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Variations
CHAPTER 1
LESSON 2
VARIATIONS
Direct
Variation
Definition 1.1 A variable y is
in direct variation with another
variable x (or y is directly
proportional to x) if y and x
are related by an equation in
the form y=kx, where k is a
nonzero constant.
Definition 1.2 The constant
k in the equation y = kx is
called the constant of
variation or the constant of
proportionality.
EXAMPLE 1
A quantity y varies directly as another
quantity x. If x = 20 then y = 8.
1.Write an equation that represents
the relationship.
2.Find the value of y when x = 125.
SOLUTION
Given that y varies directly as x, this means
that y = kx where k is a nonzero constant.
1.First, find k given that y = 8 if x = 20.
Substitute these values in y = kx to find
the value of k.
y = kx
8 = k • 20
Divide both sides of the equation by 20.
k = or
Thus, y =
20
8
5
2
χ
5
2
2. To find the value of y when x = 125,
substitute 125 for x in the equation
y =
25
y =
Hence, y = 50 when x = 125.1
50521
5
2
=///•
/
χ
5
2
DIRECT nth
POWER
VARIATION
Definition 1.3 A variable is in
direct nth power variation with
another variable x (or y is directly
proportional to the nth power of
x) if y and x are related by an
equation in the form y = kx ,ⁿ
where k is a nonzero constant
and n is a positive real number.
EXAMPLE 1
Assume that S varies directly as
the second power of r. If the
constant of variation is equal to
4π. Write an equation that
represents the relationship
between S and r.
Solution
The given relationship is a
direct nth power variation in
which n = 2 and the constant
of variation k is equal to .
Hence, the required equation
is S = r
π4
π4 2
EXAMPLE 2
Suppose L varies directly as
the cube of t. If t = 5 then
L = 162.5. If t = 2 what is
the value of L?
SOLUTION
Since L varies directly as the
cube of t, the equation is L = kt
where k is a nonzero
constant. First, find k by
substituting the values t = 5 and
L = 162.5 in the equation L =
3
3
L = kt
162.5 = k•(5)
162.5 = 125k
k = or 1.3
3
3
125
5.162
Hence, L = 1.3t . To find L when
t = 2, substitute 2 for t in the
equation.
L = 1.3t = 1.3(2) = 1.3(8) = 10.4
Thus, when t = 2, L = 10.4.
3
33
Inverse
Variation
Definition 1.4 A variable y is in
inverse nth power variation with
another variable x (or y is
inversely proportional to the nth
power of x) if the variables are
related by an equation in the form
y= , where k is a nonzero
constant and n is a positive real
number.
n
x
k
EXAMPLE
Suppose y varies inversely
as x and y = when x = 4.
Find the value of x when
y = .
5
1
35
1
SOLUTION
Since y varies inversely as x, then y =
where k is a nonzero constant. Find k by
substituting x = 4 and y = .
x
k
5
1
x
k
y =
45
1 k
=
Multiply both sides of the equation by the
LCM of 4 and 5, which is 20. Then solve
for k.
4 = 5k
k =
Therefore the equation f the inverse
variation is y = or y = .
5
4
x
5
4
x5
4
To find the value of x when y = ,
substitute for y in the equation y = .
y =
=
Multiplying both sides of the resulting
equation by 35x, which is the LCM of 35
and 5x , you get
x = 28.
35
1
35
1
x5
4
x5
4
35
1
x5
4
Thus, x = 28 when y =35
1
Combined
and Joint
Variations
In instances where a variable varies
directly or inversely, or both, with
two or more other variables in the
same formula, the relationship
among the variables is called
combined variation. When a variable
varies directly with two or more
variables, the relationship among the
variables is called a joint variation.
That is, a variable y varies
jointly as variables x and z
if y = kxz, where k is a
nonzero constant. A joint
variation is also a
combined variation.
EXAMPLE
Suppose z varies directly as
the square of t and inversely
as u. When t = 3 and u = 12,
z = 3. Find the value of z if
t = 5 and u = 15.
SOLUTION
The equation representing the variation is
z = where k is a nonzero constant still to
be found. To find the value of k, substitute
3 for t, 12 for u, and 3 for z in the equation
z = .
u
kt2
u
kt2
z =
3 =
Multiply both sides of the equation by 12.
Then divide both sides of the equation by 9.
36 = k•9
= k Therefore, z =
k = 4
12
)3( 2
k
u
kt2
9
36
u
t2
4
To find the value of z if t = 5 and u = 15,
substitute 5 for t and 15 for u in the
equation z = .
Thus, when t = 5 and u = 15.
u
t2
4
3
20
15
100
15
254
15
)5(44 22
or
u
t
z =
•
===
3
20
=z
Variations

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Variations

  • 4. Definition 1.1 A variable y is in direct variation with another variable x (or y is directly proportional to x) if y and x are related by an equation in the form y=kx, where k is a nonzero constant.
  • 5. Definition 1.2 The constant k in the equation y = kx is called the constant of variation or the constant of proportionality.
  • 6. EXAMPLE 1 A quantity y varies directly as another quantity x. If x = 20 then y = 8. 1.Write an equation that represents the relationship. 2.Find the value of y when x = 125.
  • 7. SOLUTION Given that y varies directly as x, this means that y = kx where k is a nonzero constant. 1.First, find k given that y = 8 if x = 20. Substitute these values in y = kx to find the value of k. y = kx 8 = k • 20
  • 8. Divide both sides of the equation by 20. k = or Thus, y = 20 8 5 2 χ 5 2
  • 9. 2. To find the value of y when x = 125, substitute 125 for x in the equation y = 25 y = Hence, y = 50 when x = 125.1 50521 5 2 =///• / χ 5 2
  • 11. Definition 1.3 A variable is in direct nth power variation with another variable x (or y is directly proportional to the nth power of x) if y and x are related by an equation in the form y = kx ,ⁿ where k is a nonzero constant and n is a positive real number.
  • 12. EXAMPLE 1 Assume that S varies directly as the second power of r. If the constant of variation is equal to 4π. Write an equation that represents the relationship between S and r.
  • 13. Solution The given relationship is a direct nth power variation in which n = 2 and the constant of variation k is equal to . Hence, the required equation is S = r π4 π4 2
  • 14. EXAMPLE 2 Suppose L varies directly as the cube of t. If t = 5 then L = 162.5. If t = 2 what is the value of L?
  • 15. SOLUTION Since L varies directly as the cube of t, the equation is L = kt where k is a nonzero constant. First, find k by substituting the values t = 5 and L = 162.5 in the equation L = 3 3
  • 16. L = kt 162.5 = k•(5) 162.5 = 125k k = or 1.3 3 3 125 5.162
  • 17. Hence, L = 1.3t . To find L when t = 2, substitute 2 for t in the equation. L = 1.3t = 1.3(2) = 1.3(8) = 10.4 Thus, when t = 2, L = 10.4. 3 33
  • 19. Definition 1.4 A variable y is in inverse nth power variation with another variable x (or y is inversely proportional to the nth power of x) if the variables are related by an equation in the form y= , where k is a nonzero constant and n is a positive real number. n x k
  • 20. EXAMPLE Suppose y varies inversely as x and y = when x = 4. Find the value of x when y = . 5 1 35 1
  • 21. SOLUTION Since y varies inversely as x, then y = where k is a nonzero constant. Find k by substituting x = 4 and y = . x k 5 1 x k y = 45 1 k =
  • 22. Multiply both sides of the equation by the LCM of 4 and 5, which is 20. Then solve for k. 4 = 5k k = Therefore the equation f the inverse variation is y = or y = . 5 4 x 5 4 x5 4
  • 23. To find the value of x when y = , substitute for y in the equation y = . y = = Multiplying both sides of the resulting equation by 35x, which is the LCM of 35 and 5x , you get x = 28. 35 1 35 1 x5 4 x5 4 35 1 x5 4
  • 24. Thus, x = 28 when y =35 1
  • 26. In instances where a variable varies directly or inversely, or both, with two or more other variables in the same formula, the relationship among the variables is called combined variation. When a variable varies directly with two or more variables, the relationship among the variables is called a joint variation.
  • 27. That is, a variable y varies jointly as variables x and z if y = kxz, where k is a nonzero constant. A joint variation is also a combined variation.
  • 28. EXAMPLE Suppose z varies directly as the square of t and inversely as u. When t = 3 and u = 12, z = 3. Find the value of z if t = 5 and u = 15.
  • 29. SOLUTION The equation representing the variation is z = where k is a nonzero constant still to be found. To find the value of k, substitute 3 for t, 12 for u, and 3 for z in the equation z = . u kt2 u kt2
  • 30. z = 3 = Multiply both sides of the equation by 12. Then divide both sides of the equation by 9. 36 = k•9 = k Therefore, z = k = 4 12 )3( 2 k u kt2 9 36 u t2 4
  • 31. To find the value of z if t = 5 and u = 15, substitute 5 for t and 15 for u in the equation z = . Thus, when t = 5 and u = 15. u t2 4 3 20 15 100 15 254 15 )5(44 22 or u t z = • === 3 20 =z