$ . When $x = 1$ , $y = k$ . When $x = 3$ , $y = 9k$ .

The difference is $9k - k = 8k = 32$ , so $k = 4$ and $y = 4x^2$ .

When $x = -2$ : $y = 4 \times (-2)^2 = 4 \times 4 = 16\text{.}$

$y = 16$

Which Power? Deciding from a Table

Testing a table

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

Or use multipliers: find two columns where $x$ is doubled. If $y$ is multiplied by 2, try $x$ ; by 4, try $x^2$ ; by 8, try $x^3$ ; by $\sqrt{2} \approx 1.41$ , try $\sqrt{x}$ . Then check every column.

Example 10: Which power?

Is $y$ directly proportional to $x$ , $x^2$ or $x^3$ ? Find the equation.

$x$	$2$	$4$	$5$	$8$
$y$	$5.2$	$41.6$	$81.25$	$332.8$

Solution

$x$	$2$	$4$	$5$	$8$
$y$	$5.2$	$41.6$	$81.25$	$332.8$
$\dfrac{y}{x}$	$2.6$	$10.4$	$16.25$	$41.6$
$\dfrac{y}{x^2}$	$1.3$	$2.6$	$3.25$	$5.2$
$\dfrac{y}{x^3}$	$0.65$	$0.65$	$0.65$	$0.65$

Only $\frac{y}{x^3}$ is the same in every column: $\frac{332.8}{512} = 0.65$ and the others too.

The multipliers agree: from $x = 2$ to $x = 4$ , $x$ is doubled and $y$ goes from 5.2 to 41.6, which is 8 times as large; from $x = 4$ to $x = 8$ , $y$ goes from 41.6 to 332.8, again 8 times.

$y \propto x^3$ , and $y = 0.65x^3$ .

Common Mistakes

Doubling $y$ when $x$ is doubled in $y \propto x^2$ . Then $y$ is multiplied by 4.
Taking "increased by 200%" as "multiplied by 2". It means multiplied by 3.
Calling "9 times as large" an increase of 900%. The increase is $9 - 1 = 8$ times the old value: 800%.
Multiplying before squaring. For $y = 3x^2$ at $x = 2$ : $3 \times 2^2 = 12$ , not $(3 \times 2)^2 = 36$ .
Dropping a solution. If $x$ can be negative, $x^2 = 16$ gives $x = 4$ or $x = -4$ .
Testing $\frac{y}{x^2}$ in two columns only. Every column must agree.

Practice Questions

Write the equation with $k$ first, and check each answer in it. The questions get harder as you go.

Question 1

For $y = 2x^2$ , complete a table of values for $x = 1, 2, 3, 4$ and show that $\frac{y}{x^2}$ is the same for every pair. Draw the graph of $y = 2x^2$ for $0 \le x \le 4$ .

ShowHide solution

$x$	$1$	$2$	$3$	$4$
$y$	$2$	$8$	$18$	$32$
$\dfrac{y}{x^2}$	$2$	$2$	$2$	$2$

At $x = 0$ , $y = 0$ : the curve starts at the origin.

$y = 2, 8, 18, 32$ ; $\frac{y}{x^2} = 2$ every time; the graph is the curve above, through the origin.

Question 2

$y \propto x^2$ , and $y = 20$ when $x = 2$ . Find $y$ when $x = 3$ , and the values of $x$ when $y = 125$ .

ShowHide solution

$20 = 4k$ , so $k = 5$ and $y = 5x^2$ .

When $x = 3$ : $y = 5 \times 9 = 45$ .

When $y = 125$ : $x^2 = 25$ , so $x = 5$ or $x = -5$ .

$y = 45$ ; $x = 5$ or $x = -5$

Question 3

$y \propto x^3$ , and $y = 24$ when $x = 2$ . Find $y$ when $x = 3$ , and $x$ when $y = 648$ .

ShowHide solution

$24 = 8k$ , so $k = 3$ and $y = 3x^3$ .

When $x = 3$ : $y = 3 \times 27 = 81$ .

When $y = 648$ : $x^3 = 216$ , so $x = 6$ .

$y = 81$ ; $x = 6$

Question 4

$y$ is directly proportional to $\sqrt{x}$ , and $y = 12$ when $x = 9$ . Find $y$ when $x = 16$ , and $x$ when $y = 28$ .

ShowHide solution

$12 = k\sqrt{9} = 3k$ , so $k = 4$ and $y = 4\sqrt{x}$ .

When $x = 16$ : $y = 4 \times 4 = 16$ .

When $y = 28$ : $\sqrt{x} = 7$ , so $x = 49$ .

$y = 16$ ; $x = 49$

Question 5

(a) $y \propto x^3$ , and $y = 40$ for a certain value of $x$ . Find $y$ when that value of $x$ is doubled, and when it is halved.
(b) $y \propto \sqrt{x}$ . By what number is $y$ multiplied when $x$ is multiplied by 9? By what must $x$ be multiplied to double $y$ ?

ShowHide solution

(a) Doubled: $y$ is multiplied by $2^3 = 8$ , so $y = 320$ . Halved: $y$ is multiplied by $\left(\frac{1}{2}\right)^3 = \frac{1}{8}\text{,}$ so $y = 5$ .

(b) $y$ is multiplied by $\sqrt{9} = 3$ . To multiply $y$ by 2, $\sqrt{x}$ must be multiplied by 2, so $x$ must be multiplied by $2^2 = 4$ .

(a) 320 and 5 (b) by 3; by 4

Question 6

$y \propto x^2$ . Find the percentage change in $y$ when $x$ is (a) increased by 20%, (b) decreased by 10%.

ShowHide solution

(a) $x$ is multiplied by 1.2, so $y$ is multiplied by $1.2^2 = 1.44$ .

The new $y$ is 144% of the old: an increase of 44%.

(b) $x$ is multiplied by 0.9, so $y$ is multiplied by $0.9^2 = 0.81$ .

The new $y$ is 81% of the old: a decrease of 19%.

(a) $y$ increases by 44%. (b) $y$ decreases by 19%.

Question 7

The area of a circle is directly proportional to the square of its radius. The radius is increased by 50%. Find the percentage increase in the area.

ShowHide solution

The radius is multiplied by 1.5, so the area is multiplied by $1.5^2 = 2.25$ .

The new area is 225% of the old: an increase of 125%.

The area increases by 125%.

Question 8

$y$ is directly proportional to $x^3$ . The value of $y$ when $x = 2$ is 21 more than the value of $y$ when $x = 1$ . Find $y$ when $x = 4$ .

ShowHide solution

$y = kx^3$ . At $x = 1$ , $y = k$ ; at $x = 2$ , $y = 8k$ .

$8k - k = 7k = 21$ , so $k = 3$ and $y = 3x^3$ .

When $x = 4$ : $y = 3 \times 64 = 192$ .

$y = 192$

Question 9

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

$x$	$2$	$3$	$5$	$6$
$y$	$12$	$27$	$75$	$108$

ShowHide solution

$x$	$2$	$3$	$5$	$6$
$y$	$12$	$27$	$75$	$108$
$\dfrac{y}{x}$	$6$	$9$	$15$	$18$
$\dfrac{y}{x^2}$	$3$	$3$	$3$	$3$

$\frac{y}{x^2} = 3$ in every column. Multipliers agree: from $x = 3$ to $x = 6$ , $x$ is doubled and $y$ goes from 27 to 108, four times as large.

$y \propto x^2$ , and $y = 3x^2$ .

Question 10

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

$x$	$4$	$9$	$16$	$25$
$y$	$6$	$9$	$12$	$15$

ShowHide solution

$x$	$4$	$9$	$16$	$25$
$y$	$6$	$9$	$12$	$15$
$\dfrac{y}{x}$	$1.5$	$1$	$0.75$	$0.6$
$\dfrac{y}{\sqrt{x}}$	$3$	$3$	$3$	$3$

$\sqrt{4} = 2$ , $\sqrt{9} = 3$ , $\sqrt{16} = 4$ , $\sqrt{25} = 5$ , and $\frac{y}{\sqrt{x}} = 3$ in every column.

The multipliers agree: from $x = 4$ to $x = 16$ , $x$ is multiplied by 4 and $y$ by 2.

$y \propto \sqrt{x}$ , and $y = 3\sqrt{x}$ .

Question 11

The mass of a solid metal cube is directly proportional to the cube of the length of its edge. A cube of edge 2 cm has a mass of 64 g. (a) Find the mass of a cube of edge 5 cm. (b) Find the edge of a cube of mass 216 g.

ShowHide solution

Let the mass be $m$ g for an edge of $e$ cm: $m = ke^3$ . $64 = 8k$ , so $k = 8$ and $m = 8e^3$ .

(a) $m = 8 \times 125 = 1000$ , so the mass is 1000 g, or 1 kg.

(b) $216 = 8e^3$ , so $e^3 = 27$ and $e = 3$ .

(a) 1000 g (b) 3 cm

Question 12

$y \propto x^2$ , and $y = 18$ when $x = 3$ . Siti writes: "When $x$ is doubled to 6, $y$ is doubled to 36." Find her mistake and the correct value of $y$ .

ShowHide solution

Since $y \propto x^2$ , doubling $x$ multiplies $y$ by $2^2 = 4$ , not by 2.

So $y = 18 \times 4 = 72$ . Check with $k$ : $18 = 9k$ , $k = 2$ , and $2 \times 6^2 = 72$ .

She treated $y$ as directly proportional to $x$ . The correct value is $y = 72$ .

Key Points

$y \propto x^2$ means $y = kx^2$ ; $y \propto x^3$ means $y = kx^3$ ; $y \propto \sqrt{x}$ means $y = k\sqrt{x}$ .
Find $k$ from one pair, write the equation, then find $y$ or $x$ ; $x^2 = 16$ gives $x = \pm 4$ when $x$ may be negative.
If $x$ is multiplied by $m$ , then $y$ is multiplied by $m^2$ , $m^3$ or $\sqrt{m}$ .
Turn a percentage change into a multiplier first: an increase of 200% is × 3, a decrease of 50% is × 0.5.
To decide the power from a table, test $\frac{y}{x}$ , $\frac{y}{x^2}$ , $\frac{y}{x^3}$ and $\frac{y}{\sqrt{x}}$ in every column, or use the multipliers.
The graph of $y = kx^2$ for $x \ge 0$ is a curve through the origin, not a straight line.