Section 2-9Combined and Joint Variation

Warm-upSolve for k.

1. 53 = k(2)(2)2

102. 80 =

k(3)(2)2

10

3. 107 =k(4)(2)2

104. 132.5 =

k(5)(2)2

10

Warm-upSolve for k.

1. 53 = k(2)(2)2

102. 80 =

k(3)(2)2

10

3. 107 =k(4)(2)2

104. 132.5 =

k(5)(2)2

10

k = 66.25

Warm-upSolve for k.

1. 53 = k(2)(2)2

102. 80 =

k(3)(2)2

10

3. 107 =k(4)(2)2

104. 132.5 =

k(5)(2)2

10

k = 66.25 k = 66 2/3

Warm-upSolve for k.

1. 53 = k(2)(2)2

102. 80 =

k(3)(2)2

10

3. 107 =k(4)(2)2

104. 132.5 =

k(5)(2)2

10

k = 66.25 k = 66 2/3

k = 66.875

Warm-upSolve for k.

1. 53 = k(2)(2)2

102. 80 =

k(3)(2)2

10

3. 107 =k(4)(2)2

104. 132.5 =

k(5)(2)2

10

k = 66.25 k = 66 2/3

k = 66.875 k = 66.25

QuestionWhat do you notice about the 4 warm-up problems?

What does it mean?

QuestionWhat do you notice about the 4 warm-up problems?

What does it mean?

More than one type of variation is going on!

Combined Variation

Combined Variation

When direct and inverse variation both occur in a problem

Combined Variation

When direct and inverse variation both occur in a problem

y =kxz

means y varies directly as x and inversely as z

Example 1A 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.

Example 1A 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

Example 1A 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

E =kRI

QuestionHow do you find k in a combined variation problem?

QuestionHow 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.

Example 2Matt 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?

Example 2Matt 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?E =

kRI

Example 2Matt 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?E =

kRI

2.56 = k(72)253

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

k = 8.996

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

k = 8.996 E =8.996R

I

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

k = 8.996 E =8.996R

I2.56 = 8.996R

300

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

k = 8.996 E =8.996R

I2.56 = 8.996R

300

3008.996

(2.56) = 8.996R300

i3008.996

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

k = 8.996 E =8.996R

I2.56 = 8.996R

300

3008.996

(2.56) = 8.996R300

i3008.996

R = 85.37

Example 2Matt 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?E =

kRI

2.56 = k(72)253

25372(2.56) = k(72)

253i25372

k = 8.996 E =8.996R

I2.56 = 8.996R

300

3008.996

(2.56) = 8.996R300

i3008.996

R = 85.37

Matt would have given up 85 runs

Joint Variation

Joint Variation

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

Joint Variation

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

y = kxz3 means y varies directly as x and the cube of z

QuestionHow do you find k in a joint variation problem?

QuestionHow do you find k in a joint variation problem?

The same as we did in a combined variation problem

Example 3The 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.

Example 3The 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

Example 3The 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 = kc2r

Example 3The 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 = kc2r 1500 = k(15)2 (10)

Example 3The 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 = kc2r 1500 = k(15)2 (10) k = 23

Example 3The 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 = kc2r 1500 = k(15)2 (10) k = 23

P = 23c2r

Example 3The 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 = kc2r 1500 = k(15)2 (10) k = 23

P = 23c2r P = 2

3 (20)2 (21)

Example 3The 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 = kc2r 1500 = k(15)2 (10) k = 23

P = 23c2r P = 2

3 (20)2 (21) P = 5600

Example 3The 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 = kc2r 1500 = k(15)2 (10) k = 23

P = 23c2r P = 2

3 (20)2 (21) P = 5600The power is 5600 watts

Homework

Homework

p. 125 #1 - 22