### 6.8

```§ 6.8
Modeling Using Variation
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, 4e – Slide #121
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
1) 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, 4e – Slide #122
Direct Variation
CONTINUED
45 = k(5)
Replace y with 45 and x
with 5.
Divide both sides by 9.
9=k
3) Substitute the value of k into the equation.
y = 9x
Replace k from the equation
from step 1 with 9.
y = 9(13)
y = 117
Replace x from the equation
from step 3 with 13.
Multiply.
Blitzer, Intermediate Algebra, 4e – Slide #123
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
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.
Blitzer, Intermediate Algebra, 4e – Slide #124
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
60 = k360
1/6 = k
Direct variation equation.
Replace M with 60 and E
with 360.
Divide both sides by 360.
3) Substitute the value of k into the equation.
Blitzer, Intermediate Algebra, 4e – Slide #125
Direct Variation
CONTINUED
M = kE
M = (1/6)E
Direct variation equation.
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
M = (1/6)(186)
M = 31
Equation from step 3.
Replace E with 186.
Multiply.
An object that weighs 186 pounds on the earth weighs 31 pounds
on the moon.
Blitzer, Intermediate Algebra, 4e – Slide #126
Direct Variation With Powers
Direct Variations With Powers
y varies directly as the nth power of x if there exists
some nonzero constant k such that
y  kx .
n
Blitzer, Intermediate Algebra, 4e – Slide #127
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
1) Write an equation. We know that y varies directly as the
square of x is expressed as
y  kx .
2
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, 4e – Slide #128
Direct Variation
CONTINUED
d  ks
2
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.
d  ks
2
67 . 5  k  45 
This is the equation from step 1.
2
Replace d with 67.5 and s with 45.
67 . 5  2025 k
Simplify.
0 . 033  k
Divide both sides by 2025.
3) Substitute the value of k into the equation.
2
d  ks
This is the equation from step 1.
2
d  0 . 033 s
Replace k with 0.033.
Blitzer, Intermediate Algebra, 4e – Slide #129
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.
d  0 . 033 s
2
d  0 . 033 60 
This is the equation from step 3.
2
d  0 . 033 3600
d  120
Replace s with 60.

Simplify.
Simplify.
The stopping distance for a car traveling at 60 miles per hour is
120 feet.
Blitzer, Intermediate Algebra, 4e – Slide #130
Inverse Variation
Inverse Variation
If a situation is described by an equation in the form
y
k
x
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, 4e – Slide #131
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 4 . 4  Celsius. What is the water temperature at a
depth of 5000 meters?
SOLUTION
1) Write an equation. We know that y varies inversely as x
is expressed as
y
k
.
x
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, 4e – Slide #132
Inverse Variation
CONTINUED
T 
k
D
2) Use the given values to find k. The temperature of water
in the Pacific Ocean at a depth of 1000 meters is 4 . 4  Celsius.
Substitute 1000 for D and 4.4 for T in the inverse variation
equation. Then solve for k.
T 
k
This is the equation from step 1.
D
k
4 .4 
1000
4400  k
Replace T with 4.4 and D with
1000.
Multiply both sides by 1000.
Blitzer, Intermediate Algebra, 4e – Slide #133
Inverse Variation
CONTINUED
3) Substitute the value of k into the equation.
T 
T 
k
D
4400
D
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.
T 
4400
D
This is the equation from step 3.
Blitzer, Intermediate Algebra, 4e – Slide #134
Inverse Variation
CONTINUED
T 
4400
5000
T  0 . 88
Replace D with 5000.
Simplify.
The temperature of the Pacific Ocean at a depth of 5000 meters
is 4 . 4  Celsius.
Blitzer, Intermediate Algebra, 4e – Slide #135
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 125. 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
y
kx
.
z
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, 4e – Slide #136
Direct & Inverse (Combined) Variation
CONTINUED
I 
kM
C
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.
I 
kM
C
125 
k  25 
20
2500  25 k
100  k
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, 4e – Slide #137
Direct & Inverse (Combined) Variation
CONTINUED
3) Substitute the value of k into the equation.
I 
I 
kM
C
100 M
This is the equation from step 1.
Replace k with 100.
C
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.
I 
80 
100 M
C
100  40 
This is the equation from step 3.
Replace M with 40 and I with 80.
C
Blitzer, Intermediate Algebra, 4e – Slide #138
Direct & Inverse (Combined) Variation
CONTINUED
80 
4000
Multiply.
C
80 C  4000
Multiply both sides by C.
C  50
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, 4e – Slide #139
Joint Variation
Joint Variation
If a situation is described by an equation in the form
y  kxz
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, 4e – Slide #140
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
E  kMV
36  k 8 3 
36  72 k
Translate “Kinetic energy, E,
varies jointly as the mass, M,
and the square of the velocity,
V.”
2
2
If M = 8, V = 3, then E = 36.
Simplify.
Blitzer, Intermediate Algebra, 4e – Slide #141
Joint Variation
CONTINUED
0 .5  k
E  0 . 5 MV
Divide both sides by 72.
Substitute 0.5 for k in the model
for kinetic energy.
2
E  0 . 5  4 6 
2
Find E when M = 4 and V = 6.
E  72
Simplify.
The kinetic energy is 72 ergs.
Blitzer, Intermediate Algebra, 4e – Slide #142
```