Here are the steps to solve problems involving direct variation: 1) Write an equation relating the directly proportional quantities (e.g. y = kx) 2) Use a given point to find the constant of variation k 3) Substitute k back into the original equation 4) Evaluate the equation for the desired quantity For example, to solve problem 1 on page 74: 1) Write the direct variation equation: m = kc 2) Given: m = 12 when c = 3. So, 12 = k(3). Solve for k: k = 4. 3) Substitute k into the original equation: m = 4c 4) Evaluate:

1. Section 2-1
Direct Variation

2. r Varies Directly as c:

3. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c

4. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation:

5. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation: k is a nonzero constant in
y = kxn, and n is a positive integer

6. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation: k is a nonzero constant in
y = kxn, and n is a positive integer
Direct Variation Function:

7. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation: k is a nonzero constant in
y = kxn, and n is a positive integer
Direct Variation Function: A function of the form
y = kxn with k ≠ 0 and n > 0

8. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation: k is a nonzero constant in
y = kxn, and n is a positive integer
Direct Variation Function: A function of the form
y = kxn with k ≠ 0 and n > 0
Can also be known as “directly proportional”

9. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation: k is a nonzero constant in
y = kxn, and n is a positive integer
Direct Variation Function: A function of the form
y = kxn with k ≠ 0 and n > 0
Can also be known as “directly proportional”
***The cost of gas varies directly as the amount of
gas purchased

10. r Varies Directly as c: When r gets larger, so does
c; When r gets smaller, so does c
Constant of Variation: k is a nonzero constant in
y = kxn, and n is a positive integer
Direct Variation Function: A function of the form
y = kxn with k ≠ 0 and n > 0
Can also be known as “directly proportional”
***The cost of gas varies directly as the amount of
gas purchased
The more you get, the more it costs

11. Example 1
Rewrite the statement, “The cost of gas varies
directly as the amount of gas purchased.”

12. Example 1
Rewrite the statement, “The cost of gas varies
directly as the amount of gas purchased.”
“The cost of gas is directly proportional to the
amount of gas purchased.”

13. Example 2
The weight of an object on planet P varies directly
with its weight on Earth E.
a. Write an equation relating P and E.

14. Example 2
The weight of an object on planet P varies directly
with its weight on Earth E.
a. Write an equation relating P and E.
P = kE

15. Example 2
The weight of an object on planet P varies directly
with its weight on Earth E.
a. Write an equation relating P and E.
P = kE
b. Identify the independent and dependent
variables.

16. Example 2
The weight of an object on planet P varies directly
with its weight on Earth E.
a. Write an equation relating P and E.
P = kE
b. Identify the independent and dependent
variables.
Independent: E

17. Example 2
The weight of an object on planet P varies directly
with its weight on Earth E.
a. Write an equation relating P and E.
P = kE
b. Identify the independent and dependent
variables.
Independent: E Dependent = P

18. Example 2
The weight of an object on planet P varies directly
with its weight on Earth E.
a. Write an equation relating P and E.
P = kE
b. Identify the independent and dependent
variables.
Independent: E Dependent = P
k is just a constant

19. Example 3
The ingredients for a pizza and the price are
proportional to its area. This means the quantity of
ingredients is proportional to the square of its radius.
Suppose a pizza 12 in. in diameter costs $7.00. If
the price varies directly as the square of its radius,
what would a pizza 16 in. in diameter cost? What
about an 18 in. pizza?

20. Example 3

21. Example 3
c = cost

22. Example 3
c = cost r = radius

23. Example 3
c = cost r = radius
c = kr2

24. Example 3
c = cost r = radius
c = kr2
7 = k(6)2

25. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36)

26. Example 3
c = cost r = radius
c = kr2
7= k(6)2 7 = k(36) k= 7
36

27. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2

28. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) 2

29. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) =
2 7
36 (64)

30. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) =
2 7
36 (64) ≈ $12.44

31. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) = 2 7
36 (64) ≈ $12.44
c= 7
36 (9) 2

32. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) =
2 7
36 (64) ≈ $12.44
c= 7
36 (9) =
2 7
36 (81)

33. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) =
2 7
36 (64) ≈ $12.44
c= 7
36 (9) =
2 7
36 (81) = $15.75

34. Example 3
c = cost r = radius
c = kr2
7 = k(6)2 7 = k(36) k= 7
36 c= 7
36 r 2
c= 7
36 (8) =
2 7
36 (64) ≈ $12.44
c= 7
36 (9) =
2 7
36 (81) = $15.75
A 16 in. diameter pizza would cost $12.44 and an 18
in. diameter pizza would cost $15.75.

35. Steps to solving a direct variation problem

36. Steps to solving a direct variation problem
1.Write an equation to describe the
variation

37. Steps to solving a direct variation problem
1.Write an equation to describe the
variation
2.Find k

38. Steps to solving a direct variation problem
1.Write an equation to describe the
variation
2.Find k
3.Rewrite the function using k

39. Steps to solving a direct variation problem
1.Write an equation to describe the
variation
2.Find k
3.Rewrite the function using k
4.Evaluate

40. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.

41. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.
y = kx

42. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.
y = kx
32 = k(2)

43. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.
y = kx
32 = k(2)
k = 16

44. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.
y = kx
32 = k(2)
k = 16
y = 16x

45. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.
y = kx
32 = k(2)
k = 16
y = 16x
y = 16(5)

46. Example 4
Find k if y varies directly as x where y = 32 when x = 2.
Then find y when x = 5.
y = kx
32 = k(2)
k = 16
y = 16x
y = 16(5) = 80

47. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.

48. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.
m = kn

49. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.
m = kn
48 = k(12)

50. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.
m = kn
48 = k(12)
k=4

51. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.
m = kn
48 = k(12)
k=4
m = 4n

52. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.
m = kn
48 = k(12)
k=4
m = 4n
m = 4(3)

53. Example 5
m is directly proportional to n. If m = 48 when n = 12,
find m when n = 3.
m = kn
48 = k(12)
k=4
m = 4n
m = 4(3) = 12

54. Homework

55. Homework
p. 74 #1 - 25