Variation; sections 8.1 & 8.2
Type of
Variation
phrase
simple
equation
example
practical example
The radius of the
y = kx circle lit by a car’s
light decreases (y)
Direct
as the distance
y varies
away from the
directly
“varies
garage decreases
with x
directly with”
(x).
The brightness of
a car’s lights
y = k/x
“varies
increases (y) as
Inverse
inversely
y varies the distance away
with”
from the garage
inversely
decreases (x).
with x
The heat loss
“varies
through a glass
jointly
window (y) varies
y
=
kxz
(directly)
jointly with the
with”
area of the window
Joint
y varies
(x) and the
jointly
“depends
temperature
with
x
upon both . .
difference (z)
and
z
.”
between inside
and outside.
The radius of the
y = kx/z circle lit by a car’s
“varies
light decreases (y)
directly with
as the distance
y varies
Combined
x and
away from the
directly
inversely
garage decreases
with x
with z”
(x), but the
and
inversely nervousness of the
“is directly
proportional
to”
more
complicated
equations
y = kx2
y varies
directly with
x2
y = k/x3
y varies
inversely with
x3
y = kx3z2
y varies jointly
with x3 and z2
k x
y 4
z
y varies
directly with
the sqaure
root of x and
with z
new driver
inversely with
increases (z) (he’s
z4
afraid he’s going to
hit the door)!!!!!!
k is the constant of variation in all the above equations. It has a
numerical value that will be different in different situations.
To solve the problems, you will follow these steps:
Step 1: Write the equation in general terms as in the 3rd column above
– don’t forget the k.
Step 2: Use the data given to sub in and solve for k.
Step 3: Use the equation in Step 1 to fill in k from step 2.
Step 4: Fill Step 3’s equation in with the second set of data; solve for
the only variable remaining.
Example 1: If a varies directly as b and a = 75 when b = 40, find a when
b = 12.
Step 1: a = kb
Step 2: 75 = k(40)
k = 15/8
Step 3: a = 15/8 b
Step 4: a = (15/8)(12)
a = 45/2 
Example 2:
The stretch in a wire under a given tension varies directly as the length
of the wire and inversely as the square of its diameter. A wire having
length 2 m and diameter 1.5 mm stretches 1.2 mm, If a second wire of
the same material has length 3 m and diameter 2 mm, find the amount
of stretch.
let stretch = S; length = L, and diameter = D
Step 1: S = kL/D2
Step 2: 1.2 = k(2)/1.52
1.2 = {k(2)}/(9/4)
12/10 = k(8/9)
k = 27/20
Step 3: S = (27/20)L/D2
Step 4: S = (27/20)(3)/22
S = 81/80 = 1.0125 mm

Note that on this problem, different size units were given (some meters,
some millimeters). I did NOT need to convert in this case to all of one
type of unit. Each data point was consistently one or the other and k
was calculated as such.