Proportion Basics

Definition 1. The quantity y is said to be directly proportional to the quantity x if there exists a nonzero constant k such that

y = k · x.

In this case, we write y ∝ x.

Problem 1. Show that the area A of a circle is directly proportional to its squared-radius r².

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Solution. Since $A = \pi r^2$ and $\pi \neq 0$ , we have $A \propto r^2$ .

Problem 2. Show that the volume V of a sphere is directly proportional to its cubed-radius r³.

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Solution. Since $V = \frac 43 \pi r^3$ and $\frac 43 \pi \neq 0$ , we have $V \propto r^3$ .

Problem 3. Show that the following properties hold:

For any quantity x, x ∝ x.
For quantities x, y, if y ∝ x, then x ∝ y.
For quantities x, y, z, if y ∝ x and z ∝ y, then z ∝ x.

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Solution. Since multiplying by 1 does not change a number,

$x = 1 \cdot x \quad \Rightarrow \quad x \propto x.$

If $y \propto x$ , then there exists some nonzero $k$ such that $y = k \cdot x$ . Since $1/k \neq 0$ ,

$\begin{aligned} y = k \cdot x \quad &\Rightarrow \quad x = \frac 1k \cdot y \\ &\Rightarrow \quad x \propto y. \end{aligned}$

Given a quantity x, let x₁, x₂ denote specific instances of x.

Problem 4. Suppose there exists an integer n such that y ∝ xⁿ. Show that

$\displaystyle \frac{y_2}{y_1} = \left( \frac{x_2}{x_1} \right)^n.$

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Solution. Since $y \propto x^n$ , there exists $k \neq 0$ such that $y = k \cdot x^n$ . As specific instances,

$y = k \cdot x_1^n,\quad y = k \cdot x_2^n.$

Therefore,

$\begin{aligned} \frac{y_2}{y_1} &= \frac{ k \cdot x_2^n }{ k \cdot x_1^n } = \frac{ x_2^n }{ x_1^n } = \left( \frac{ x_2 }{ x_1 } \right)^n. \end{aligned}$

Problem 5. Given that y ∝ x² and x increased by 50%, determine the percentage change in y.

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Solution. Since $x$ increased by 50%, we have

$x_2 = x_1 + 0.5 x_1 = 1.5 x_1.$

Therefore, $x_2/x_1 = 1.5$ . By Problem 4,

$\displaystyle \frac{y_2}{y_1} =(1.5)^2 = 2.25.$

Therefore, $y_2 = 2.25y_1$ . Calculating the percentage change yields

$\begin{aligned} \frac{y_2 - y_1}{y_1} &= \frac{2.25y_1 - y_1}{y_1} \\ &= \frac{1.25y_1}{y_1} \\ &= 1.25 \\ &= 125\%. \end{aligned}$

Problem 6. Given that y ∝ x³ and x decreased by 10%, determine the percentage change in y.

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Solution. Since $x$ decreased by 10%, we have

$x_2 = x_1 - 0.1 x_1 = 0.9 x_1.$

Therefore, $x_2/x_1 = 0.9$ . By Problem 4,

$\displaystyle \frac{y_2}{y_1} =(0.9)^3 = 0.729.$

Therefore, $y_2 = 0.729y_1$ . Calculating the percentage change yields

$\begin{aligned} \frac{y_2 - y_1}{y_1} &= \frac{0.729y_1 - y_1}{y_1} \\ &= \frac{-0.271y_1}{y_1} \\ &= -0.271 \\ &= -27.1\%. \end{aligned}$

Problem 7. Given that y ∝ xⁿ and x increases by r, determine the relative change in y.

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Solution. Since $x$ increases by $r$ , we have

$x_2 = x_1 + rx_1 = (1+r)x_1.$

Therefore, $x_2/x_1 = 1 + r$ . By Problem 4,

$\displaystyle \frac{y_2}{y_1} = (1+r)^n.$

Therefore, $y_2 = (1+r)^n y_1$ . Calculating the relative change yields

$\begin{aligned} \frac{y_2 - y_1}{y_1} &= \frac{(1+r)^n y_1 - y_1}{y_1} \\ &= \frac{((1+r)^n - 1)y_1}{y_1} \\ &= (1+r)^n - 1. \end{aligned}$

Definition 2. The quantity y is said to be inversely proportional to the quantity x if

y ∝ 1/x.

Problem 8. Under Definition 2, show that xy is constant.

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Solution. By definition, there exists $k \neq 0$ such that

$\displaystyle y = k \cdot \frac 1x \quad \Rightarrow \quad xy = k.$

Therefore, $xy$ equals the constant $k$ .

Problem 9. Fix positive integers m, n. Let x, y, z be quantities such that

y is inversely proportional to x^m,
x is inversely proportional to zⁿ.

Show that y is directly proportional to z^α for some integer α.

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Solution. By definition, there exist constants $k_1 \neq 0$ and $k_2 \neq 0$ such that

$\displaystyle y = k_1 \cdot \frac 1{x^m},\quad x = k_2 \cdot \frac 1{z^n}.$

Substituting the second equation into the first,

$\begin{aligned} y &= k_1 \cdot \frac 1{ \left( k_2 \cdot \frac 1{z^n} \right)^m } \\ &= k_1 \cdot \frac 1{ k_2^m \cdot \frac 1{ z^{mn} } } \\ &= k_1 \cdot \frac{ z^{mn} }{ k_2^m } \\ &= \frac{ k_1 }{k_2^m} \cdot z^{mn}.\end{aligned}$