Inverse Variation

1.2

 Inverse Variation

 
Definition inverse variation
 
In inverse variation, variable \(y\) increases when the variable \(x\) decreases at the same rate, and vice versa. This relation can be written as \(y\) varies inversely as \(x\)
 
In general,For an inverse variation, \(y\) varies inversely as \(x^n\) can be written as \(x^n\) 
 

\(\begin{aligned}y\propto \frac{1}{x^n}\end{aligned}\hspace{1mm}\text{(variation relation)}\) or

\(\begin{aligned} y=\frac{k}{x^n} \end{aligned} \hspace{1mm} \text{(equation relation)}\)

 
where \(\begin{aligned} n=1,2,3,\frac{1}{2},\frac{1}{3} \end{aligned}\) and \(k\) is a constant.
 
Example 2
 

Given \(y=3\) when \(x=7\).

Express \(y\) in terms of \(x\) if 

a) \(y\) varies inversely as \(x\).
b) \(y\) varies inversely as \(x^2\).
 
Solution:
 

\(\begin{aligned}a)\hspace{1mm}& y\propto \frac{1}{x}\implies y = \frac{k}{x} \dots (1) \end{aligned}\)

Substitute \(y=3\) and \(x=7\) into \((1)\):

\(\begin{aligned}3&=\frac{k}{7}\implies k=(3)(7)\\\\&=21.\\\\&\therefore y=\dfrac{21}{x}. \end{aligned}\)

\(\begin{aligned}b)\hspace{1mm}& y\propto \frac{1}{x^2}\implies y = \frac{l}{x^2} \dots (2) \end{aligned}\)

Substitute \(y=3\) and \(x=7\) into \((2)\):

\(\begin{aligned}3&=\frac{l}{7^2}\implies k=(3)(49)\\\\&=147.\\\\ &\therefore y=\dfrac{147}{x^2}. \end{aligned}\)

 

Inverse Variation

1.2

 Inverse Variation

 
Definition inverse variation
 
In inverse variation, variable \(y\) increases when the variable \(x\) decreases at the same rate, and vice versa. This relation can be written as \(y\) varies inversely as \(x\)
 
In general,For an inverse variation, \(y\) varies inversely as \(x^n\) can be written as \(x^n\) 
 

\(\begin{aligned}y\propto \frac{1}{x^n}\end{aligned}\hspace{1mm}\text{(variation relation)}\) or

\(\begin{aligned} y=\frac{k}{x^n} \end{aligned} \hspace{1mm} \text{(equation relation)}\)

 
where \(\begin{aligned} n=1,2,3,\frac{1}{2},\frac{1}{3} \end{aligned}\) and \(k\) is a constant.
 
Example 2
 

Given \(y=3\) when \(x=7\).

Express \(y\) in terms of \(x\) if 

a) \(y\) varies inversely as \(x\).
b) \(y\) varies inversely as \(x^2\).
 
Solution:
 

\(\begin{aligned}a)\hspace{1mm}& y\propto \frac{1}{x}\implies y = \frac{k}{x} \dots (1) \end{aligned}\)

Substitute \(y=3\) and \(x=7\) into \((1)\):

\(\begin{aligned}3&=\frac{k}{7}\implies k=(3)(7)\\\\&=21.\\\\&\therefore y=\dfrac{21}{x}. \end{aligned}\)

\(\begin{aligned}b)\hspace{1mm}& y\propto \frac{1}{x^2}\implies y = \frac{l}{x^2} \dots (2) \end{aligned}\)

Substitute \(y=3\) and \(x=7\) into \((2)\):

\(\begin{aligned}3&=\frac{l}{7^2}\implies k=(3)(49)\\\\&=147.\\\\ &\therefore y=\dfrac{147}{x^2}. \end{aligned}\)