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Inverse Variation With Exponents

Direct and Inverse Variation - Concepts

Direct Variation

Inverse Variation

Time and Work

Direct Variation With Exponents

Inverse Variation With Exponents

Class - 9th IMO Subjects

Concept Explanation

Inverse Variation With Exponents

Inverse Variation:

Two terms are said to be in inverse variation if increase or decrease of term will result in the decrease or increase of the other term respectively.

For example:let us consider the equation

$y=frac{5}{x}$

$y;=kfrac{1}{x};where ;k=5$

$y;alpha;frac{1}{x}$

Now let us calculate the value of y for different values of the x

x	1	2	3	4	5
y	5	2.5	1.66	1.25	1

If we graph y against x we get the graph below

Inverse Variation With Exponents:

Two terms are said to be in inverse variation with exponents if increase or decrease of term will result in the exponential decrease or increase of the other term respectively.

For example:let us consider the equation

$y=frac{5}{x^2}$

$y;=kfrac{1}{x^2};where ;k=5$

$y;alpha;frac{1}{x^2}$

Now let us calculate the value of y for different values of the $x^2$ . Here we include in the table a row for the values of $x^2$ :

x	1	2	3	4	5
$x^{2}$	1	4	9	16	25
y	5	1.25	0.56	0.31	0.20

If we plot the value y against x we get the graph below. From the graph we can infer that there is an steep fall in the value of y when the value of x increases

Illustration: Suppose y is inversely proportional to the square of the x , and that y =36 when x = 5

(a) find y when x = 15 (b) given $x > 0$ , find x when y = 49 .

Solution: According to the question it is given that

$y propto frac{1}{x^{2}}$

(a) It is given that when x= 5 the value of y = 36 . To find the value of y when x = 15

x	5	15
y	36	?

we see that the new value of x is obtained when x is multiplied by 3

$therefore x^{2};is ;multiplied; by;3^2=9$

$Rightarrow frac{1}{x^{2}} ;is ;multiplied; by;frac{1}{9}$

$Rightarrow y ;is ;multiplied; by;frac{1}{9};;;;[as;y;alpha;frac{1}{x^2}]$

$Rightarrow y = 36timesfrac{1}{9 } = 4$

(b) It is given that when x= 5 the value of y = 36 . To find the value of y when x = 15

x	5	?
y	36	49

we see that the new value of y is obtained when present value of y is multiplied by 49 and divided by 36

$Rightarrow y ;is; multiplied; by; frac{49}{36}$

$Rightarrow frac{1}{x^2};is;multiplied;by; frac{49}{36};;;;left [ as....ypropto frac{1}{x^{2}} right ]$

$Rightarrow x^2;is;multiplied;by; frac{36}{49}$

$Rightarrow x;is;multiplied;by;sqrt{frac{36}{49}} = frac{6}{7}$ $left [ as...............x> 0 right ]$

$Rightarrow x = 5timesfrac{6}{7} = frac{30}{7}$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

$y^{4}propto frac{1}{x^{3}}$ , in this expression which power of x is in inverse relation with the fourth power of y.
A. 4	B. 3	C. 2	D. 1
Right Option : B
View Explanation

Question : 2

Suppose that y varies inversely as x ² and that y = 10 when $x = frac{5}{2}$ . Find the equation connecting x and y.
A. $yx^{2} = frac{125}{2}$	B. $yx= frac{125}{2}$	C. $yx^{2} = 125$	D. $yx^{2} = frac{125}{22}$
Right Option : A
View Explanation

Question : 3

If y varies inversely as $x ^{2}$ , and the constant of variation is k = , what is y when x = 10?
A. $y = frac{1}{4}$	B. $Y = frac{1}{5}$	C. Y = 4	D. Y = 5
Right Option : A
View Explanation

Chapters

Polynomials II

Direct and Inverse Variation

Linear Equations in One Variable

Introduction to Trigonometry

Introduction to Euclids Geometry

Herons Formula Complete

Polynomial I

Surface Area and Volume II

Surface Area and Volume I

Coordinate Geometry

Constructions

Areas Of Parallelograms And Triangles II

Areas Of Parallelograms And Triangles I

Linear Equations in Two Variables

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Inverse Variation With Exponents

Inverse Variation With Exponents

Inverse Variation:

Inverse Variation With Exponents:

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Archit Segal

Eshan Arora

Barkha Arora

Hiya Gupta

Om Umang

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Manish Kumar

Aman Kumar Shrivastava

Prisha Gupta

Yatharthi Sharma