SOLVING VARIATION PROBLEMS

SOLVING VARIATION PROBLEMS

DIRECT VARIATION

It is a special function which can be expressed as the equation y = kx where k is a constant. The equation y = kx is read “y varies directly as x” or “y is proportional to x. The constant k is called the the constant of variation or constant of proportionality.

Illustration Number One :

The circumference ( C ) of a circle varies directly as the diameter ( d ). The direct variation is written as C = ∏ d . The constant of variation is ∏ .

Illustration Number Two :

A teacher makes $10 per hour. The total wage of the teacher is directly proportional to the number of hours (h ) worked. The equation of variation is W = 10h. The constant of proportionality is 10.

A direct variation equation can also be written in the form y = kx^n, where n is a positive number. For example the equation y = kx^2 is read ” y varies directly as the square of x..

Illustration Number Three :

The area ( A ) of a circle varies directly as the square of a radius ( r ) of the circle.

The direct variation equation is A = ∏ r ^2. The constant of variation is ∏.

Sample Problem Number One :

Given that V varies directly as r and that V = 50 when r = 5 , find the constant of

Variation and the equation of variation.

First, write the basic direct variation equation: V = kr

Then, replace V and r by the given values : 50 = k * 5

Solve for k : (1/5) 50 = 5k (1/5 ==è k = 10

Write the direct variation equation by substituting the values of k into the basic direct

variation equation: V = 10r

Sample Problem Number Two :

The tension ( T ) in a spring varies directly as the distance ( x ) it is stretched.

If T = 20 lbs. when x = 5 inches. Find T when x = 10 inches.

Write the basic direct variation equation : T = kx

Replace T and x by the given value then solve for k :

20 = 5 * k =è (1/5) 20 = 5k ( 1/5) ===è k = 4

Write the direct variation equation by substituting the value of k into the basic direct

Variation equation : T = 4x

To find T when x = 10 inches. Substitute 10 for x in the equation to solve for T.

T = 4 (10) = 40 lbs.

INVERSE VARIATION

It is a function which can be expressed as the equation y = k/x where k is a constant. The equation y = k/x is read ” y varies inversely as x” or “y is inversely proportional to x .” In general, an inverse variation equation can be written y = k/x^n

where n is a positive number .

Illustration Number Four :

The equation y = k/x^2 is read “y varies inversely as the square of x.”

Given that P varies inversely as the square of x and that P = 10 when x = 2,

Find the variation constant and the equation of variation :.

Set the inverse variation equation : P = k/x^2

Substitute the given values to corresponding variables in the equation and solve for k :

10 = k/2^2 ==è ( 4 ) 10 = k/4 (4) =è k = 40

The constant of variation is 40. The inverse variation equation is P = 40/x^2

Sample Problem Number Three :

The length ( L ) of a rectangle with fixed area is inversely proportional to the width.

If L = 10 W = 4, find the length when w = 7.

Write inverse variation equation : L = k/W

Substitute the given values to the equation and solve for k :

10 = k/4 =è k = 40

L = 40/7 or L = 5 and 5/7

JOINT VARIATION

It is a variation wherein a variable varies directly as the product of two or more

other variables. A joint variation can be expressed as the equation Z = kXY, where

K is a constant . The equation is read as “Z varies jointly as X and Y.

Illustration Number Five :

The area ( A ) of a triangle varies jointly as the base and the height. The joint variation equation is written as A = ½ bh. The constant of variation is ½.

COMBINED VARIATION

It is a variation wherein two or more types of variation occurs at the same time.

For example in Physics, the volume ( V ) of a gas varies directly as the temperature

( T ) and inversely as the pressure ( P ). The combind variation equation is written as

V = k T/ P

Sample Problem Number Four :

The pressure P of a gas varies directly as the temperature T and inversely as the volume ( V ). When T = 50 degrees and V = 200 in ^3 P = 30lb/in. Find the pressure of a gas when T = 70 degrees and V = 300 in^3.

Write first the basic combined variation equation : P = kT/V

Replace the variables by the given values then solve for k :

30 = k (50)/ 200

30 * 200 = 50*k

(1/50) 6000 = 50 k (1/50)

k = 6000/50 = 120

P = 120 T/V =è P = 120 (70)/300

(1/300) 300 P = 8400 (1/300)

P = 8400/300 = 28 lb/ in.

SOURCE : INTERMEDIATE ALGEBRA

AN APPLIED APPROACH By Aufmann/Barker