September 11, 2015

2­2 Direct Variation

Definition: direct variation

A direct variation is a function in the form y = kx.

When two variable quantities have a constant (unchanged) ratio, their relationship is called a direct variation.

September 11, 2015

The constant of variation is k, the coefficient of x.

y = kx

y=mx+bk = m = slope = rate of change

An equation is a direct variation if:

• it passes through (0, 0)

• is in the form y = kx.

September 11, 2015

Many direct variation word problems are expressed as:

"...y varies directly as x..."

y varies or changes because it is the dependent variable.

What does a direct variation graph look like?

k=3

enter y = kx into thecalculator

k=­7

k=1/2

k=­7k=3

k=1/2

k=3k=­7

k=1/2

September 11, 2015

Is the Equation a Direct Variation?

If it is, find the constant of variation.5x + 2y = 0

Is the Equation a Direct Variation?

If it is, find the constant of variation.

5x + 2y = 9 Solve for y.

y = ­5/2x +4.5

No, it’s not a direct variation.

It’s not in the form y = kx.

September 11, 2015

Is the Equation a Direct Variation?

If it is, find the constant of variation.

7y = 2x

Yes, it is a direct variation.The constant of variation, k, is

What is the equation of the direct variation passing through (6, 2)?

September 11, 2015

Writing an Equation Given a Point Write an equation of the direct variation that includes the point (4, ­3).

y = kx­3 = k(4)

Substitute the value of k into the original formula.

y = kxy = kx

Real­World Problem Solving

Your distance from lightning varies directly with the time it takes you to hear thunder. If you hear thunder 10 seconds after you see the lightning, you are about 2 miles from the lightning. Write an equation for the relationship between time and distance.

September 11, 2015

Relate:

The distance varies directly with the time. When x = 10, y = 2.

Let x = seconds. (Independent)

Let y = miles. (Dependent)

Real­World Problem SolvingThe force you must apply to lift an object varies directly with the object’s weight. You would need to apply 0.625 lb of force to a windlass to lift a 28­lb weight. How much force would you need to lift 100 lb?

Relate: A force of 0.625 lifts 28 lb. What lifts 100 lb? (You can also use a proportion.)

You need about 2.2 lb of force to lift 100 lb.

September 11, 2015

Practice Problems

1. 8x + 4y = 12 2. 8x + 2y = 0

.

Is each equation a direct variation? If it is, find the constant of variation.

No Yes, k = ­4

Practice Problems

3. (3, 2) and (6, y) 4. (–2, 8) and (x, 12)

The ordered pairs in each exercise are for the same direct variation. Find each missing value.

­34

September 11, 2015

Practice ProblemsI. Charles’s Law states that at constant pressure, the volume of a fixed amount of gas varies directly with its temperature measured in degrees Kelvin. A gas has a volume of 250 mL at 300° K.

a. Write an equation for the relationship between volumeand temperature.

b. What is the volume if the temperature increases to 420°K?

V = (5/6)T; 350mL

Practice ProblemsII. Your percent grade varies directly with the number of correct answers. You got a grade of 80 when you had 20 correct answers.

a. Write an equation for the relationship between percent grade and number of correct answers.

b. What would your percent grade be with 24 correct answers?

G = 4A; 96%

September 11, 2015

Practice ProblemsIII. The amount of simple interest earned in a savings account varies directly with the amount of money in the savings account.You have $1000 in your savings account and earn $50 in simple interest. How much interest would you earn if you had $1500 in your savings account?

$75

Extension...

September 11, 2015

Practice makes perfect...

Assignment Jp. 66 #1­9, 16­19

September 11, 2015