int math 2 section 6-8 1011

Section 6-8Direct Variation

Wednesday, April 11, 2012

Essential Question

✦ How do you solve problems involving direct variation and direct square variation functions?

✦ Where you’ll see this:

✦ Finance, biology, physics, business


Vocabulary1. Direct Variation:

2. Constant of Variation:

3. Direct Square Variation:


Vocabulary1. Direct Variation: A function where when one value

gets larger, so does the other; occurs where only multiplication is being applied to the independent variable

2. Constant of Variation:





2. Constant of Variation: A ratio that is found in a direct variation problem by dividing a range value (y) by its corresponding domain value (x)





2. Constant of Variation: A ratio that is found in a direct variation problem by dividing a range value (y) by its corresponding domain value (x)

3. Direct Square Variation: A direct variation problem that produces a parabola due to squaring the independent variable


Direct Variation


Direct Variation

y = kx


Direct Variation

y = kx

k ≠ 0


Direct Variation

y = kx

k ≠ 0

(0, 0)


Direct Variation

y = kx

k ≠ 0

(0, 0)

Line


Example 1A law discovered by Jacques Charles, a French scientist of the late 18th and early 19th centuries, states that for a

fixed amount of gas, the volume varies directly as the temperature. If a gas has a volume of 559 mL at 40°C,

what is the volume at 25°C?





Direct variation:





Direct variation: y = kx





Direct variation: y = kxvolume temperature





Direct variation: y = kxvolume temperaturey = x =






y = kx






y = kx

559 = k(40)






y = kx

559 = k(40)40 40






y = kx

559 = k(40)40 40k = 13.975






y = kx

559 = k(40)40 40k = 13.975

y = 13.975x






y = kx

559 = k(40)40 40k = 13.975

y = 13.975x

y = 13.975(25)






y = kx

559 = k(40)40 40k = 13.975

y = 13.975x

y = 13.975(25)y = 349.375 mL


Direct Square Variation



y = kx2



y = kx2

k ≠ 0



y = kx2

k ≠ 0

(0, 0)



y = kx2

k ≠ 0

(0, 0)

Parabola


Example 2The distance it takes to stop after you have applied the

brakes of a car varies directly as the square of the speed of the car. A car going 64 km/h stops in 27 m. If you are

going 90 km/h, how far will the car travel after you have applied the brakes?





Direct variation:





Direct variation: y = kx2





Direct variation: y = kx2 distance speed





Direct variation: y = kx2 distance speedy = x =





Direct variation: y = kx2 distance speedy = x =27 m = .027 km






y = kx227 m = .027 km






y = kx227 m = .027 km

.027 = k(64)2






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096

k = .000006591796875






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096

k = .000006591796875

k ≈ 6.59 ×10−6






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096

k = .000006591796875

k ≈ 6.59 ×10−6

y ≈ 6.59 ×10−6 x2






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096

k = .000006591796875

k ≈ 6.59 ×10−6

y ≈ 6.59 ×10−6 x2

y ≈ 6.59 ×10−6(90)2






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096

k = .000006591796875

k ≈ 6.59 ×10−6

y ≈ 6.59 ×10−6 x2

y ≈ 6.59 ×10−6(90)2

y ≈ .053379 km






y = kx227 m = .027 km

.027 = k(64)2

.027 = k(4096)4096 4096

k = .000006591796875

k ≈ 6.59 ×10−6

y ≈ 6.59 ×10−6 x2

y ≈ 6.59 ×10−6(90)2

y ≈ .053379 km

y ≈ 53.379 m


Example 3Assume y varies directly as x.

a. When x = 10, y = 15. Find y when x = 15.




y = kx




y = kx

15 = k(10)




y = kx

15 = k(10)

k =

32




y = kx

15 = k(10)

k =

32

y =

32

x




y = kx

15 = k(10)

k =

32

y =

32

x

y =

32

(15)




y = kx

15 = k(10)

k =

32

y =

32

x

y =

32

(15)

y =

452



b. When x = 3, y = 12. Find y when x = 21.




y = kx




y = kx

12 = k(3)




y = kx

12 = k(3)

k = 4




y = kx

12 = k(3)

k = 4

y = 4x




y = kx

12 = k(3)

k = 4

y = 4x

y = 4(21)




y = kx

12 = k(3)

k = 4

y = 4x

y = 4(21)

y = 84


Problem Set


Problem Set

p. 278 #1-23 odd

"The conventional view serves to protect us from the painful job of thinking." – John Kenneth Galbraith


int math 2 section 6-8 1011

Education

direct variation problemthat

constant of variation

direct variationy

direct variationwednesday

adirect variation problem

jacques charles

french scientistof

theindependent variable2