Applying Direct Variation

Direct variation can be used to model real-world situations.

Many real-world situations can be modeled by direct variation. In geometry, the perimeter $P$ of a square is proportional to the side length $s$ of the square by the formula $P=4s$ . The formula $d=rt$ describes how distance $d$ is proportional to time $t$ by the constant of proportionality $r$ , which represents a constant rate of travel. When shopping for several same-price items, the total cost $C$ is proportional to the number of items $n$ by the price $p$ of each item, or $C = pn$ .

Sales tax is another example of direct variation. The amount of sales tax on an item is proportional to the price. If the sales tax rate is

r

, then the amount of sales tax on an item with price

x

rx

. The total cost

y

of the item with tax is:

\begin{aligned}y&=x + rx\\y&=(1+r)x\end{aligned}

Another application of direct variation is converting units, such as distance, volume, or currency. The constant of variation is the conversion factor.

Applying Direct Variation to Sales Tax

Graph and then determine the exact total cost of an item with a price of $5 after a sales tax of 10%.

The total cost

y

of an item with price

x

after a sales tax rate of

r

is represented by the sales tax equation:

y=(1 + r)x

For a 10% sales tax, the value of

r

is 0.10, which can also be written as 0.1. Substitute 0.1 for

r

\begin{aligned}y &= (1 + 0.1)x\\y &= 1.1x\end{aligned}

Determine two points on the line of the equation.

The graph of a direct variation passes through the point

(0, 0)

and

(1, k)

, where

k

is the constant of proportionality. In this situation,

k

is 1.1.

\begin{aligned} y&=kx\\y&=1.1x\end{aligned}

So, the line of the equation goes through the points

(0, 0)

and

(1, 1.1)

Plot the points and determine the slope, or

k

, of the equation. The graph of the equation goes through the origin with a slope of 1.1.

From the graph, when the price of the item

x

is $5, the total cost

y

after tax is between $5 and $6. Substitute 5 for

x

in the equation to identify the exact total cost.

\begin{aligned}y &= 1.1x\\ &=1.1(5)\\&=5.5\end{aligned}

The exact total cost is $5.50.

Applying Direct Variation to Currency

Currency in U.S. dollars varies directly with the value in other currencies. On a given day, 8 euros are exchanged for 9.25 U.S. dollars. How many euros would be exchanged for 50 U.S. dollars?

Write the direct variation equation. Let

y

represent the number of euros and

x

represent the number of U.S. dollars.

y=kx

Substitute the given values of

x

and

y

into the equation, and solve for

k

\begin{aligned}8&=k(9.25)\\ \frac{8}{9.25}&=k\\ 0.865&=k\end{aligned}

This means that the exchange rate is about 0.865 euro per U.S. dollar.

Substitute the value of

k

into the direct variation equation.

y=0.865x

The number of U.S. dollars is given, which is represented by

x

. Substitute 50 for the value of

x

into the equation.

\begin{aligned}y&=0.865(50)\\y&=43.25\end{aligned}

At the given exchange rate, 43.25 euros would be exchanged for 50 U.S. dollars.

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Applying Direct Variation