Download solving a joint variation problem

10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
3.6 - 1
3.6 Variation
Direct Variation
Inverse Variation
Combined and Joint Variation
3.6 - 2
Direct Variation
When one quantity is a constant
multiple of another quantity, the two
quantities are said to vary directly. For
example, if you work for an hourly wage
of $6, then [pay] = 6 [hours worked].
Doubling the hours doubles the pay.
Tripling the hours triples the pay, and so
on. This is stated more precisely as
follows.
3.6 - 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.
3.6 - 4
Solving Variation Problems
Step 1 Write the general relationship among the
variables as an equation. Use the constant
k.
Step 2 Substitute given values of the variables and
find the value of k.
Step 3 Substitute this value of k into the equation
from Step 1, obtaining a specific formula.
Step 4 Substitute the remaining values and solve
for the required unknown.
3.6 - 5
Example 1
SOLVING A DIRECT VARIATION
PROBLEM
The area of a rectangle varies directly as its
length. If the area is 50 m2 when the
length is 10 m, find the area when the length
is 25 m.
Solution
Step 1 Since the area varies directly
as the length,
A  kL,
where A represents the area of the rectangle, L is
the length, and k is a nonzero constant.
3.6 - 6
Example 1
SOLVING A DIRECT VARIATION
PROBLEM
The area of a rectangle varies directly as its
length. If the area is 50 m2 when the
length is 10 m, find the area when the length
is 25 m.
Solution
Step 2 Since A = 50 when L = 10, the
equation A = kL becomes
50  10k
k  5.
3.6 - 7
Example 1
SOLVING A DIRECT VARIATION
PROBLEM
The area of a rectangle varies directly as its
length. If the area is 50 m2 when the
length is 10 m, find the area when the length
is 25 m.
Solution
Step 3 Using this value of k, we can express
the relationship between the area and
the length as
Direct variation
A  5L. equation.
3.6 - 8
Example 1
SOLVING A DIRECT VARIATION
PROBLEM
The area of a rectangle varies directly as its
length. If the area is 50 m2 when the
length is 10 m, find the area when the length
is 25 m.
Solution
Step 4 To find the area when the length is 25,
we replace L with 25.
A  5L  5(25)  125
The area of the rectangle is 125 m2 when
the length is 25 m.
3.6 - 9
Direct Variation as nth Power
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
y  kx .
n
3.6 - 10
Inverse Variation as nth
Power
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

, and y varies inversely
If n = 1, then
x
as x.
3.6 - 11
Example 2
SOLVING AN INVERSE VARIATION
PROBLEM
In a certain manufacturing process, the cost
of producing a single item varies inversely as
the square of the number of items produced.
If 100 items are produced, each costs $2.
Find the cost per item if 400 items are
produced.
3.6 - 12
Example 2
SOLVING AN INVERSE VARIATION
PROBLEM
Solution
Step 1 Let x represent the number of items
produced and y represent the cost per
item. Then for some nonzero constant
k,
k
y  2.
x
y varies inversely as the
square of x.
3.6 - 13
Example 2
SOLVING AN INVERSE VARIATION
PROBLEM
Solution
Step 2
k
2
2
100
Substitute, y = 2 when x = 100.
k  20,000
Solve for k.
3.6 - 14
Example 2
SOLVING AN INVERSE VARIATION
PROBLEM
Solution
Step 3 The relationship between x and y is
20,000
y
.
2
x
Step 4 When 400 items are produced, the
cost per item is
20,000 20,000
y

 .125, or 12.5 cents.
2
2
x
400
3.6 - 15
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
y  kx n z m .
3.6 - 16
Caution Note that and in the
expression “y varies jointly as x and z”
translates as the product y = kxz. The
word “and” does not indicate addition
here.
3.6 - 17
Example 3
SOLVING A JOINT VARIATION
PROBLEM
The area of a triangle varies jointly as the
lengths of the base and the height. A
triangle with base 10 ft and height 4 ft has
area 20 ft2. Find the area of a triangle with
base 3 ft and height 8 ft.
3.6 - 18
Example 3
SOLVING A JOINT VARIATION
PROBLEM
Solution
Step 1 Let A represent the area, b the base,
and h the height of the triangle.
Then for some number k,
A  kbh.
A varies jointly as b
and h.
3.6 - 19
Example 3
SOLVING A JOINT VARIATION
PROBLEM
Solution
Step 2 Since A is 20 when b is 10 and h is 4,
20  k (10)(4)
1
 k.
2
3.6 - 20
Example 3
SOLVING A JOINT VARIATION
PROBLEM
Solution
Step 3 The relationship among the variables
is the familiar formula for the area of
a triangle,
1
A  bh.
2
3.6 - 21
Example 3
SOLVING A JOINT VARIATION
PROBLEM
Solution
Step 4 When b = 3 ft and h = 8 ft,
1
2
A  (3)(8)  12 ft .
2
3.6 - 22
Example 4
SOLVING A COMBINED
VARIATION PROBLEM
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 .60 m, find the number of vibrations
per second when the tension is 196 kg and
the length is .65 m.
3.6 - 23
Example 4
SOLVING A COMBINED
VARIATION PROBLEM
Solution Let n represent the number of
vibrations per second, T represent the tension,
and L represent the length of the string. Then,
from the information in the problem, write the
variation equation. (Step 1)
k T
n
L
n varies directly as the square root of
T and inversely as L.
3.6 - 24
Example 4
SOLVING A COMBINED
VARIATION PROBLEM
Solution Substitute the given values for n, T,
and L to find k. (Step 2)
k 225
5
.60
Let n = 5, T = 225, L = .60.
3  k 225
Multiply by .60.
3  15k
1
k   .2
5
225  15
Divide by 15.
3.6 - 25
Example 4
SOLVING A COMBINED
VARIATION PROBLEM
Solution Substitute for k to find the
relationship among the variables (Step 3).
.2 T
n
L
3.6 - 26
Example 4
SOLVING A COMBINED
VARIATION PROBLEM
Solution Now use the second set of values for
T and L to find n. (Step 4)
.2 196
n
 4 .3
.65
Let T = 196, L = .65.
The number of vibrations per second is
approximately 4.3.
3.6 - 27