§ 6.8
Modeling Using Variation
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.
y  kx
y varies directly as x
y  kx .
n
y
k
varying directly as a power of x
y varies inversely as x
x
y  kxz
y varies jointly as x and z
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.8
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
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. (For example: Purchase 5 pounds of steak for $45.)
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
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.8
Given equation.
Direct Variation
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.
For example: $9 a
pound.
y = 9x
Replace k from the equation from step
1 with 9.
4) Answer the problem’s question.
For example: How much will it cost
for 13 pounds?
y = 9(13)
Replace x from the equation from step
3 with 13.
Will cost
y = 117
Multiply.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.8
$117.
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, 5e – Slide #6 Section 6.8
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, 5e – Slide #7 Section 6.8
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, 5e – Slide #8 Section 6.8
Direct Variation
Check Point 1 on page 465
The number of gallons of water, W used when taking a shower
varies directly as the time, t, in minutes, in the shower. A shower
lasting 5 minutes uses 30 gallons of water. How much water is used
in a shower lasting 11 minutes?
SOLUTION
1) Write an equation. We know that y varies directly as x is
expressed as
y = kx.
W = kt.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.8
Direct Variation
CONTINUED
2) Use the given values to find k. A shower lasting 5 minutes uses
30 gallons of water. Use the given values to find k.
W = kt
30 = k5
6=k
Direct variation equation.
Replace W with 30 and t
with 5.
Divide both sides by 5.
3) Substitute the value of k into the equation.
W = 6t
Replace k with 6.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.8
Direct Variation
CONTINUED
4) Answer the problem’s question. Find W when t is 11 minutes.
W = 6t
W = (6)(11)
Equation from step 3.
Replace t with 11.
W = 66
Multiply.
An 11 minute shower will use 66 gallons of water.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.8
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
y  kx .
n
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.8
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, 5e – Slide #13 Section 6.8
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, 5e – Slide #14 Section 6.8
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, 5e – Slide #15 Section 6.8
Direct Variation
Check Point 2 on page 467
The distance required to stop a car varies directly as the square of its speed. If it
requires 200 feet to stop a car traveling 60 miles per hour, how many feet are
required to stop a car traveling 100 miles per hour?
SOLUTION
1) Write an equation.
y  kx .
2
s  kv
2
By changing letters, we can write an equation that describes the following
English statement: Stopping distance, s, varies directly as the square of the
speed of the car, v.
2


200

k
60
2) Use the given values to find k.
200  k  3600
200
3600
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.8
k
k 
1
18
Direct Variation
CONTINUED
3) Substitute the value of k into the equation.
s  kv
s
1
2
This is the equation from step 1.
v
2
18
4) Answer the problem’s question: find s when v=100.
s
100
2
Replace v with 100.
18
s  556
Simplify.
About 556 feet are required to stop a car traveling 100 miles per hour.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.8
Inverse Variation
And now we consider 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, 5e – Slide #18 Section 6.8
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, 5e – Slide #19 Section 6.8
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, 5e – Slide #20 Section 6.8
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, 5e – Slide #21 Section 6.8
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 . 88  Celsius.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.8
Inverse Variation
Check Point 3 on page 468
The length of a violin string varies inversely as the frequency of its vibrations. A
violin string 8 inches long vibrates at a frequency of 640 cycles per second. What is
the frequency of a 10-inch string?
SOLUTION
1) Write an equation. We know that y varies inversely as x is expressed as
y
k
f 
.
k
x
l
By changing letters, we can write an equation that describes the following English
statement: Frequency, f. varies inversely as the water’s depth, D.
2) Use the given values to find k.
640 
k
5120  k
8
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.8
Inverse Variation
CONTINUED
3) Substitute the value of k into the equation.
f 
k
This is the equation from step 1.
l
f 
5120
l
4) Answer the problem’s question: find f when l=10.
f 
5120
10
f  512
Replace v with 100.
Simplify.
A string length of 10 inches will vibrate at 512 cycles per second.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.8
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,
k
S 
P
We can combine these two variation equations into one equation:
S 
kA
P
This is an example of combined variation.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.8
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, 5e – Slide #26 Section 6.8
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, 5e – Slide #27 Section 6.8
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, 5e – Slide #28 Section 6.8
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, 5e – Slide #29 Section 6.8
Direct & Inverse (Combined) Variation
Check Point 4 on page 470
The number of minutes needed to solve an exercise set of variation problems varies
directly as the number of problems and inversely as the number of people working to
solve the problems. It takes 4 people 32 minutes to solve 16 problems. How many
minutes will it tak 8 people to solve 24 problems?
SOLUTION
1) Write an equation. y varies directly as x and inversely as z
m=the number of minutes needed to solve an exercise set (32)
p =the number of people working on the problems (4)
x= the number if problems in the exercise set (16)
2) Use the given values to find k.
32 
k 16
32  4 k
4
k 8
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 6.8
y
kx
z
m 
kx
p
.
Direct & Inverse (Combined) Variation
CONTINUED
3) Substitute the value of k into the equation.
m 
kx
.
z
m
8x
This is the equation from step 1.
Replace k with 8
p
4) Answer the problem’s question. Find m when p=8 and x = 24
m 
8  24 
Replace M with 40 and I with 80.
8
m  24
It will take 24 minutes for 8 people to solve 24 problems.
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 6.8
Joint Variation
Finally, we have 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, 5e – Slide #32 Section 6.8
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, 5e – Slide #33 Section 6.8
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
E  72
Find E when M = 4 and V = 6.
Simplify.
The kinetic energy is 72 ergs.
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 6.8
Joint Variation
Check Point on page 471
The volume of a cone, V, varies jointly as its height, h, ad the square of its radius r. A
cone with a radius measuring 6 feet and a height measuring 10 feet has a volume of
120pi cubic feet. Find the volume of a cone having a radius of 12 feet and a height of
2 feet.
SOLUTION
V  khr
120   k 10 6 

k
V 
2
  2  12
3
2
V  96 
3
Blitzer, Intermediate Algebra, 5e – Slide #35 Section 6.8
2
DONE