Lesson 5: Inverse Variation

New Concept

This lesson introduces inverse variation, a relationship where one quantity decreases as another increases, following the form $y = k/x$. You will learn to identify this pattern, find its formula, and solve real-world application problems.

What’s next

Now, let's put this concept into action. You'll analyze worked examples and then tackle a sequence of practice cards to build your problem-solving skills.

Property

$y$ varies inversely with $x$ if

where $k$ is a positive constant. This relationship implies that the product of the variables is constant: $xy = k$.

Property

If you know that two variables vary inversely and have one corresponding pair of values, you can find the constant of variation, $k$. Given a point $(x_1, y_1)$, you can calculate $k = x_1 y_1$ and write the specific formula $y = \frac{k}{x}$.

Examples

• The current, $I$, in a circuit varies inversely with resistance, $R$. An iron with 12 ohms of resistance draws 10 amps. First, find $k = I \cdot R = 10 \cdot 12 = 120$. The formula is $I = \frac{120}{R}$.

• Using the formula $I = \frac{120}{R}$, how much current is drawn by a device with 20 ohms of resistance? Substitute $R=20$ to get $I = \frac{120}{20} = 6$ amps.

Property

$y$ varies inversely with $x^n$ if

where $k$ and $n$ are positive constants. This means the product $yx^n$ is constant.

Property

If $y$ varies inversely with $x^n$ (so $y = \frac{k}{x^n}$), and you multiply $x$ by a factor of $c$, the new value of $y$ is the old value multiplied by $\frac{1}{c^n}$.
For $y = \frac{k}{x}$, multiplying $x$ by $c$ changes $y$ to $\frac{1}{c}y$.
For $y = \frac{k}{x^2}$, multiplying $x$ by $c$ changes $y$ to $\frac{1}{c^2}y$.

Examples

• The force $F$ needed to loosen a bolt varies inversely with wrench length $l$, so $F = \frac{k}{l}$. If you double the length of the wrench ($l \to 2l$), the force required is halved ($F \to \frac{1}{2}F$).

• An object's weight $w$ varies inversely with the square of its distance $d$ from Earth's center ($w = \frac{k}{d^2}$). If a satellite moves to an orbit 3 times farther away, its weight becomes $\frac{1}{3^2} = \frac{1}{9}$ of its original weight.