1. § 6.8
Modeling Using Variation

2. Variation
Certain situations occur so frequently in applied situations that they are given special names. Variation formulas show how one quantity changes in relation to other quantities. Quantities can vary directly, inversely, or jointly.
In this section, we look at situations that can be modeled by each of these kinds of variation.
First, we will consider direct variation.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.8

3. Solving Variation Problems
Direct Variation
Direct Variation
If a situation is described by an equation in the form
y = kx
where k is a constant, we say that y varies directly as x. The number k is called the constant of variation.
Solving Variation Problems
1) Write an equation that describes the given English statement.
2) Substitute the given pair of values into the equation in step 1 and find the value of k.
3) Substitute the value of k into the equation in step 1.
4) Use the equation from step 3 to answer the problem’s question.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.8

4. Direct Variation
EXAMPLE
Use the 4-step procedure for solving the variation problem.
y varies directly as x. y = 45 when x = 5. Find y when x = 13.
SOLUTION
Write an equation. We know that y varies directly as x is expressed as
y = kx.
2) Use the given values to find k.
y = kx
Given equation.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.8

5. Direct Variation 45 = k(5) Replace y with 45 and x with 5. 9 = k
CONTINUED
45 = k(5)
Replace y with 45 and x with 5.
9 = k
Divide both sides by 9.
3) Substitute the value of k into the equation.
y = 9x
Replace k from the equation from step 1 with 9.
4) Answer the problem’s question.
y = 9(13)
Replace x from the equation from step 3 with 13.
y = 117
Multiply.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.8

6. Direct Variation
EXAMPLE
An object’s weight on the moon, M, varies directly as its weight on Earth, E. Neil Armstrong, the first person to step on the moon on July 20, 1969, weighed 360 pounds on Earth (with all of his equipment on) and 60 pounds on the moon. What is the moon weight of a person who weighs 186 pounds on Earth?
SOLUTION
Write an equation. We know that y varies directly as x is expressed as
y = kx.
By changing letters, we can write an equation that describes the following English statement: Moon weight, M, varies directly as Earth weight, E.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.8

7. Direct Variation
CONTINUED
M = kE.
2) Use the given values to find k. We are told that an object with an earth weight of 360 pounds has a moon weight of 60 pounds. Substitute 360 for E and 60 for M in the direct variation equation. Then solve for k.
M = kE
Direct variation equation.
60 = k360
Replace M with 60 and E with 360.
1/6 = k
Divide both sides by 360.
3) Substitute the value of k into the equation.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.8

8. Direct Variation M = kE Direct variation equation. M = (1/6)E
CONTINUED
M = kE
Direct variation equation.
M = (1/6)E
Replace k with 1/6.
4) Answer the problem’s question. How much does an object with a 186 pound earth weight weigh on the moon? Substitute 186 for E in the preceding equation and then solve for M.
M = (1/6)E
Equation from step 3.
M = (1/6)(186)
Replace E with 186.
M = 31
Multiply.
An object that weighs 186 pounds on the earth weighs 31 pounds on the moon.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.8

9. Direct Variations With Powers
Direct Variation With Powers
Next, we consider y varying directly as a power of x:
Direct Variations With Powers
y varies directly as the nth power of x if there exists some nonzero constant k such that
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.8

10. Direct Variation
EXAMPLE
On a dry asphalt road, a car’s stopping distance varies directly as the square of its speed. A car traveling at 45 miles per hour can stop in 67.5 feet. What is the stopping distance for a car traveling at 60 miles per hour?
SOLUTION
Write an equation. We know that y varies directly as the square of x is expressed as
By changing letters, we can write an equation that describes the following English statement: Distance, d, varies directly as the square of the speed, s.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.8

11. Direct Variation
CONTINUED
2) Use the given values to find k. We are told that a car traveling at 45 miles per hour can stop in 67.5 feet. Substitute 45 for s and 67.5 for d. Then solve for k.
This is the equation from step 1.
Replace d with 67.5 and s with 45.
Simplify.
Divide both sides by 2025.
3) Substitute the value of k into the equation.
This is the equation from step 1.
Replace k with
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.8

12. Direct Variation
CONTINUED
4) Answer the problem’s question. What is the stopping distance for a car traveling at 60 miles per hour? Substitute 60 for s in the preceding equation.
This is the equation from step 3.
Replace s with 60.
Simplify.
Simplify.
The stopping distance for a car traveling at 60 miles per hour is 120 feet.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.8

13. Inverse Variation Inverse Variation
And now we consider inverse variation…
Inverse Variation
If a situation is described by an equation in the form
where k is a constant, we say that y varies inversely as x. The number k is called the constant of variation.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.8

14. Inverse Variation
EXAMPLE
The water temperature of the Pacific Ocean varies inversely as the water’s depth. At a depth of 1000 meters, the water temperature is Celsius. What is the water temperature at a depth of 5000 meters?
SOLUTION
Write an equation. We know that y varies inversely as x is expressed as
By changing letters, we can write an equation that describes the following English statement: Temperature, T, varies inversely as the water’s depth, D.
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.8

15. Inverse Variation
CONTINUED
2) Use the given values to find k. The temperature of water in the Pacific Ocean at a depth of 1000 meters is Celsius. Substitute 1000 for D and 4.4 for T in the inverse variation equation. Then solve for k.
This is the equation from step 1.
Replace T with 4.4 and D with 1000.
Multiply both sides by 1000.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.8

16. Inverse Variation 3) Substitute the value of k into the equation.
CONTINUED
3) Substitute the value of k into the equation.
This is the equation from step 1.
Replace k with 4400.
4) Answer the problem’s question. We need to find the temperature of the water at a depth of 5000 meters. Substitute 5000 for D.
This is the equation from step 3.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.8

17. Inverse Variation Replace D with 5000. Simplify.
CONTINUED
Replace D with 5000.
Simplify.
The temperature of the Pacific Ocean at a depth of 5000 meters is Celsius.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.8

18. Combined Variation
In combined variation, direct and inverse variation occur at the same time.
For example, as the advertising budget, A, of a company increases, its monthly
Sales, S, also increase. Then S increases directly as A and
S = kA
But by contrast, as the price of the company’s product P increases, the
company’s monthly sales decrease. Therefore, monthly sales S vary inversely
as the price P of the product. That is,
We can combine these two variation equations into one equation:
This is an example of combined variation.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 6.8

19. Direct & Inverse (Combined) Variation
EXAMPLE
One’s intelligence quotient, or IQ, varies directly as a person’s mental age and inversely as that person’s chronological age. A person with a mental age of 25 and a chronological age of 20 has an IQ of What is the chronological age of a person with a mental age of 40 and an IQ of 80?
SOLUTION
1) Write an equation. We know that y varies directly as x and inversely as z is expressed as
By changing letters, we can write an equation that describes the following English statement: IQ, I, varies directly as the mental age, M, and inversely as chronological age, C.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 6.8

20. Direct & Inverse (Combined) Variation
CONTINUED
2) Use the given values to find k. A person with a mental age of 25 and a chronological age of 20 has an IQ of Substitute 125 for I, 25 for M, and 20 for C.
This is the equation from step 1.
Replace I with 125, M with 25, and C with 20.
Multiply both sides by 20.
Divide both sides by 25.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 6.8

21. Direct & Inverse (Combined) Variation
CONTINUED
3) Substitute the value of k into the equation.
This is the equation from step 1.
Replace k with 100.
4) Answer the problem’s question. We need to determine the chronological age of a person who has a mental age of 40 and an IQ of 80. Substitute 40 for M and 80 for I.
This is the equation from step 3.
Replace M with 40 and I with 80.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 6.8

22. Direct & Inverse (Combined) Variation
CONTINUED
Multiply.
Multiply both sides by C.
Divide both sides by 80.
When a person has a mental age of 40 and an IQ of 80, his/her chronological age is 50 years old.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.8

23. Joint Variation Joint Variation Finally, we have joint variation.
If a situation is described by an equation in the form
where k is a constant, we say that y varies jointly as x and z. The number k is called the constant of variation.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.8

24. Joint Variation
EXAMPLE
Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 8 grams and velocity of 3 centimeters per second has a kinetic energy of 36 ergs. Find the kinetic energy for a mass of 4 grams and velocity of 6 centimeters per second.
SOLUTION
Translate “Kinetic energy, E, varies jointly as the mass, M, and the square of the velocity, V.”
If M = 8, V = 3, then E = 36.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.8

25. Joint Variation Divide both sides by 72.
CONTINUED
Divide both sides by 72.
Substitute 0.5 for k in the model for kinetic energy.
Find E when M = 4 and V = 6.
Simplify.
The kinetic energy is 72 ergs.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.8