 Variations

 Direct variation

 Definition of direct variation Direct variation explains the relationship between two variables, such that when variable $$y$$ increases, then variable $$x$$ also increases at the same rate and vice versa. This relation can be written as $$y$$ varies directly as $$x$$ . In general, for a direct variation, $$y$$ varies directly as $$x^n$$ can be written as \begin{aligned}x\propto x^n\end{aligned}\hspace{1mm}\text{(variation relation)} or \begin{aligned} x=kx^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 1 Given $$m=12$$ when $$n=3$$. Express $$m$$ in terms of $$n$$ if a) $$m$$ varies directly as $$n$$. b) $$m$$ varies directly as $$n^3$$. Solution: a) $$n\implies m = kn \dots (1)$$. Substitute $$m=12$$ and $$n=3$$ into $$(1)$$ $$12=k(3)\implies k=\dfrac{12}{3}=4$$. $$\therefore m=4n$$. b) $$m\propto n^3\implies m = ln^3 \dots (2).$$ Substitute $$m=12$$ and $$n=3$$ into $$(2)$$: $$12=l(3)^3\implies l=\dfrac{12}{27}=\dfrac{4}{9}$$ $$\therefore m=\dfrac{4}{9}n^3.$$

 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}

 Joint Variation

 Definition joint variation In general, for a combined variation, $$y$$ varies directly as $$x^m$$ and inversely as $$z^n$$ can be written as \begin{aligned}x\propto\frac{x^m}{z^n}\end{aligned}\hspace{1mm}\text{(variation relation)} or \begin{aligned} x=\frac{kx^m}{z^n} \end{aligned} \hspace{1mm} \text{(equation relation)} such that \begin{aligned} m&=1,2,3,\frac{1}{2},\frac{1}{3},\hspace{1mm} \\\\n&=1,2,3,\frac{1}{2},\frac{1}{3} \end{aligned} and $$k$$ is a constant.
 Example 3 Given that $$y$$ varies directly as then square of $$x$$  varies inversely as square root of $$z$$. If $$y=8$$ when $$x=4$$ and $$z=36$$, express $$y$$ in terms of $$x$$ and $$z$$. Solution: \begin{aligned}\hspace{1mm}& y\propto \frac{x^2}{\sqrt{2}}\implies y = \frac{kx^2}{\sqrt{2}} \dots (1). \end{aligned} Substitute $$y=8$$, $$x=4$$, and $$z=36$$ into $$(1)$$: \begin{aligned}8&=\frac{k4^2}{\sqrt{36}}\implies k=\frac{(8)(6)}{16}\\\\&=3.\\\\ &\therefore y=\frac{3x^2}{\sqrt{z}}. \end{aligned}