Lesson 5.8 : Modeling Using Variation

New Concept

Variation models how quantities change in relation to one another. You will learn to solve problems involving direct ($y=kx^n$), inverse ($y=\frac{k}{x^n}$), and joint variation by identifying the constant of variation, $k$.

What’s next

Next, you'll tackle interactive examples for each variation type. Then, you'll apply these skills in a series of challenge problems to solidify your understanding.

Property

If $x$ and $y$ are related by an equation of the form

then we say that the relationship is direct variation and $y$ varies directly with, or is proportional to, the $n$th power of $x$. In direct variation relationships, there is a nonzero constant ratio $k = \dfrac{y}{x^n}$, where $k$ is called the constant of variation, which helps define the relationship between the variables.

Examples

• The variable $a$ varies directly with the square of $b$. If $a=32$ when $b=4$, we find $k$ using $a=kb^2$, so $32=k(4^2)$ which gives $k=2$. To find $a$ when $b=5$, we calculate $a=2(5^2)=50$.

• A person's weekly earnings $E$ vary directly with the hours $h$ they work. If they earn 300 dollars for 20 hours, then $E=kh$ gives $300=k(20)$, so $k=15$. For 35 hours, they would earn $E=15(35)=525$ dollars.

Property

If $x$ and $y$ are related by an equation of the form

where $k$ is a nonzero constant, then we say that $y$ varies inversely with the $n$th power of $x$. In inversely proportional relationships, or inverse variations, there is a constant multiple $k = x^n y$.

Examples

• The time $t$ to complete a journey varies inversely with speed $s$. If a trip takes 4 hours at 60 mph, we find $k$ from $t=k/s$, so $4=k/60$, which gives $k=240$. At 80 mph, the trip would take $t=240/80=3$ hours.

• The intensity of a signal $I$ varies inversely with the square of the distance $d$. If the intensity is 100 units at 3 meters, then $I=k/d^2$ gives $100=k/3^2$, so $k=900$. At 10 meters, the intensity is $I=900/10^2=9$ units.

Property

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if $x$ varies directly with both $y$ and $z$, we have $x = kyz$. If $x$ varies directly with $y$ and inversely with $z$, we have $x = \dfrac{ky}{z}$. Notice that we only use one constant in a joint variation equation. To solve, first write the equation showing the relationship, then use a complete set of values to find the constant $k$. Finally, use the full equation to find the missing value.

Examples

• The variable $A$ varies jointly with $b$ and $h$. If $A=20$ when $b=5$ and $h=8$, we find $k$ from $A=kbh$, so $20=k(5)(8)$, which gives $k=0.5$. To find $A$ when $b=6$ and $h=10$, we calculate $A=0.5(6)(10)=30$.