The document contains examples and explanations of combined and joint variation. It begins with four warm-up problems that demonstrate combined variation, where the dependent variable varies directly with one independent variable and inversely with another. It explains that to find the constant of variation k, one variable is held constant while solving for k. The document then provides an example of finding k and solving a combined variation word problem. It also defines and provides an example of joint variation, where the dependent variable varies directly with two or more independent variables. It explains finding k is done in the same way as in combined variation problems.

1. Section 2-9
Combined and Joint Variation

2. Warm-up
Solve for k.
2 2
k(2)(2) k(3)(2)
1. 53 = 2. 80 =
10 10
2 2
k(4)(2) k(5)(2)
3. 107 = 4. 132.5 =
10 10

3. Warm-up
Solve for k.
2 2
k(2)(2) k(3)(2)
1. 53 = 2. 80 =
10 10
k = 66.25
2 2
k(4)(2) k(5)(2)
3. 107 = 4. 132.5 =
10 10

4. Warm-up
Solve for k.
2 2
k(2)(2) k(3)(2)
1. 53 = 2. 80 =
10 10
k = 66.25 k = 66 2/3
2 2
k(4)(2) k(5)(2)
3. 107 = 4. 132.5 =
10 10

5. Warm-up
Solve for k.
2 2
k(2)(2) k(3)(2)
1. 53 = 2. 80 =
10 10
k = 66.25 k = 66 2/3
2 2
k(4)(2) k(5)(2)
3. 107 = 4. 132.5 =
10 10
k = 66.875

6. Warm-up
Solve for k.
2 2
k(2)(2) k(3)(2)
1. 53 = 2. 80 =
10 10
k = 66.25 k = 66 2/3
2 2
k(4)(2) k(5)(2)
3. 107 = 4. 132.5 =
10 10
k = 66.875 k = 66.25

7. Question
What do you notice about the 4 warm-up problems?
What does it mean?

8. Question
What do you notice about the 4 warm-up problems?
What does it mean?
More than one type of variation is going on!

9. Combined Variation

10. Combined Variation
When direct and inverse variation both occur in a problem

11. Combined Variation
When direct and inverse variation both occur in a problem
kx means y varies directly as x and inversely as z
y=
z

12. Example 1
A baseball pitcher’s ERA varies directly as the number of earned
runs allowed and inversely as the number of innings pitched. Write a
model for the situation.

13. Example 1
A baseball pitcher’s ERA varies directly as the number of earned
runs allowed and inversely as the number of innings pitched. Write a
model for the situation.
E = ERA; R = runs; I = innings

14. Example 1
A baseball pitcher’s ERA varies directly as the number of earned
runs allowed and inversely as the number of innings pitched. Write a
model for the situation.
E = ERA; R = runs; I = innings
kR
E=
I

15. Question
How do you find k in a combined variation problem?

16. Question
How do you find k in a combined variation problem?
Keep all but one of the independent variables constant (like we
saw in the warm-up). Choose a value for the other independent
variables and plug in.

17. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?

18. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR
E=
I

19. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72)
E= 2.56 =
I 253

20. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72

21. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72
k = 8.996

22. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72
8.996R
k = 8.996 E =
I

23. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72
8.996R 8.996R
k = 8.996 E = 2.56 =
I 300

24. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72
8.996R 8.996R
k = 8.996 E = 2.56 =
I 300
300 8.996R 300
(2.56) = i
8.996 300 8.996

25. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72
8.996R 8.996R
k = 8.996 E = 2.56 =
I 300
300
(2.56) =
8.996R 300
i R = 85.37
8.996 300 8.996

26. Example 2
Matt Mitarnowski had an ERA of 2.56, having given up 72
earned runs in 253 innings. How many earned runs would he have
given up if he had pitched 300 innings if his ERA was the same?
kR k(72) 253 k(72) 253
E= 2.56 = (2.56) = i
I 253 72 253 72
8.996R 8.996R
k = 8.996 E = 2.56 =
I 300
300
(2.56) =
8.996R 300
i R = 85.37
8.996 300 8.996
Matt would have given up 85 runs

27. Joint Variation

28. Joint Variation
When the dependent variable varies directly as two or more
independent variables, but there is no inverse variation

29. Joint Variation
When the dependent variable varies directly as two or more
independent variables, but there is no inverse variation
y = kxz 3
means y varies directly as x and the cube of z

30. Question
How do you find k in a joint variation problem?

31. Question
How do you find k in a joint variation problem?
The same as we did in a combined variation problem

32. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.

33. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance

34. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 2

35. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 1500 = k(15) (10)
2 2

36. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 1500 = k(15) (10) k = 3
2 2 2

37. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 1500 = k(15) (10) k = 3
2 2 2
P= 2
3
2
cr

38. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 1500 = k(15) (10) k = 3
2 2 2
P= 2
3
2
cr P= 2
3
2
(20) (21)

39. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 1500 = k(15) (10) k = 3
2 2 2
P= 2
3
2
cr P= 2
3
2
(20) (21) P = 5600

40. Example 3
The power in an electric circuit varies jointly as the square of the
current and the resistance. The power is 1500 watts when the current
is 15 amps and the resistance is 10 ohms. Find the power when the
current is 20 amps and the resistance is 21 ohms.
P = Power; c = current; r = resistance
P = kc r 1500 = k(15) (10) k = 3
2 2 2
P= 2
3
2
cr P= 2
3
2
(20) (21) P = 5600
The power is 5600 watts

41. Homework

42. Homework
p. 125 #1 - 22