Soalan 8:
Rajah 4 menunjukkan satu lengkung dan tangen kepada lengkung itu pada titik A.
Cari
(a) nilai p, [1 markah]
(b) luas rantau berwarna, [4 markah]
(c) isi padu janaan, dalam sebutan π, apabila rantau berlorek dikisarkan melalui 180o pada paksi-y. [4 markah]
Penyelesaian:
(a)
$$ \begin{aligned} & y=x^2-9 \ldots (1)\\ & \frac{x}{5}-\frac{y}{10}=1 \\ & 2 x-y=10 \\ & 2 x-10=y \ldots (2) \end{aligned} $$
$$ \begin{aligned} &\text { (1) = (2), }\\ &\begin{aligned} x^2-9 & =2 x-10 \\ x^2-2 x-9+10 & =0 \\ x^2-2 x+1 & =0 \\ (x-1)(x-1) & =0 \\ x-1 & =0 \\ x & =1 \end{aligned}\\ &\therefore p=1 \end{aligned} $$
(b)
$$ \begin{aligned} & y=x^2-9 \\ & y=(x+3)(x-3) \end{aligned} $$
$$ \begin{array}{rlrl} \text { Apabila } y=0, & (x+3)(x-3)=0 & \\ x+3=0, & x-3=0 \\ x=-3, & x=3 \end{array} $$
$$ \begin{aligned} \text { Luas } & =\left|\int_1^3\left(x^2-9\right) \mathrm{d} x\right| \\ & =\left|\left[\frac{1}{3} x^3-9 x\right]_1^3\right| \\ & =\left|\left[\frac{1}{3}(3)^3-9(3)\right]-\left[\frac{1}{3}(1)^3-9(1)\right]\right| \\ & =\left|-18-\left(-\frac{26}{3}\right)\right| \\ & =\left|-\frac{28}{3}\right| \\ & =\frac{28}{3} \text { unit }^2 \end{aligned} $$
(c)
$$ \begin{aligned} &\text { Dari (2), } y=2 x-10 \text { apabila } x=0\\ &\begin{aligned} & y=2(0)-10 \\ & y=-10 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & y=2 x-10 \text { apabila } y=0 \\ & \quad 0=2 x-10 \\ & -2 x=-10 \\ & x=5 \end{aligned} $$
$$ \begin{aligned} \text { Isi padu kon } & =\frac{1}{3} \pi r^2 h \\ & =\frac{1}{3} \pi(5)^2(10) \\ & =83 \frac{1}{3} \pi \text { unit }^3 \end{aligned} $$
$$ \begin{aligned} & \text { Dari (1), } y=x^2-9 \text { apabila } x=0\\ &\begin{aligned} & y=(0)^2-9 \\ & y=-9 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Luas lengkung }\\ &\begin{aligned} & =\pi \int_{-9}^0 x^2 d y=\pi \int_{-9}^0(y+9) d y \\ & =\pi\left[\frac{1}{2} y^2+9 y\right]_{-9}^0 \\ & =\pi\left[\frac{1}{2}(0)^2+9(0)\right]-\pi\left[\frac{1}{2}(-9)^2+9(-9)\right] \\ & =0-\left(-\frac{81}{2} \pi\right) \\ & =40 \frac{1}{2} \pi \text { unit }^3 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & V_{\text {lorekan dikisar } 180^{\circ}}=\frac{1}{2}(\text { Isi padu kon }- \text { Isi padu lengkung }) \\ & =\frac{1}{2}\left(83 \frac{1}{3} \pi-40 \frac{1}{2} \pi\right) \\ & =\frac{1}{2}\left(42 \frac{5}{6} \pi\right) \text { unit }^3 \\ & =21 \frac{5}{12} \pi \text { unit }^3 \end{aligned} $$
Rajah 4 menunjukkan satu lengkung dan tangen kepada lengkung itu pada titik A.
Cari(a) nilai p, [1 markah]
(b) luas rantau berwarna, [4 markah]
(c) isi padu janaan, dalam sebutan π, apabila rantau berlorek dikisarkan melalui 180o pada paksi-y. [4 markah]
Penyelesaian:
(a)
$$ \begin{aligned} & y=x^2-9 \ldots (1)\\ & \frac{x}{5}-\frac{y}{10}=1 \\ & 2 x-y=10 \\ & 2 x-10=y \ldots (2) \end{aligned} $$
$$ \begin{aligned} &\text { (1) = (2), }\\ &\begin{aligned} x^2-9 & =2 x-10 \\ x^2-2 x-9+10 & =0 \\ x^2-2 x+1 & =0 \\ (x-1)(x-1) & =0 \\ x-1 & =0 \\ x & =1 \end{aligned}\\ &\therefore p=1 \end{aligned} $$
(b)
$$ \begin{aligned} & y=x^2-9 \\ & y=(x+3)(x-3) \end{aligned} $$
$$ \begin{array}{rlrl} \text { Apabila } y=0, & (x+3)(x-3)=0 & \\ x+3=0, & x-3=0 \\ x=-3, & x=3 \end{array} $$
$$ \begin{aligned} \text { Luas } & =\left|\int_1^3\left(x^2-9\right) \mathrm{d} x\right| \\ & =\left|\left[\frac{1}{3} x^3-9 x\right]_1^3\right| \\ & =\left|\left[\frac{1}{3}(3)^3-9(3)\right]-\left[\frac{1}{3}(1)^3-9(1)\right]\right| \\ & =\left|-18-\left(-\frac{26}{3}\right)\right| \\ & =\left|-\frac{28}{3}\right| \\ & =\frac{28}{3} \text { unit }^2 \end{aligned} $$
(c)
$$ \begin{aligned} &\text { Dari (2), } y=2 x-10 \text { apabila } x=0\\ &\begin{aligned} & y=2(0)-10 \\ & y=-10 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & y=2 x-10 \text { apabila } y=0 \\ & \quad 0=2 x-10 \\ & -2 x=-10 \\ & x=5 \end{aligned} $$
$$ \begin{aligned} \text { Isi padu kon } & =\frac{1}{3} \pi r^2 h \\ & =\frac{1}{3} \pi(5)^2(10) \\ & =83 \frac{1}{3} \pi \text { unit }^3 \end{aligned} $$
$$ \begin{aligned} & \text { Dari (1), } y=x^2-9 \text { apabila } x=0\\ &\begin{aligned} & y=(0)^2-9 \\ & y=-9 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Luas lengkung }\\ &\begin{aligned} & =\pi \int_{-9}^0 x^2 d y=\pi \int_{-9}^0(y+9) d y \\ & =\pi\left[\frac{1}{2} y^2+9 y\right]_{-9}^0 \\ & =\pi\left[\frac{1}{2}(0)^2+9(0)\right]-\pi\left[\frac{1}{2}(-9)^2+9(-9)\right] \\ & =0-\left(-\frac{81}{2} \pi\right) \\ & =40 \frac{1}{2} \pi \text { unit }^3 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & V_{\text {lorekan dikisar } 180^{\circ}}=\frac{1}{2}(\text { Isi padu kon }- \text { Isi padu lengkung }) \\ & =\frac{1}{2}\left(83 \frac{1}{3} \pi-40 \frac{1}{2} \pi\right) \\ & =\frac{1}{2}\left(42 \frac{5}{6} \pi\right) \text { unit }^3 \\ & =21 \frac{5}{12} \pi \text { unit }^3 \end{aligned} $$