Lesson 8.9: Use Direct and Inverse Variation

New Concept

This lesson explores how quantities are proportionally related. You'll master two types: direct variation, where values change together ($y=kx$), and inverse variation, where one increases as the other decreases ($y=k/x$).

What’s next

Now, you'll apply these formulas. Get ready for interactive examples and practice problems to master solving for the constant of variation, $k$.

Property

For any two variables $x$ and $y$, $y$ varies directly with $x$ if

The constant $k$ is called the constant of variation. When two quantities are related by a proportion, we say they are proportional to each other.

To solve direct variation problems:

1. Write the formula for direct variation: $y = kx$.
2. Substitute the given values for the variables.
3. Solve for the constant of variation, $k$.
4. Write the equation that relates $x$ and $y$ using the value of $k$.

Examples

• If $y$ varies directly with $x$, and $y=45$ when $x=9$, find the equation. We use $y=kx$, so $45=k(9)$, which gives $k=5$. The equation is $y=5x$.
• The cost of juice, $C$, varies directly with the number of bottles, $n$. If 4 bottles cost 12 dollars, how much would 7 bottles cost? The relation is $C=kn$. We find $k$ from $12=k(4)$, so $k=3$. The equation is $C=3n$. For 7 bottles, the cost is $C=3(7)=21$ dollars.
• The distance, $d$, an ant crawls varies directly with time, $t$. If it crawls 120 cm in 3 minutes, how far can it crawl in 10 minutes? The formula is $d=kt$. Substituting gives $120=k(3)$, so $k=40$. The equation is $d=40t$. In 10 minutes, it crawls $d=40(10)=400$ cm.

Property

When one variable varies directly with the square of another variable, the equation of direct variation is:

The process to solve is the same as linear direct variation, but you must square the $x$ variable before multiplying by the constant of variation, $k$.

Examples

• The distance, $d$, an object falls varies directly with the square of the time, $t$. A rock falls 125 meters in 5 seconds. Write the equation relating $d$ and $t$. We use $d=kt^2$. So, $125=k(5^2)$, which gives $125=25k$, and $k=5$. The equation is $d=5t^2$.
• The area, $A$, of a circle varies directly as the square of the radius, $r$. A circle with a radius of 10 cm has an area of $100\pi$ cm$^2$. What is the area of a circle with a radius of 3 cm? From $A=kr^2$, we get $100\pi = k(10^2)$, so $k=\pi$. The equation is $A=\pi r^2$. For $r=3$, the area is $A=\pi(3^2)=9\pi$ cm$^2$.
• The power, $P$, generated by a windmill varies directly with the square of the wind speed, $w$. A 10 mph wind generates 200 watts. How much power is generated by a 15 mph wind? The formula is $P=kw^2$. So, $200=k(10^2)$, which gives $k=2$. The equation is $P=2w^2$. For a 15 mph wind, $P=2(15^2)=450$ watts.

Explanation

In this relationship, one quantity grows much faster than the other. As one variable increases, the other increases by the square of that amount. It's an accelerated growth, like how the area of a square skyrockets when you increase its side length.

Property

For any two variables $x$ and $y$, $y$ varies inversely with $x$ if

The constant $k$ is called the constant of variation. The word ‘inverse’ in inverse variation refers to the multiplicative inverse, as the equation can be seen as $y = k \cdot \frac{1}{x}$.

To solve inverse variation problems:

1. Write the formula for inverse variation: $y = \frac{k}{x}$.
2. Substitute the given values for the variables.
3. Solve for the constant of variation, $k$.
4. Write the equation that relates $x$ and $y$.

Examples

• If $y$ varies inversely with $x$, and $y=6$ when $x=5$, find the equation relating them. Using $y=\frac{k}{x}$, we substitute to get $6=\frac{k}{5}$, which means $k=30$. The equation is $y=\frac{30}{x}$.
• The time, $t$, it takes to drive a certain distance varies inversely with speed, $s$. If the trip takes 3 hours at 60 mph, how long would it take at 90 mph? The relation is $t=\frac{k}{s}$. We find $k$ from $3=\frac{k}{60}$, so $k=180$. The equation is $t=\frac{180}{s}$. At 90 mph, the time is $t=\frac{180}{90}=2$ hours.
• The number of workers, $w$, needed to finish a job varies inversely with the number of days, $d$. If 4 workers can finish in 9 days, how many workers are needed to finish in 3 days? The formula is $w=\frac{k}{d}$. We find $k$ from $4=\frac{k}{9}$, so $k=36$. The equation is $w=\frac{36}{d}$. For 3 days, we need $w=\frac{36}{3}=12$ workers.