Section 3.6Section 3.6

VariationVariation

Direct VariationDirect Variation

If a situation gives rise to a linear function If a situation gives rise to a linear function ff((xx) ) = = kxkx, or , or yy = = kxkx, where , where kk is a positive constant, is a positive constant, we say that we have we say that we have direct variationdirect variation, or that , or that yy varies directly as varies directly as xx,, or that or that yy is directly is directly proportional to proportional to xx..

The number The number kk is called the is called the variation variation constant,constant, or or constant of proportionalityconstant of proportionality..

Direct VariationDirect Variation

The graph of The graph of yy = = kxkx, , k k > 0, always goes > 0, always goes through the origin and rises from left to right. through the origin and rises from left to right.

As As xx increases, increases, yy increases; that is, the increases; that is, the function is increasing on the interval (0,function is increasing on the interval (0,). ).

The constant The constant kk is also the slope of the line. is also the slope of the line.

Direct VariationDirect Variation ExampleExample: : Find the variation constant and an equation of Find the variation constant and an equation of

variation in which variation in which yy varies directly as varies directly as xx, and , and yy = 42 = 42 when when xx = 3. = 3.

SolutionSolution: We know that (3, 42) is a solution of : We know that (3, 42) is a solution of yy = = kxkx.. yy = = kxkx

4242 = k = k 33

== k k

14 =14 = k k The variation constant 14, is the rate of change of The variation constant 14, is the rate of change of yy with respect to with respect to xx. . The equation of variation is The equation of variation is y y = 14= 14xx..

42

3

ApplicationApplication Example:Example: WagesWages. . A cashier earns an hourly wage. If the cashier worked 18 A cashier earns an hourly wage. If the cashier worked 18

hours and earned $168.30, how much will the cashier earn if hours and earned $168.30, how much will the cashier earn if she works 33 hours? she works 33 hours?

SolutionSolution:: We can express the amount of money earned as a We can express the amount of money earned as a function of the amount of hours worked.function of the amount of hours worked.

ff((hh) = ) = khkh ff(18) = (18) = kk 18 18 $168.30 = $168.30 = kk 18 18 $9.35 = $9.35 = kk The hourly wage is the variation constant. The hourly wage is the variation constant.

Next, we use the equation to find how much the cashier will Next, we use the equation to find how much the cashier will earn if she works 33 hours.earn if she works 33 hours. ff(33) = $9.35(33)(33) = $9.35(33) = $308.55= $308.55

Inverse VariationInverse Variation

If a situation gives rise to a function If a situation gives rise to a function ff((xx) = ) = kk//xx, or , or yy = = kk//x, x, where where kk is a positive constant, is a positive constant, we say that we have we say that we have inverse variationinverse variation, or that , or that yy varies inversely as varies inversely as xx,, or that or that yy is is inversely proportional toinversely proportional to xx..

The number The number kk is called the is called the variation variation constant,constant, or or constant of proportionalityconstant of proportionality..

Inverse VariationInverse Variation

For the graph For the graph yy = = k/xk/x, , kk 0, as 0, as xx increases, increases, yy decreases; that is, the function is decreases; that is, the function is decreasing on the interval (0, decreasing on the interval (0, ).).

Inverse VariationInverse Variation ExampleExample: : Find the variation constant and an equation of variation in Find the variation constant and an equation of variation in

which which yy varies inversely as varies inversely as xx, and , and yy = 22 when = 22 when xx = 0.4. = 0.4.

SolutionSolution::

The variation constant is 8.8. The variation constant is 8.8. The equation of variation is The equation of variation is yy = 8.8/ = 8.8/xx..

220.4

(0.4)22

8.8

ky

xk

k

k

ApplicationApplication ExampleExample: : RoadRoad ConstructionConstruction.. The time The time tt required to do a job varies inversely as the number of required to do a job varies inversely as the number of

people people PP who work on the job (assuming that all work at the who work on the job (assuming that all work at the same rate). If it takes 180 days for 12 workers to complete a same rate). If it takes 180 days for 12 workers to complete a job, how long will it take 15 workers to complete the same job?job, how long will it take 15 workers to complete the same job?

SolutionSolution: : We can express the amount of time required, in We can express the amount of time required, in

days, as a function of the number of people working.days, as a function of the number of people working.

tt varies inversely as varies inversely as PP

This is the variation constant.This is the variation constant.

( )

(12)12

18012

2160

kt P

Pk

t

k

k

Application continuedApplication continuedThe equation of variation is The equation of variation is tt((PP) = 2160 / ) = 2160 / P.P.

Next we computeNext we compute t t(15).(15).

It would take 144 days for 15 people to complete the It would take 144 days for 15 people to complete the same job.same job.

2160( )

2160(15)

15144

t PP

t

t

Combined VariationCombined Variation

Other kinds of variation:Other kinds of variation:

yy varies varies directly as the directly as the nnth power of th power of xx if there if there is some positive constant is some positive constant kk such that . such that .

yy varies varies inversely as the inversely as the nnth power ofth power of xx if if there is some positive constant there is some positive constant kk such that such that . .

yy varies varies jointly as jointly as xx and and zz if there is some if there is some positive constant positive constant kk such that such that yy = = kxzkxz..

ny kx

n

ky

x

ExampleExample The illuminance of a light (The illuminance of a light (EE) varies directly with the intensity () varies directly with the intensity (II) )

of the light and inversely with the square distance (of the light and inversely with the square distance (DD) from the ) from the light. At a distance of 10 feet, a light meter reads 3 units for a light. At a distance of 10 feet, a light meter reads 3 units for a 50-cd lamp. Find the illuminance of a 27-cd lamp at a distance 50-cd lamp. Find the illuminance of a 27-cd lamp at a distance of 9 feet.of 9 feet.

Solve for Solve for kk..

Substitute the second set of Substitute the second set of data into the equation.data into the equation.

The lamp gives an The lamp gives an

illuminance illuminance reading of 2 units.reading of 2 units.

2

2

2

503

106

6 27

92

IE k

D

k

k

E

E