Variations (6.16.3)
Understand direct variations (direct proportions) and inverse variations (inverse proportions), and their applications to solving reallife problems

Theory

ExamplesIt is given that y varies inversely as \(3{x^2} + 1\), and \(y = 30\) when \(x = 3\).
Find the value of y when \(x = \frac{1}{2}\). 
Solutions\(y \propto \frac{1}{{3{x^2} + 1}}\)
\(y = \frac{k}{{3{x^2} + 1}}\), where k is a nonzero constant
When \(x = 3\), \(y = 30\).
\(\begin{array}{ccccc}\therefore {\rm{ }}30 = \frac{k}{{3{{(3)}^2} + 1}}\\k = 30(27 + 1)\\k = 840\\\therefore {\rm{ }}y = \underline{\underline {\frac{{840}}{{3{x^2} + 1}}}} \end{array}\)
When \(x = \frac{1}{2}\),
\(\begin{array}{c}y = \frac{{840}}{{3{{({\textstyle{1 \over 2}})}^2} + 1}}\\ = \frac{{840}}{{{\textstyle{3 \over 4}} + 1}}\\ = \underline{\underline {480}} \end{array}\)
Understand the graphs of direct and inverse variations

Theory

Examples

Solutions
Understand joint and partial variations, and their applications to solving reallife problems

Theory

Examplesz varies directly as \({x^2}\) and inversely as y where \(y \ne 0\). \(z = 12\) when \(x = 3\) and \(y = 6\).
(a) Express z in terms of x and y.
(b) Find the value of z when \(x = 2\) and \(y = \frac{1}{3}\).
(c) If x decreases by 40% and y increases by 50%, find the percentage change in z.

Solutionsa) Let \(z = \frac{{k{x^2}}}{y}\), where k is a nonzero constant.
When \(x = 3\) and \(y = 6\), \(z = 12\).
\(\begin{array}{ccccc}\therefore {\rm{ }}12 = \frac{{k{{(3)}^2}}}{6}\\12 = k \times \frac{9}{6}\\12 = \frac{{3k}}{2}\\k = 8\\\therefore {\rm{ }}z = \frac{{8{x^2}}}{y}\end{array}\)
(b) When \(x = 2\) and \(y = \frac{1}{3}\),
\(\begin{array}{c}z = \frac{{8{{(2)}^2}}}{{{\textstyle{1 \over 3}}}}\\ = \underline{\underline {96}} \end{array}\)
(c) Let \({x_1}\) and \({y_1}\) be the original values of x and y respectively.
\(\text{Original value of z}
= \frac{{8{x_1}^2}}{{{y_1}}}\)
\(\begin{array}{l}\text{New value of x}
= {x_1}(1  40\% )\\ = 0.6{x_1}\end{array}\)
\(\begin{array}{l}\text{New value of y}
= {y_1}(1 + 50\% )\\ = 1.5{y_1}\end{array}\)
\(\begin{array}{l}\text{New value of z }
= \frac{{8{{(0.6{x_1})}^2}}}{{1.5{y_1}}}\\= \frac{{1.92{x_1}^2}}{{{y_1}}}\end{array}\)
Percentage change in z
\(\begin{array}{l} = \frac{{{\textstyle{{1.92{x_1}^2} \over {{y_1}}}}  {\textstyle{{8{x_1}^2} \over {{y_1}}}}}}{{{\textstyle{{8{x_1}^2} \over {{y_1}}}}}} \times 100\% \\ = \frac{{1.92  8}}{8} \times 100\% \\ = \underline{\underline {  {\rm{ }}76\% }} \end{array}\)