Transcript Section 2.6 - math-clix

Section 2.6
Variation and
Applications
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.
Objectives

Find equations of direct, inverse, and combined
variation given values of the variables.

Solve applied problems involving variation.
Direct Variation
If a situation gives rise to a linear function
f(x) = kx, or y = kx, where k is a positive constant,
we say that we have direct variation, or that y
varies directly as x, or that y is directly
proportional to x. The number k is called the
variation constant, or constant of
proportionality.
Direct Variation
The graph of y = kx, k > 0, always goes through the origin
and rises from left to right. As x increases, y increases;
that is, the function is increasing on the interval (0,). The
constant k is also the slope of the line.
y  kx, k  0
Example
Find the variation constant and an equation of variation in
which y varies directly as x, and y = 42 when x = 3.
We know that (3, 42) is a solution of y = kx.
y = kx
42 = k  3
42
k
3
14 = k
The variation constant 14, is the rate of change of y with
respect to x. The equation of variation is y = 14x.
Example
Wages. A cashier earns an hourly wage. If the cashier worked 18
hours and earned $168.30, how much will the cashier earn if she
works 33 hours?
We can express the amount of money earned as a function of the
amount of hours worked.
I(h) = kh
I(18) = k  18
$168.30 = k  18
$9.35 = k The hourly wage is the variation constant.
Next, we use the equation to find how much the cashier will
earn if she works 33 hours.
I(33) = $9.35(33)
= $308.55
Inverse Variation
If a situation gives rise to a function f(x) = k/x, or y = k/x,
where k is a positive constant, we say that we have
inverse variation, or that y varies inversely as x, or that
y is inversely proportional to x. The number k is called
the variation constant, or constant of proportionality.
For the graph y = k/x, k  0, as x increases, y decreases;
that is, the function is decreasing on the interval (0, ).
Inverse Variation
For the graph y = k/x, k  0, as x increases, y decreases;
that is, the function is decreasing on the interval (0, ).
y
k
, k 0
x
Example
Find the variation constant and an equation of variation in
which y varies inversely as x, and y = 22 when x = 0.4.
k
y
x
k
22 
0.4
(0.4)22  k
8.8  k
The variation constant is 8.8. The equation of variation is
y = 8.8/x.
Example
Road Construction. The time t required to do a job
varies inversely as the number of people P who work on
the job (assuming that they all work at the same rate). If
it takes 180 days for 12 workers to complete a job, how
long will it take 15 workers to complete the same job?
We can express the amount of time required, in days,
as a function of the number of people working.
k
P
k
t (12) 
12
t ( P) 
k
12
2160  k
180 
Example continued
The equation of variation is t(P) = 2160/P.
Next we compute t(15).
2160
t ( P) 
P
2160
t (15) 
15
t  144
It would take 144 days for 15 people to complete the
same job.
Combined Variation
Other kinds of variation:
y varies directly as the nth power of x if there is some
positive constant k such that y  kx n .
y varies inversely as the nth power of x if there is some
k
positive constant k such that y  n .
x
y varies jointly as x and z if there is some positive
constant k such that y = kxz.
Example
Find the equation of variation in which y varies directly as
the square of x, and y = 12 when x = 2.
y  kx 2
12  k  2
12  4k
3k
Thus y = 3x2.
2
Example
Find the equation of variation in which y varies jointly as x
and z and inversely as the square of w, and y = 105 when
x = 3, z = 20, and w = 2.
xz
y k 2
w
3  20
105  k  2
2
105  15k
7k
Thus y  7
xz
7 xz
or y  2
2
w
w
Example
The luminance of a light (E) varies directly with the intensity (I) of
the light and inversely with the square distance (D) from the light.
At a distance of 10 feet, a light meter reads 3 units for a 50-cd
lamp. Find the luminance of a 27-cd lamp at a distance of 9 feet.
I
Ek 2
D
k  50
3
102
6k
Substitute the second set of data into the equation.
The lamp gives an luminance reading of 2 units.
6  27
E 2
9
E2