Transcript Document

3.7: Modeling Using Variation
Direct Variation
Certain formulas 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.
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.
3.7: Modeling Using Variation
Direct Variation
Our work up to this point provides a step-by-step procedure for solving
variation problems. This procedure applies to direct variation problems as well
as to the other kinds of variation problems that we will discuss.
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.
3.7: Modeling Using Variation
Direct Variation
Direct variation with powers is modeled by polynomial functions.
Direct Variation with Powers
y varies directly as the nth power of x if there exists some nonzero constant k
such that
y = kx n.
3.7: Modeling Using Variation
EXAMPLE:
Solving a Direct Variation
Problem
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 1
Write an equation.
expressed as
We know that y varies directly as x is
y = kx.
By changing letters, we can write an equation that describes the following
English statement: Garbage production, G, varies directly as the population, P.
G = kP
more
3.7: Modeling Using Variation
EXAMPLE:
Solving a Direct Variation
Problem
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 2 Use the given values to find k. Allegheny County has a population
of 1.3 million and creates 26 million pounds of garbage weekly. Substitute 26
for G and 1.3 for P in the direct variation equation. Then solve for k.
G
26
26
1.3
20
= kP
= k  1.3
k  1.3
=
1.3
=k
Divide both sides by 1.3.
Simplify.
more
3.7: Modeling Using Variation
EXAMPLE:
Solving a Direct Variation
Problem
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 3 Substitute the value of k into the equation.
G = kP
Use the equation from step 1.
G = 20P
Replace k, the constant of variation, with 20.
more
3.7: Modeling Using Variation
EXAMPLE:
Solving a Direct Variation
Problem
The amount of garbage, G, varies directly as the population, P. Allegheny
County, Pennsylvania, has a population of 1.3 million and creates 26 million
pounds of garbage each week. Find the weekly garbage produced by New
York City with a population of 7.3 million.
Solution
Step 4 Answer the problem's question. New York City has a population
of 7.3 million. To find its weekly garbage production, substitute 7.3 for P in
G = 20P and solve for G.
G = 20P
Use the equation from step 3.
G = 20(7.3) Substitute 7.3 for P.
G = 146
The weekly garbage produced by New York City weighs approximately 146
million pounds.
3.7: Modeling Using Variation
Inverse Variation
When two quantities vary inversely, one quantity increases as the other
decreases, and vice versa. Generalizing, we obtain the following statement.
Inverse Variation
If a situation is described by an equation in the form
k
x
where k is a constant, we say that y varies inversely as x. The number k is
called the constant of variation.
y=
We use the same procedure to solve inverse variation problems as we did to
solve direct variation problems.
3.7: Modeling Using Variation
EXAMPLE:
Solving an Inverse Variation
Problem
To continue making money, the number of new songs, S, a rock band needs to
record each year varies inversely as the number of years, N, the band has been
recording. After 4 years of recording, a band needs to record 15 new songs per
year to be profitable. After 6 years, how many new songs will the band need to
record in order to make a profit in the seventh year?
Solution
Step 1 Write an equation.
expressed as
We know that y varies inversely as x is
k
,
x
By changing letters, we can write an equation that describes the following
English statement: The number of new songs each year, S, varies inversely as
the number of years, N.
k
S =
N
y=
more
3.7: Modeling Using Variation
EXAMPLE:
Solving an Inverse Variation
Problem
To continue making money, the number of new songs, S, a rock band needs to
record each year varies inversely as the number of years, N, the band has been
recording. After 4 years of recording, a band needs to record 15 new songs per
year to be profitable. After 6 years, how many new songs will the band need to
record in order to make a profit in the seventh year?
Solution
Step 2 Use the given values to find k. After 4 years of recording, the
band needs to record 15 new songs. Substitute 15 for S and 4 for N in the
inverse variation equation. Then solve for k.
k
S =
N
k
15 =
4
k
15  4 =  4 Multiply both sides by 4.
4
60 = k
Simplify.
more
3.7: Modeling Using Variation
EXAMPLE:
Solving an Inverse Variation
Problem
To continue making money, the number of new songs, S, a rock band needs to
record each year varies inversely as the number of years, N, the band has been
recording. After 4 years of recording, a band needs to record 15 new songs per
year to be profitable. After 6 years, how many new songs will the band need to
record in order to make a profit in the seventh year?
Solution
Step 3 Substitute the value of k into the equation.
k
N
60
S =
N
S =
Use the equation from step 1.
Replace k , the constant of variation, with 60.
more
3.7: Modeling Using Variation
EXAMPLE:
Solving an Inverse Variation
Problem
To continue making money, the number of new songs, S, a rock band needs to
record each year varies inversely as the number of years, N, the band has been
recording. After 4 years of recording, a band needs to record 15 new songs per
year to be profitable. After 6 years, how many new songs will the band need to
record in order to make a profit in the seventh year?
Solution
Step 4 Answer the problem's question. We need to find how many new
songs will the band need to record after 6 years in order to make a profit in the
seventh year. Substitute 6 for N in the equation from step 3 and solve for S.
S =
60 60
=
= 10
N
6
The band will need to record 10 new songs after 6 years.
3.7: Modeling Using Variation
Joint Variation
Joint Variation
Joint variation is a variation in which a variable varies directly as the product
of two or more other variables. Thus, the equation y = kxz is read "y varies
jointly as x and z."
3.7: Modeling Using Variation
EXAMPLE:
Modeling Centrifugal Force
The centrifugal force, C, of a body moving in a circle varies jointly with the
radius of the circular path, r, and the body's mass, m, and inversely with the
square of the time, t, it takes to move about one full circle. A 6-gram body
moving in a circle with radius 100 centimeters at a rate of 1 revolution in 2
seconds has a centrifugal force of 6000 dynes. Find the centrifugal force of an
18-gram body moving in a circle with radius 100 centimeters at a rate of 1
revolution in 3 seconds.
more
3.7: Modeling Using Variation
EXAMPLE:
Modeling Centrifugal Force
Solution
C =
6000 =
40 =
C =
=
=
krm
t2
k (100)(6)
22
k
40rm
t2
40(100)(18)
32
8000
Translate "Centrifugal force, C, varies jointly with radius,
r, and mass, m, and inversely with the square of time, t."
If r = 100, m = 6, and t = 2, then C = 6000.
Solve for k.
Substitute 40 for k in the model for centrifugal force.
Find C when r = 100, m = 18, and t = 3.
The centrifugal force is 8000 dynes.