Partial Variation deals with quantity or variable that depends on another independent variable where one of the independent variable remains the same always and the other changes according to the quantity of the independent variable used.
For example in the electricity bill, the total amount depends on two quantities rental charge and the number of wants consumed by the customer. The rental change is always a constant. The bill can be in the forms.
T = a + bw, where a and b are constants problems on partial variation always result in the solution of a pair of simultaneous equations. The equations are then solved to find the constant.
Having known this fact, let start solving problems.
Example 1:
The force E needed to make a machine pull a load is partly constant and partly varies as the load to be pulled itself. When the load is 20g, the force needed is 1.4N, whereas the force needed for 30g load is 2N. Find the
- Law connecting the load and the force
- Force for the load of 50g
Solution
This means that the force is constant while the load varies. The two constants are a and b
Force = E = a (Force is constant and ExL, E = bL (the force varies with the load)
Adding them together yields
E = a + bL
Where a and b are constant
When L = 20g, E = 1.4N
1.4 = a + 20b —– (1)
Where E = 2N, L = 30g
2 = a + 30b —– (2)
Solve equation 1 and two simultaneously by elimination method
Subtract equation 1 from 2
2 – 1.4 = 30b – 20b
0.6/10 = 10b/10
b = 0.06
Substitute b = 0.06 into equation 1
1.4 = a + 20b
1.4 = a + 20 x 0.06
1.4 = a + 1.2
a = 0.2
- The law connecting the load and the force is by substituting the value of a and b into the initial equation
The law is
E = 0.2 + 0.06L
- Force (E) when load (L) = 50g
E = 0.2 + 0.06 x 50
E = 0.2 + 3
E = 3.2N
Example 2:
S is partly constant and partly varies with T. where S = 530, T = 1600, when S = 730, T = 3600, Find:
- The formula connecting S and T
- S when T = 1300
Solution
S = a – b, S x T, 8 =
For S x T, remove the sign of proportionality X and put a constant b
S = bT — (2)
Add equation 1 and 2
S = a + bT —– (3)
When S = 530, T = 1600
530 = a + 1600b — (4)
When S = 730, T = 3600
730 = a + 3600b —- (5)
Subtract equation 4 from 5
730 – 530 = 3600b – 1600b
200/200 = 2000/2000
b = 1/10 = 0.1
Substitute b = 0.1 into equation 4
530 = a + 1600 x 0.1
530 = a + 160
a = 530 – 160
a = 370 and b = 1/10 = 0.1
- The formula connecting S and T
Is S = 370 + 1/10 T
- S When T = 1300
S = 370 + 0.1 x 1300
S = 370 + 130
S = 500
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Example 3:
The charge C of a telephone company is partly constant and partly varies as the number of units of call U. the cost of 50 units is N2500 and the cost of 120 units is N3,000. Find.
- The formula connecting C and U
- U when C N4,000
Solution
C = charge, U = number of units
C = a and C x U
C = bu
The equation then becomes
C = a + bu—-(1)
When U = 50, C = N2500
2500 = a + 50b – (2)
When C = N3000, U = 120
3000 = a + 120b —- (3)
Subtract equation 2 from 3
3000 – 2500 – a – a + 120b – 50b
500/70 = 70b/70
b = 500/70 = 50/7
Substitute b = 50/7 into equation 2
2500 = a + 50 x 50/7
2500 = a + 2500/7
2500 = a + 2500/7
Multiply both sides of the equation by 7 to clear the fraction
7 x 2500 – 7xa + 7(2500)/7
17500 = 7a + 2500
Collect like terms
17500 – 2500 = 7a
1500/7 = 7a/7
a = 15000/7
- The formula connecting C and U is by substituting a = 15000/7 and b = 50/7into the initial formula (equation 1)
C= a + bU
C = 15000/7 + 50U/7
Multiply each term by 7 to clear the fraction
7xC = 7(15000)/7 + 500 x 7/7
7C = 15000 x 50U
Therefore, the formula connecting C and U is
7C = 15000 + 50U
- U when C = N4000
Substitute C = N4000 into the formula connecting C and U
7×4000 = 15000 + 50U
28,000 = 15000 + 50U
Collect like terms
28000-15000 = 50U
13000/50 = 50U/50
U = 260 units
Example 4:
The resistance r to the motion of a car is partly constant and partly proportional to the square of the speed V. when the sped is 30km/h, the resistance is 190 Newton and when the speed is 50km/h, the resistance is 350 Newton
Find for what speed the resistance is 302.5 Newton
Solution
Resistance = r
Speed = V
r = a and r x r2
r = br2
the equation becomes
r = a + br2 —– (1)
when V = 30km/h, r = 190newton
190 = a + (30)2b
190 = a + 900b —- (2)
When r = 50km/h, r = 350newton
350 = a + (50)2b
350 = a + 2500b —- (3)
Subtract equation 2 from equation 3
350 – 190 = a – a + 2500b – 900b
160 = 1600b
b = 160/16000 = 1/10 = 0.1
Substitute b = 0.1 into equation 2
190 = a + 900 x 0.1
190 = a + 90
Collect like terms
a = 190 – 90
When r = 302.5 newton
302.5 = 100 + 0.1r2
302.5 – 100 = 0.1r2
202.5/0.1 = 0.1r2 /0.1
r2 = 2025
Take square root of both sides
r2×1/2 =
r = 45km/h
In conclusion, some believe partial variation is a very difficult aspect of variation. But today after this lesson I believe you can prove them wrong.
The key thing you need to know about partial variation is that there are two constant and it always leads to a pair of the simultaneous equation which requires using either substitution or elimination method to solve.
Just follow the pattern we have established today, I believe you will not have any challenge on this topic.
Correction to example 4.
The law connecting r and V should be : r=a+bV² √ and not r=a+br² ×
Help me solve x= 1/y. x=1/2. When y=60 find y when x=5
The cost of paintings room consiste of two parts. One part is directly proportional to the area and the other to the square of the length of the room. The cost required to paints room that measures 20m by 30m is N250.what is the cost of a table 60meters square?