Theorem of Joint Variation

Here we will discuss about the Theorem of Joint Variation with the detailed explanation.

The theorem of joint variation can be established by stating the relationship between three variables which are separately in direct variation with each other.


Theorem of Joint Variation: If x ∝ y when z is constant and x ∝ z when y is constant, then x ∝ yz when both y and z vary.

Proof: 

Since x ∝ y when z is constant.

Therefore x = ky where k = constant of variation and is independent to the changes of x and y that means the value of K doesn’t change for any value of X and Y. 


Again, x ∝ z when y is constant. 

or, ky ∝ z when y is constant (By putting ky in place of x we get). 

or, k ∝ z (y is constant). 

or, k = mz where m is a constant which is independent to the changes of k and z that means the value of m doesn’t change for any value of k and z.

Now, the value of k is independent to the changes of x and y. Hence, the value of m is independent to the changes of x, y and z.

Therefore x = ky = myz (since, k = mz)

where m is a constant whose value does not depend on x, y and z.

Therefore x ∝ yz when both y and z vary. 

Note: (i) The above theorem can be extended for a longer number of variables. For example, if A ∝ B when C and D are constants, A ∝ C when B and D are constants and A ∝ D when B and C are constants, thee A ∝ BCD when B, C and D all vary. 


(ii) If x ∝ y when z is constant and x ∝ 1/Z when y is constant, then x ∝ y when both y and z vary.


So in this theorem we use the principle of direct variation to prove that how joint variation works for to establish a correlation among more than two variables.

For solving a problems related to the theory of joint variation first we need to solve by following steps.

1. Build the correct equation by adding a constant and relate the variables. 

2. We need to determine the value of the constant from the given data. 

3. Substitute the value of the constant in the equation.

4. Put the values of variables for required situation and determine the answer.


Now we will see some problems and solutions related to the theorem of joint variation:

1. The variable x is in joint variation with y and z. When the values of y and z are 2 and 3, x is 16. What is the value of x when y = 8 and z =12?

The equation for the given problem of joint variation is

x = Kyz where K is the constant.

For the given data

16 = K × × 3

or, K = \(\frac{8}{3}\)

So substituting the value of K the equation becomes

x = \(\frac{8yz}{3}\)

Now for the required condition

x = \(\frac{8 × 8 × 12}{3}\) = 256

Hence the value of x will be 256.


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?

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

Some Useful Results:

Theorem of Joint Variation

(i) If A ∝ B, then B ∝ A.

(ii) If A ∝ B and B∝ C, then A ∝ C.


(iii) If A ∝ B, then Aᵇ ∝ Bᵐ where m is a constant.

(iv) If A ∝ BC, then B ∝ A/C and C ∝ A/B.

(v) If A ∝ C and B ∝ C, then A + B ∝ C and AB ∝ C²

(vi) If A ∝ B and C ∝ D, then AC ∝ BD and A/C ∝ B/D



Now we are going to proof the useful results with step-by-step detailed explanation

Proof: (i) If A ∝ B, then B ∝ A.

Since, A ∝ B Therefore A = kB, where k = constant.

or, B = 1/K ∙ A Therefore B ∝ A. (since,1/K = constant)


Proof: (ii) If A ∝ B and B ∝ C, then A ∝ C.

Since, A ∝ B Therefore A = mB where, m = constant

Again, B ∝ C Therefore B = nC where n= constant.

Therefore A= mB = mnC = kC where k = mn = constant, as m and n are both Constants.

Therefore A ∝ C.

Proof: (iii) If A ∝ B, then Aᵇ ∝ Bᵐ where m is a constant.

Since A ∝ B Therefore A = kB where k= constant.

Aᵐ = KᵐBᵐ = n ∙ Bᵐ where n = kᵐ = constant, as k and m are both constants.

Therefore Aᵐ ∝ Bᵐ.

Results (iv), (v) and (vi) can be deduced by similar procedure.

Summarisation:

(i) If A varies directly as B, then A ∝ B or, A = kB where k is the constant of variation. Conversely, if A = kB i.e., A/B = k where k is a constant, then A varies directly as B.

(ii) If A varies inversely as B, then A ∝ 1/B or, A= m ∙ 1/B or, AB= m where m = constant of variation. Conversely, if AB = k (a constant), then A varies inversely as B.

(iii) If A varies jointly as B and C, then A ∝ BC or A = kBC where k = constant of variation.



 Variation




11 and 12 Grade Math 

From Theorem of Joint Variation to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Multiplying 3-Digit Number by 1-Digit Number | Three-Digit Multiplicat

    Oct 22, 24 03:26 PM

    Multiplying 3-Digit Number by 1-Digit Number
    Here we will learn multiplying 3-digit number by 1-digit number. In two different ways we will learn to multiply a two-digit number by a one-digit number. 1. Multiply 201 by 3 Step I: Arrange the numb…

    Read More

  2. Word Problems on Multiplication |Multiplication Word Problem Worksheet

    Oct 22, 24 01:23 AM

    Multiplication Word Problem
    Word problems on multiplication for fourth grade students are solved here step by step. Problem Sums Involving Multiplication: 1. 24 folders each has 56 sheets of paper inside them. How many sheets of…

    Read More

  3. Worksheet on Word Problems on Multiplication | Multiplication Problems

    Oct 22, 24 12:31 AM

    In worksheet on word problems on multiplication, all grade students can practice the questions on word problems involving multiplication. This exercise sheet on word problems on multiplication

    Read More

  4. Multiplying 2-Digit Number by 1-Digit Number | Multiply Two-Digit Numb

    Oct 21, 24 03:38 PM

    Multiplying 2-Digit Number by 1-Digit Number
    Here we will learn multiplying 2-digit number by 1-digit number. In two different ways we will learn to multiply a two-digit number by a one-digit number. Examples of multiplying 2-digit number by

    Read More

  5. Multiplication Table of 4 |Read and Write the Table of 4|4 Times Table

    Oct 21, 24 02:26 AM

    Multiplication Table of Four
    Repeated addition by 4’s means the multiplication table of 4. (i) When 5 candle-stands having four candles each. By repeated addition we can show 4 + 4 + 4 + 4 + 4 = 20 Then, four 5 times

    Read More