Transcript Direct & Inverse Variation Problems.

Direct & Inverse Variation
Problems.
Example 1.
If W varies directly with F and when W = 24 , F = 6 . Find the
value W when F = 10.
W  4 F
Solution.
W 
F
W  kF
When W = 24, F = 6
 24  k 6
k 6  24
k4
When F = 10 W = ?
W  4  10
W  40
Example 2.
If g varies directly with the square of h and when g = 100 ,
h = 5 . Find the value h when g = 64.
g  4h
Solution.
g h
2
When g = 64 , h = ?
g  kh
2
When g = 100 , h = 5
100  k 25
k 25  100
k 4
2
64  4h2
4h2  64
h2  16
h  16
h = 4 or h = - 4
Example 3.
If d varies inversely with w and when d = 3 , w = 9 . Find
the value d when w = 3.
Solution.
1
d
w
k
d
w
27
d
w
When w = 3 d = ?
27
d
3
When d = 3 , w = 9
k
3
9
k = 27
d=9
Example 4 .
If r varies inversely with the square root of f and when
r = 32 , f = 16. Find f when r = 32.
Solution.
r
1
f
r
k
f
When r = 32 , f= 16.
k
16
32 
32 
k
4
k = 128
128
r
f
When r = 32 , f = ?
32 
128
f
32 f  128
f 
128
32
f 4
f  16
Example 5 .
If t varies jointly with m and b and t = 80 when
and b = 5. Find t when m = 5 and b = 8 .
Solution.
t  mb
m=2
t  8mb
When m = 5 , b = 8 , t = ?
t  kmb
t  8 5 8
When t = 80 , m = 2 and b = 5
t  320
80  k  2  5
10 k  80
k8
Example 6 .
c varies directly with the square of m and inversely with w.
c = 9 when m = 6 and w = 2 . Find c when m = 10 and
w=4.
Solution.
m2
c
w
c
km2
w
When c = 9 , m = 6 and w = 2
9
k 62
2
36 k 9

2
1
K=½
m2
c
2w
When m = 10 , w = 4
and c = ?
102
=12.5
c
2 4
Examination Questions.
Example 1.
The time,T minutes ,taken for a stadium to empty varies
directly as the number of spectators , S, and inversely
as the number of open exits, E.
(a) Write down a relationship connecting T,S and E.
It takes 12 minutes for a stadium to empty when there are
20 000 spectators and 20 open exits.
(b) How long does it take the stadium to empty when there
are 36 000 spectators and 24 open exits ?
Solution.
6S
T
500 E
(a) T S
E
KS
T
E
K is the constant of variation.
Now S = 36 000
and E = 24 .
(b)
6  36000
T
500  24
T = 12 , S = 20 000 and E = 20
Substitute to find the value of K.
12 K 20000

1
20
Cross multiply.
20 000 K = 20 x 12
240
6
K

20000 500
Substitute.
T = 18 minutes
Example 2.
The number of letters, N , which can be typed on a sheet
of paper varies inversely as the square of the size, S , of
the letters used.
(a) Write down a relationship connecting N and S .
(b) The size of the letters used is doubled.
What effect does this have on the number of letters which
can be typed on the sheet of paper ?
Solution.
(a)
(b)
Letter size = 2S
1
N 2
S
K
N 
(2 S )2
K
N 2
S
N
K
4S 2
By doubling the
size of letters
the number of
letters is
quartered.
Example 3.
A frictional force is necessary for a car to round a bend.
The frictional force , F kilonewtons , varies directly as the
square of the car’s speed , V metres per second, and inversely
as the radius of the bend, R metres.
(a) Write down a relationship between F, V and R.
A frictional force of 20 kilonewtons is necessary for
a car , travelling at a given speed , to round a bend.
(b) Find the frictional force necessary for the same
car , travelling at twice the given speed , to round the
same bend.
Solution.
(a)
V2
F
R
KV 2
F
R
(b)
Let the speed = 2V
K ( 2V )2
F 
R
K 4V 2
F 
R
4 KV 2
F
R
By doubling the speed
the frictional force F
required to round the
bend becomes 4 times
greater.
F = 4 x 20 = 80