Application of Linear Law

Example:

The variables \(x\) and \(y\) are related by the equation \(y=3x^2-\dfrac{q}{x}\), where \(q\) is a constant.

A straight line is obtained by plotting \(xy\) against \(x^3\), as shown in the diagram.

Find the value of \(h\) and of \(q\).

Solution:

For linear graph, \(Y=mX+c\),

\(X\) represents variables on the horizontal axis,

\(Y\) represents variables on the vertical axis,

\(m\) represents gradient and

\(c\) represents \(Y\)-intercept.

We can find the gradient of graph, \(m\) using this formula

\(m=\dfrac{y_2-y_1}{x_2-x_1}\),

Then,

\(\begin{aligned} m&=\dfrac{13-1}{h-0} \\\\ &=\dfrac{12}{h}.\end{aligned}\)

Based on the graph, we know that the \(Y\)-intercept , \(c=9\).

Then,

\(\begin{aligned} Y&=mX+c \\\\ xy&=\dfrac{12}{h}x^3+1\\\\ y&=\dfrac{12}{h} \dfrac{x^3}{x}+\dfrac{1}{x}\\\\ y&=\dfrac{12}{h} x^2+\dfrac{1}{x}. \end{aligned}\)

We compare the equation that we get with the given equation.

\(\begin{aligned} 3x^2-\dfrac{q}{x} =\dfrac{12}{h}x^2+\dfrac{1}{x} \\\\ \end{aligned}\)

Coefficient of \(\boldsymbol{x^2}\)

\(\begin{aligned} 3x^2&=\dfrac{12}{h}x^2 \\\\ 3&=\dfrac{12}{h} \\\\ 3h&=12 \\\\ h&=\dfrac{12}{3} \\\\ &=4. \end{aligned}\)

Coefficient of \(\boldsymbol{\dfrac{1}{x}}\)

\(\begin{aligned} -\dfrac{q}{x} &=\dfrac{1}{x} \\\\ -q&=1 \\\\ q&=-1. \end{aligned}\)