Many formulas in science say how one quantity changes with another. Variation is the language for the simplest of these. You will rearrange formulas as in Solving for a Variable and recognize the shapes of the Parent Functions $y = x$ and $y = \frac{1}{x}$.
Key idea
Direct: $y = kx$ ($y$ doubles when $x$ doubles). Inverse: $y = \frac{k}{x}$ ($y$ halves when $x$ doubles). Joint: $y = kxz$. "Inversely as the square" means $y = \frac{k}{x^2}$. In every case, use one known pair to find $k$, then use $k$ to answer the question.
Why it works
Direct variation means the ratio $\frac{y}{x}$ stays fixed, so $k$ is that ratio, and the graph is a line through the origin. Inverse variation means the product $xy$ stays fixed, so $k$ is that product, and the graph is a stretched copy of $\frac{1}{x}$. Once you know $k$, the relationship is completely known.
Worked examples
Worked example: Inverse variation
The time to drive a fixed distance varies inversely with speed. At 60 mph it takes 3 hours. How long at 45 mph?
- $t = \frac{k}{s}$, so $k = 60 \cdot 3 = 180$ (the distance).
- $t = \frac{180}{45} = 4$.
Answer: 4 hours
Worked example: Inverse square
Light intensity varies inversely with the square of distance. At 2 m it is 90 units. Find it at 6 m.
- $I = \frac{k}{d^2}$, so $k = 90 \cdot 4 = 360$.
- $I = \frac{360}{36} = 10$. (Triple the distance, one ninth the light.)
Answer: 10 units
Worked example: Joint variation
$y$ varies jointly with $x$ and $z$, and $y = 60$ when $x = 3$ and $z = 4$. Find $y$ when $x = 5$ and $z = 2$.
- $60 = k \cdot 3 \cdot 4$, so $k = 5$.
- $y = 5 \cdot 5 \cdot 2 = 50$.
Answer: $50$
Common mistake
Using the direct-variation idea for an inverse problem: "45 is less than 60, so the time is less." Ask which way the quantities move. For inverse variation the product stays the same: $60 \cdot 3 = 45 \cdot t$.