Variations
Direct Variations
y  4 x
y varies directly as x.
y is directly proportional to x.
Constant of proportionality is 4.
y  4  x is an example of a direct variation.
Inverse Variations
y
3
x
y varies inversely as x.
Constant of proportionality is 3.
y
3
is an example of an inverse variation.
x
Joint Variations
y  6 x z
y varies jointly as x and z.
Constant of proportionality is 6.
y  6  x  z is an example of a joint variation.
Combined Variations
y
7 x
z
y varies directly as x and inversely as z.
Constant of proportionality is 7.
y
7 x
is an example of a combined variation.
z
Examples
C  2  r
Circumference (C ) varies directly as radius (r ).
Constant of proportionality is 2 .
C  2  r is an example of a direct variation.
A    r2
Area (A) varies directly as the square of the radius (r 2 ).
Constant of proportionality is  .
A    r 2 is an example of a direct variation.
4
V    r3
volume of sphere
3
Volume (V ) varies directly as the cube of the radius (r 3 ).
4
Constant of proportionality is  .
3
4
V    r 3 is an example of a direct variation.
3
Suppose y varies directly as x. When x is 5, y is 20. Find y when x is 45.
Solution:
Let k be the constant of proportionality.
Since y varies directly as x, y   constant of proportionality   x.
Or y  k  x
When x is 5, y is 20.
y kx
20  k  5
20 k  5

5
5
4k
Hence, y  k  x  4  x
Find y when x is 45.
y  4 x
y  4  45  180
Thus, y  180 when x  45.
Suppose y varies inversely as x. When x is 4, y is 12. Find y when x is 10.
Solution:
Let k be the constant of proportionality.
Since y varies inversely as x, y 
Or y 
k
x
When x is 4, y is 12
k
y
x
k
12 
4
k
12(4)  ( 4)
4
48  k
k 48
Hence, y  
x x
Find y when x is 10.
48
y
x
48
y
10
y  4.8
Thus, y  4.8 when x  10.
 constant of proportionality 
x
.
Suppose y varies directly as x and inversely as z. When x is 4 and z  10, y is 12.
Find y when x is 10 and z  2.
Solution:
Let k be the constant of proportionality.
Since y varies directly as x and inversely as z , y 
Or y 
kx
z
When x is 4 and z  10, y is 12
kx
y
z
k (4)
12 
(10)
k (4)
12(10) 
(10)
(10)
120  4k
120 4k

4
4
30  k
kx 30 x
Hence, y 

z
z
Find y when x is 10 and z  2.
30 x
y
z
30(10)
y
2
y  150
Thus, y  150 when x is 10 and z  2.
 constant of proportionality   x
z
.
Suppose y varies directly as x and inversely as the square of z.
When x is 2 and z  8, y is 12. Find y when x is 15 and z  4.
Solution:
Let k be the constant of proportionality.
Since y varies directly as x and inversely as the square of z ,
y
 constant of proportionality   x
Or y 
z2
.
kx
z2
When x is 2 and z  8, y is 12.
kx
y 2
z
k (2)
12 
(8) 2
k (2)
12 
64
k (2)
12(64) 
(64)
(64)
768  2k
768 2k

2
2
384  k
kx 384 x
Hence, y  2  2
z
z
Find y when x is 15 and z  4.
384 x
y 2
z
384(15)
y
(4) 2
y  360
Thus, y  360 when x is 15 and z  4.
Suppose y varies jointly as x and the square root z.
When x is 2 and z  4, y is 12. Find y when x is 15 and z  9.
Solution:
Let k be the constant of proportionality.
Since y varies jointly as x and the square root z.
y   constant of proportionality   x  z
Or y  kx z
When x is 2 and z  4, y is 12.
y  kx z
12  k (2) 4
12  k (2)(2)
12  4k
12 4k

4
4
3k
Hence, y  kx z  3 x z
Find y when x is 15 and z  9.
y  3x z
y  3(15) 9
y  3(15)(3)  135
Thus, y  135 when x is 15 and z  9.