Joint Variation

If more than two variables are related directly or one variable changes with the change product of two or more variables it is called as joint variation.

If X is in joint variation with Y and Z, it can be symbolically written as X α YZ. If Y is constant also then X is in direct variation with Z. So for joint variation two or more variables are separately in direct variation. So joint variation is similar to direct variation but the variables for joint variation are more than two.

Equation for a joint variation is X = KYZ where K is constant.

One variable quantity is said to vary jointly as a number of other variable quantities, when it varies directly as their product. If the variable A varies directly as the product of the variables B, C and D, i.e., if.A ∝ BCD or A = kBCD (k = constant ), then A varies jointly as B, C and D. 

For solving a problems related to joint variation first we need to build the correct equation by adding a constant and relate the variables. After that we need determine the value of the constant. Then substitute the value of the constant in the equation and by putting the values of variables for required situation we determine the answer.

We know, area of a triangle = ½ × base × altitude. Since ½ is a constant, hence area of a triangle varies jointly as its base and altitude. 

A is said to vary directly as B and inversely as C if A ∝ B ∙ \(\frac{1}{C}\) or A = m ∙ B ∙ \(\frac{1}{C}\) (m = constant of variation) i.e., if A varies jointly as B and \(\frac{1}{C}\).


If x men take y days to plough z acres of land, then x varies directly as z and inversely as y.


1. The variable x is in joint variation with y and z. When the values of y and z are 4 and 6, x is 16. What is the value of x when y = 8 and z =12?

Solution:

The equation for the given problem of joint variation is

x = Kyz where K is the constant.

For the given data

16 = K × 4 × 6

or, K = \(\frac{4}{6}\).

So substituting the value of K the equation becomes

x = \(\frac{4yz}{6}\)

Now for the required condition

x = \(\frac{4 × 8 × 12}{6}\)

   = 64

Hence the value of x will be 64.


2. A is in joint variation with B and square of C. When A = 144, B = 4 and C = 3. Then what is the value of A when B = 6 and C = 4?

Solution:

From the given problem equation for the joint variation is

A = KBC2

From the given data value of the constant K is

K = \(\frac{BC^{2}}{A}\)

K = \(\frac{4 × 3^{2}}{144}\)

   = \(\frac{36}{144}\)

    = \(\frac{1}{4}\).

Substituting the value of K in the equation

A = \(\frac{BC^{2}}{4}\)

A = \(\frac{6 × 4^{2}}{4}\)

   = 24


3. The area of a triangle is jointly related to the height and the base of the triangle. If the base is increased 10% and the height is decreased by 10%, what will be the percentage change of the area?

Solution:

We know the area of triangle is half the product of base and height. So the joint variation equation for area of triangle is A = \(\frac{bh}{2}\) where A is the area, b is the base and h is the height.

Here \(\frac{1}{2}\) is the constant for the equation.

Base is increased by 10%, so it will be b x \(\frac{110}{100}\) = \(\frac{11b}{10}\).

Height is decreased by 10%, so it will be h x \(\frac{90}{100}\) = \(\frac{9h}{10}\).

So the new area after the changes of base and height is

\(\frac{\frac{11b}{10} \times \frac{9h}{10}}{2}\)

= (\(\frac{99}{100}\))\(\frac{bh}{2}\) = \(\frac{99}{100}\)A.

So the area of the triangle is decreased by 1%.


4. A rectangle’s length is 6 m and width is 4 m. If length is doubled and width is halved, how much the perimeter will increase or decrease?

Solution:

Formula for the perimeter of rectangle is P = 2(l + w) where P is perimeter, l is length and w is width.

This is joint variation equation where 2 is constant.

So P = 2(6 + 4) = 20 m

If length is doubled, it will become 2l.

And width is halved, so it will become \(\frac{w}{2}\).

So the new perimeter will be P = 2(2l + \(\frac{w}{2}\)) = 2(2 x 6 + \(\frac{4}{2}\)) = 28 m.

So the perimeter will increase by (28 - 20) = 8 m.

 Variation







11 and 12 Grade Math 

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