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5.8 Modeling Using Variation
Chapter 5 Polynomial and Rational Functions
Concepts and Objectives
⚫ The objectives of this section are
⚫ Solve direct variation problems.
⚫ Solve inverse variation problems.
⚫ Solve problems involving joint variation.
Direct Variation
⚫ y varies directly as x, or y is directly proportional to x,
if there exists a nonzero real number k, called the
constant of variation, such that
y = kx.
⚫ Let n be a positive real number. Then y varies directly
as the nth power of x, or y is directly proportional to
the nth power of x, if there exists a nonzero real
number k such that n
y kx
=
Inverse Variation
⚫ y varies inversely as x, or y is inversely proportional
to x, if there exists a nonzero real number k, such that
⚫ Let n be a positive real number. Then y varies inversely
as the nth power of x, or y is inversely proportional to
the nth power of x, if there exists a nonzero real
number k such that
k
y
x
=
n
k
y
x
=
Solving Variation Problems
1. Write the general relationship among the variables as
an equation. Use the constant k.
2. Substitute given values of the variables and find the
value of k.
3. Substitute this value and the remaining values into the
original equation and solve for the unknown.
Examples
1. If y varies directly as x, and y = 24 when x = 8, find y
when x = 12.
2. If y varies inversely as x, and y = 7 when x = 4, find y
when x = 14.
Examples
1. If y varies directly as x, and y = 24 when x = 8, find y
when x = 12.
2. If y varies inversely as x, and y = 7 when x = 4, find y
when x = 14.
( )
24 8
k
=
3
k =
( )( )
3 12
y =
36
y =
7
4
k
=
28
k =
28
14
y =
2
y =
Step 1
Step 2
Step 3
Joint Variation
⚫ Let m and n be real numbers. Then y varies jointly as
the nth power of x and the mth power of z if there exists
a nonzero real number k such that
⚫ Note: If n or m is negative, then the variable is said to
vary inversely.
n m
y kx z
=
Example
3. If u varies jointly as v and w, and u = 48 when v = 12 and
w = 8, find u when v = 10 and w = 6.
4. If z varies directly as x and inversely as y2, and z = 8
when x = 6 and y = 3, find z when x = 10 and y = 4.
Example
3. If u varies jointly as v and w, and u = 48 when v = 12 and
w = 8, find u when v = 10 and w = 6.
4. If z varies directly as x and inversely as y2, and z = 8
when x = 6 and y = 3, find z when x = 10 and y = 4.
( )( )
48 12 8
k
=
0.5
k =
( )( )( )
0.5 10 6
u =
30
u =
( )
2
6
8
3
k
=
12
k =
( )( )
2
12 10
4
z =
7.5
z =
u kvw
=
2
kx
z
y
=
Example
5. The number of vibrations per second (the pitch) of a
steel guitar string varies directly as the square root of
the tension and inversely as the length of the string. If
the number of vibrations per second is 5 when the
tension is 225 kg and the length is .6 m, find the number
of vibrations per second when the tension is 196 kg and
the length is .65 m.
Example
5. The number of vibrations per second (the pitch) of a
steel guitar string varies directly as the square root of
the tension and inversely as the length of the string. If
the number of vibrations per second is 5 when the
tension is 225 kg and the length is .6 m, find the number
of vibrations per second when the tension is 196 kg and
the length is .65 m.
Let n represent the number of vibrations per second, T
represent the tension, and L represent the length of the
string.
Example (cont.)
Let n represent the number of vibrations per second, T
represent the tension, and L represent the length of the
string.
k T
n
L
=
225
5
.6
k
=
15
5
.6
.2
k
k
=
=
.2 196
4.3
.65
n = 
Classwork
⚫ College Algebra 2e
⚫ 5.8: 4-22 (even); 5.6: 30-34, 40-50 (even); 5.5: 56-70
(even)
⚫ 5.8 Classwork Check
⚫ Quiz 5.6

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5.8 Modeling Using Variation

  • 1. 5.8 Modeling Using Variation Chapter 5 Polynomial and Rational Functions
  • 2. Concepts and Objectives ⚫ The objectives of this section are ⚫ Solve direct variation problems. ⚫ Solve inverse variation problems. ⚫ Solve problems involving joint variation.
  • 3. Direct Variation ⚫ y varies directly as x, or y is directly proportional to x, if there exists a nonzero real number k, called the constant of variation, such that y = kx. ⚫ Let n be a positive real number. Then y varies directly as the nth power of x, or y is directly proportional to the nth power of x, if there exists a nonzero real number k such that n y kx =
  • 4. Inverse Variation ⚫ y varies inversely as x, or y is inversely proportional to x, if there exists a nonzero real number k, such that ⚫ Let n be a positive real number. Then y varies inversely as the nth power of x, or y is inversely proportional to the nth power of x, if there exists a nonzero real number k such that k y x = n k y x =
  • 5. Solving Variation Problems 1. Write the general relationship among the variables as an equation. Use the constant k. 2. Substitute given values of the variables and find the value of k. 3. Substitute this value and the remaining values into the original equation and solve for the unknown.
  • 6. Examples 1. If y varies directly as x, and y = 24 when x = 8, find y when x = 12. 2. If y varies inversely as x, and y = 7 when x = 4, find y when x = 14.
  • 7. Examples 1. If y varies directly as x, and y = 24 when x = 8, find y when x = 12. 2. If y varies inversely as x, and y = 7 when x = 4, find y when x = 14. ( ) 24 8 k = 3 k = ( )( ) 3 12 y = 36 y = 7 4 k = 28 k = 28 14 y = 2 y = Step 1 Step 2 Step 3
  • 8. Joint Variation ⚫ Let m and n be real numbers. Then y varies jointly as the nth power of x and the mth power of z if there exists a nonzero real number k such that ⚫ Note: If n or m is negative, then the variable is said to vary inversely. n m y kx z =
  • 9. Example 3. If u varies jointly as v and w, and u = 48 when v = 12 and w = 8, find u when v = 10 and w = 6. 4. If z varies directly as x and inversely as y2, and z = 8 when x = 6 and y = 3, find z when x = 10 and y = 4.
  • 10. Example 3. If u varies jointly as v and w, and u = 48 when v = 12 and w = 8, find u when v = 10 and w = 6. 4. If z varies directly as x and inversely as y2, and z = 8 when x = 6 and y = 3, find z when x = 10 and y = 4. ( )( ) 48 12 8 k = 0.5 k = ( )( )( ) 0.5 10 6 u = 30 u = ( ) 2 6 8 3 k = 12 k = ( )( ) 2 12 10 4 z = 7.5 z = u kvw = 2 kx z y =
  • 11. Example 5. The number of vibrations per second (the pitch) of a steel guitar string varies directly as the square root of the tension and inversely as the length of the string. If the number of vibrations per second is 5 when the tension is 225 kg and the length is .6 m, find the number of vibrations per second when the tension is 196 kg and the length is .65 m.
  • 12. Example 5. The number of vibrations per second (the pitch) of a steel guitar string varies directly as the square root of the tension and inversely as the length of the string. If the number of vibrations per second is 5 when the tension is 225 kg and the length is .6 m, find the number of vibrations per second when the tension is 196 kg and the length is .65 m. Let n represent the number of vibrations per second, T represent the tension, and L represent the length of the string.
  • 13. Example (cont.) Let n represent the number of vibrations per second, T represent the tension, and L represent the length of the string. k T n L = 225 5 .6 k = 15 5 .6 .2 k k = = .2 196 4.3 .65 n = 
  • 14. Classwork ⚫ College Algebra 2e ⚫ 5.8: 4-22 (even); 5.6: 30-34, 40-50 (even); 5.5: 56-70 (even) ⚫ 5.8 Classwork Check ⚫ Quiz 5.6