Mathematical Variations – Definition
In mathematics, Variation is defined and expressed as Equation of mathematical relationship between two variables (like Direct, Inverse, Joint, Combined variation).
Mathematical Direct Variations
When value of A varies directly as B, it can be written as A α B (pronounced as A varies with B). A = K×B, or K = A/B.
Direct Variation – Properties : In direct variation, as one variable increases, the other variable also increases.
Direct Variation – Example : Joy gets wages ( W ) on basis of Hours (H) Worked. So, His total Wages ( W ) vary directly with Hours (H) worked. It may be expressed in mathematical relation as W α H. We can write as W=K×H (where K is the constant of variation). Here, the value of W would change, as H changes, but the value of K would remain constant. Let us assume, the value of K is 100. Then the equation would be W=100× H
If the worked are plotted on the x axis against wages on the y axis, the resulting line is a straight line showing direct variation.
Mathematical Inverse Variation
If A varies Inversely to B, then A varies directly as the reciprocal of B.
A ∝1/B . Hence, A= K× 1/B=K/B , or K=A×B
Inverse Variation – Example : Speed (S) is inversely proportional to Time (T).
So, S∝ 1/T, or S=K×1/T. Or, S=K/T
Mathematical Joint Variation
A is said to varies jointly as B, C, D…. if A varies directly as the product of B, C, D ….
i.e A ∝ (B, C, D … … …). So, A = K (B, C, D … … …)
Example: The volume of cuboid (V) varies directly as the product of length (L), breadth (B), height(H)
i.e. A ∝ LBH . So, V = K×(L×B×H). Or, K =K/LxBxH
Mathematical Combined Variation
Combined Variation is a more relationship between three variables, where one variable (x) varies directly to one variable (y) and inversely to another (z)
Combined Variation is denoted as x ∝ k.y/z, So, x = kxy/z
The force of attraction F of a body varies directly as its mass m times a constant k and inversely as the square of the distance d between the bodies. The equation is expressed as F = ,
So, if F equals 100 Newtons, m equals 8 kg, and d equals 5 meters,
then the equation is 100 = . So, 8k=2500, or k==312.5.
Suppose the mass is 10 kg and the distance is 15 meters.
Then F = 312.5 × = = 13.89 Newtons.
Mathematical Partial Variation
When two variables are related by a formula or a variable is related by the sum of two or more variables then it is called as partial variation. For example, X = KY + C (where K and C are constants). This is example of equation straight line
Mathematical Variations Properties
- If A ∝ B, then B ∝ A
- If A ∝ B and B ∝ C, then A ∝ C
- If A ∝ B and B ∝ C, then A-B ∝ C
- If A ∝ C and B ∝ C , then A-B ∝ C
- If A ∝ C and B ∝ C, then √(A×B)=C
- If A ∝ B, then An ∝ Bn.
- If A ∝ B and C ∝ D, then (A×C) ∝ (B×D) and =
- If A ∝ BC, then B ∝ and C ∝
Mathematical Variations Examples
Variation – Example-1: If a + b ∝ a – b, prove that (i) a ∝ b
Since a + b ∝ a-b, then a + b = k (a-b)=ka-kb, ka-a=kb +b. Or, (k-1) a = (k+1) b
or, = . Let us assume m=, So, =m. So, a=mb (m is constant, as k is constant).
So, a ∝b (proved)
Variation – Example-2 : If (x2+y2) ∝ (x2-y2), prove that x ∝ y
(x2+y2) ∝ (x2-y2). So, (x2+y2) = k(x2-y2). Or, (x2+y2) = kx2-ky2. Or, x2 – kx2 = -k2y2-y2.
Or, x2(1-k)=y2(-k2-1), or, =. Let us assume m2=.
So, =m2, or =m. or, x=my. So, x ∝ y (proved)
Variation – Example-3 : If a² ∝ bc, b² ∝ ca and c² ∝ ab, then find the relation between the three constants of variation.
Since, a² ∝ bc. So, a² = k×b×c …….(1) [where k is constant of variation]
Again, b² ∝ ca, So, Therefore, b² = l×c×a ……. (2) [where l is constant of variation]
Again, c² ∝ ab, So, c² = m×a×b ……. (3) [where m is constant of variation]
Multiplying both sides of (1), (2) and (3) we get, a²×b²×c² = (k×b×c) × (l×c×a) ×(m×a×b)
= (k×l×m) × (a²×b²×c²). or, k×l×m = =1,
So. k×l×m =1, which is the required relation between the three constants of variation.
Mathematical Joint Variations Examples
Variation (Joint Variation) – Example- 1 : A car takes 3 Hrs running at 60 kmph. How long would it take running at 40 kmph?
Let T =Time (in Hrs) taken to cover the distance (S), V= Speed of the car, So, S= V×T (S is constant, V and T are variables).
Here S = VT = 60 x 3 = 180 km.
So at a speed of the car is 40 kmph :
S = VT, or, 180 = 40 x T, or, T =180/40 = 9/2. Hrs = 4 hrs 30 mins.
Variation (Joint Variation) – Example- 2: The area of a triangle is jointly related to the height and the base of the triangle. If the base is increased 20% and the height is decreased by 10%, what will be the percentage change of the area?
Area of triangle is half the product of base and height.
So the joint variation equation for area of triangle is A = where A=Area, b=base and h is the height. Here ½ is the constant for the equation.
If Base is increased by 20%, so the increased Base will be = b x =
If Height is decreased by 10%, so decreased Height will be = h x = .
So the new area after the changes of base and height is = == ×=×A
So the area of the triangle is changed by 8%.
Variation (Joint Variation) – Example- 3 : A rectangle’s length is doubled and width is halved, how much the area will increase or decrease?
We know, A = L×W. So, A ∝ LW, (where A=Area, L=Length, W=Width). Here the constant=1
When Length (L) is doubled, changed length would be = 2L
When Width (W) is halved, changed Width would be = .
So, the changed area will be =2L × = WL =A, which is same as the previous value of A.
So, the changed area of the rectangle would remain same. So, there would be no change in Area of the rectangle in this case.
Mathematical Partial Variations Examples
Variation (Partial Variation) – Example- 1 : Rita goes to a fair. The entry fee to the fair is Rs.10, and charge for Ride is proportional to the number of rides enjoyed. Rita spent Rs 24 for 7 Rides. Find how much she would have to spend for 10 Rides.
Here we can express the Partial Variation relationship as C=E+n×F (where C is the total Cost of Ride, E is Entry fee (which is the constant part, n= number of rides, which is variable, F=Cost per Ride, which is also constant part)
So, for 7 Rides : 24=10+7×F (where F is the variable cost per Ride). So, 7F=24-10=14. Or F= =2
Now for 10 Rides, Cost = 10+10×2=10+20=30
Variation (Partial Variation) – Example- 2: The Taxi Fare is partially fixed (Rs 30) and partially directly proportion to kilometres travelled. If the Taxi Fare for travelling 10 km is Rs 160, what would be fare for a distance of 20 km ?
Let us express the Formula for Fare as F=C+DR (Where F is the total Fare, C is the Fixed part of the Fare, D is the Distance in km travelled (variable part) and R is the rate per km (fixed part).
So, 160=30+10R, So, 10R= 160-30=130. So, R=13.
So, the fare for 20 km would be F=30+20R = 30+13×20= 30+260=Rs. 290