§ 6.8 modeling using variation. blitzer, intermediate algebra, 5e – slide #2 section 6.8...

§ 6.8

Modeling Using Variation

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.8

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.


Direct Variation

Direct VariationIf 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 Problems1) 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.


Direct Variation

EXAMPLEEXAMPLE

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.

SOLUTIONSOLUTION

1) Write an equation. We know that y varies directly as x is expressed as

2) Use the given values to find k.

y = kx.

y = kx Given equation.


Direct Variation

3) Substitute the value of k into the equation.

45 = k(5) Replace y with 45 and x with 5.

CONTINUECONTINUEDD

9 = k Divide both sides by 9.

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.


Direct Variation

EXAMPLEEXAMPLE

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?

SOLUTIONSOLUTION

1) 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.


Direct Variation

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.CONTINUECONTINUE

DD

M = kE Direct variation equation.

60 = k360 Replace M with 60 and E with 360.

1/6 = k Divide both sides by 360.



Direct Variation

CONTINUECONTINUEDD

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 = kE Direct variation equation.

M = (1/6)E Replace k with 1/6.

M = (1/6)E Equation from step 3.

M = (1/6)(186) Replace E with 186.

M = 31

An object that weighs 186 pounds on the earth weighs 31 pounds on the moon.

Multiply.


Direct Variation With Powers

Direct Variations With Powersy varies directly as the nth power of x if there exists some nonzero constant k such that

.nkxy

Next, we consider y varying directly as a power of x:


Direct Variation

EXAMPLEEXAMPLE

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?

SOLUTIONSOLUTION

1) 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.

.2kxy


Direct Variation

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.

2ksd CONTINUECONTINUE

DD

2ksd This is the equation from step 1. 2455.67 k Replace d with 67.5 and s with 45.

k20255.67 Simplify.k033.0 Divide both sides by 2025.

3) Substitute the value of k into the equation.2ksd This is the equation from step 1.

2033.0 sd Replace k with 0.033.


Direct Variation

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.

CONTINUECONTINUEDD

This is the equation from step 3. 2033.0 sd

Replace s with 60. 260033.0d

Simplify. 3600033.0d

Simplify.120d

The stopping distance for a car traveling at 60 miles per hour is 120 feet.


Inverse Variation

Inverse VariationIf 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.

x

ky

And now we consider inverse variation…


Inverse Variation

EXAMPLEEXAMPLE

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?

SOLUTIONSOLUTION

1) 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.

.x

ky

4.4


Inverse Variation

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.

D

kT

CONTINUECONTINUEDD

4.4

D

kT

Replace T with 4.4 and D with 1000.1000

4.4k

Multiply both sides by 1000.k4400


Inverse Variation



CONTINUECONTINUEDD

D

kT

Replace k with 4400.D

T4400

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.D

T4400

Inverse Variation

CONTINUECONTINUEDD

Replace D with 5000.5000

4400T

Simplify.88.0T

The temperature of the Pacific Ocean at a depth of 5000 meters is Celsius.


4.4

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 = kABut by contrast, as the price of the company’s product P increases, the company’s monthly sales decrease. Therefore, monthly sales S vary inverselyas the price P of the product. That is,

P

kS

We can combine these two variation equations into one equation:

P

kAS

This is an example of combined variation.


Direct & Inverse (Combined) Variation

EXAMPLEEXAMPLE

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 125. What is the chronological age of a person with a mental age of 40 and an IQ of 80?

SOLUTIONSOLUTION

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.

.z

kxy



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 125. Substitute 125 for I, 25 for M, and 20 for C.


CONTINUECONTINUEDD

Replace I with 125, M with 25, and C with 20.

C

kMI

C

kMI

20

25125

k

Multiply both sides by 20.k252500

Divide both sides by 25.k100





CONTINUECONTINUEDD

C

kMI

Replace k with 100.C

MI

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.C

MI

100

Replace M with 40 and I with 80. C

4010080



CONTINUECONTINUEDD

When a person has a mental age of 40 and an IQ of 80, his/her chronological age is 50 years old.

Multiply.C

400080

Multiply both sides by C.400080 C

Divide both sides by 80.50C


Joint Variation

Joint VariationIf 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.

kxzy

Finally, we have joint variation.


Joint Variation

EXAMPLEEXAMPLE

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.

SOLUTIONSOLUTIONTranslate “Kinetic energy, E, varies jointly as the mass, M, and the square of the velocity, V.”

2kMVE

If M = 8, V = 3, then E = 36. 23836 k

Simplify.k7236


Joint Variation

25.0 MVE Substitute 0.5 for k in the model for kinetic energy.

CONTINUECONTINUEDD

2645.0E Find E when M = 4 and V = 6.

72E Simplify.

The kinetic energy is 72 ergs.

Divide both sides by 72.k5.0

§ 6.8 modeling using variation. blitzer, intermediate algebra, 5e – slide #2 section 6.8...

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