Sometimes y is not directly proportional to x itself but to a power or a root of x. If y is directly proportional to x2, write y=kx2. The quantity multiplied by k is x2, not x. **Statement** | **Equation** | **Constant ratio** y is directly proportional to x2 | y=kx2 | yx2=k y is directly proportional to x3 | y=kx3 | yx3=k y is directly proportional to sqrt(x) | y=ksqrt(x) | ysqrt(x)=k y is directly proportional to (x+1) | y=k(x+1) | yx+1=k sqrt(y) is directly proportional to x | sqrt(y)=kx | sqrt(y)x=k y2 is directly proportional to x3 | y2=kx3 | y2x3=k The proportional quantity can be a power or a root of y as well as of x. To name the two proportional quantities in an equation, rearrange it into a constant ratio: from y2=4x3, y2x3=4, so y2 is directly proportional to x3. Why: plot y=2x2 against x and you get a curve, so y is not directly proportional to x. The ratios yx are 2, 4 and 6. Plot the same values against x2 and you get a straight line through O with gradient 2, because yx2=2 for every point. Method - Write the equation with exactly the power or root stated, for example y=kx2. - Substitute the given pair, working out the power first, and solve for k. - Rewrite the equation with the value of k. - Substitute and solve. When you take a square root, write +-; reject a value only if the context needs a positive quantity. Example The area, A cm2, of a printed logo is directly proportional to the square of its width, w cm. When w=4, A=12. Find A when w=10, and the width when A=48. A=kw2 12=k(42)=16k k= 34 When w=10: A= 34(102)=75. When A=48: 48= 34w2, so w2=64 and w=+-8. A width is positive, so w=8. Second example: sqrt(y) is directly proportional to x, and y=36 when x=3. Then sqrt(36)=k(3), so k=2 and sqrt(y)=2x. Squaring both sides gives y=4x2 (for x 0), so y is also directly proportional to x2. When x=5, y=4(52)=100. Check: a common error is to find k from 12=4k, which ignores the square. Here doubling the width from 4 to 8 multiplies the area by 4: 12 x 4=48. With a root of y, find k first and square at the end.

Other forms of direct proportion

Secondary 2

Sometimes $y$ is not directly proportional to $x$ itself but to a power or a root of $x$. If $y$ is directly proportional to $x^2$, write $y=kx^2$. The quantity multiplied by $k$ is $x^2$, not $x$.

Statement	Equation	Constant ratio
$y$ is directly proportional to $x^2$	$y=kx^2$	$\dfrac{y}{x^2}=k$
$y$ is directly proportional to $x^3$	$y=kx^3$	$\dfrac{y}{x^3}=k$
$y$ is directly proportional to $\sqrt{x}$	$y=k\sqrt{x}$	$\dfrac{y}{\sqrt{x}}=k$
$y$ is directly proportional to $(x+1)$	$y=k(x+1)$	$\dfrac{y}{x+1}=k$
$\sqrt{y}$ is directly proportional to $x$	$\sqrt{y}=kx$	$\dfrac{\sqrt{y}}{x}=k$
$y^2$ is directly proportional to $x^3$	$y^2=kx^3$	$\dfrac{y^2}{x^3}=k$

The proportional quantity can be a power or a root of $y$ as well as of $x$. To name the two proportional quantities in an equation, rearrange it into a constant ratio: from $y^2=4x^3$, $\dfrac{y^2}{x^3}=4$, so $y^2$ is directly proportional to $x^3$.

Why: plot $y=2x^2$ against $x$ and you get a curve, so $y$ is not directly proportional to $x$. The ratios $\dfrac{y}{x}$ are $2$, $4$ and $6$.

Plot the same values against $x^2$ and you get a straight line through $O$ with gradient $2$, because $\dfrac{y}{x^2}=2$ for every point.

Method

Write the equation with exactly the power or root stated, for example $y=kx^2$.
Substitute the given pair, working out the power first, and solve for $k$.
Rewrite the equation with the value of $k$.
Substitute and solve. When you take a square root, write $\pm$; reject a value only if the context needs a positive quantity.

Example

The area, $A\text{ cm}^2$, of a printed logo is directly proportional to the square of its width, $w$ cm. When $w=4$, $A=12$. Find $A$ when $w=10$, and the width when $A=48$.

$\begin{aligned} A&=kw^2 \\ 12&=k(4^2)=16k \\ k&=\dfrac{3}{4} \end{aligned}$

When $w=10$: $A=\dfrac{3}{4}(10^2)=75$.

When $A=48$: $48=\dfrac{3}{4}w^2$, so $w^2=64$ and $w=\pm8$. A width is positive, so $w=8$.

Second example: $\sqrt{y}$ is directly proportional to $x$, and $y=36$ when $x=3$. Then $\sqrt{36}=k(3)$, so $k=2$ and $\sqrt{y}=2x$. Squaring both sides gives $y=4x^2$ (for $x\ge0$), so $y$ is also directly proportional to $x^2$. When $x=5$, $y=4(5^2)=100$.

Check: a common error is to find $k$ from $12=4k$, which ignores the square. Here doubling the width from $4$ to $8$ multiplies the area by $4$: $12\times4=48$. With a root of $y$, find $k$ first and square at the end.

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Statement	Equation	Constant ratio
\(y\) is directly proportional to \(x^2\)	\(y=kx^2\)	\(\dfrac{y}{x^2}=k\)
\(y\) is directly proportional to \(x^3\)	\(y=kx^3\)	\(\dfrac{y}{x^3}=k\)
\(y\) is directly proportional to \(\sqrt{x}\)	\(y=k\sqrt{x}\)	\(\dfrac{y}{\sqrt{x}}=k\)
\(y\) is directly proportional to \((x+1)\)	\(y=k(x+1)\)	\(\dfrac{y}{x+1}=k\)
\(\sqrt{y}\) is directly proportional to \(x\)	\(\sqrt{y}=kx\)	\(\dfrac{\sqrt{y}}{x}=k\)
\(y^2\) is directly proportional to \(x^3\)	\(y^2=kx^3\)	\(\dfrac{y^2}{x^3}=k\)