### Variations- (6.1-6.3)

#### Understand direct variations (direct proportions) and inverse variations (inverse proportions), and their applications to solving real-life problems

• It is given that y varies inversely as $$3{x^2} + 1$$, and $$y = 30$$ when $$x = 3$$.
Find the value of y when $$x = \frac{1}{2}$$.
• $$y \propto \frac{1}{{3{x^2} + 1}}$$
$$y = \frac{k}{{3{x^2} + 1}}$$, where k is a non-zero 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 joint and partial variations, and their applications to solving real-life problems

• z 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.
• a) Let $$z = \frac{{k{x^2}}}{y}$$, where k is a non-zero 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
$$\begin{array}{l} = \frac{{{\textstyle{{1.92{x_1}^2} \over {{y_1}}}} - {\textstyle{{8{x_1}^2} \over {{y_1}}}}}}{{{\textstyle{{8{x_1}^2} \over {{y_1}}}}}} \times 100\% \\ = \frac{{1.92 - 8}}{8} \times 100\% \\ = \underline{\underline { - {\rm{ }}76\% }} \end{array}$$