Inverse Variation With Exponents


 
 
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

Inverse variation can be written as 

Right Option : C
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Question : 2

If y varies inversely as x ^{2} , and the constant of variation is k = , what is y when x = 10?

Right Option : A
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Explanation
Question : 3

 Suppose y is inversely proportional to the square of the x , and that y =36  when x = 5. Given  x > 0, find x when y = 49 . 

Right Option : D
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