Soalan 8:
Rajah 8 menunjukkan lengkung y = f(x) bersilang dengan garis lurus y = -2x + 14 pada titik A.
Diberi bahawa fungsi kecerunan lengkung itu ialah x/2, cari
(a) persamaan lengkung itu, [3 markah]
(b) luas rantau yang dibatasi oleh lengkung, garis lurus x = 4, paksi-x dan paksi-y, [3 markah]
(c) isi padu janaan, dalam sebutan π, apabila rantau berlorek dikisarkan melalui 360° pada paksi-y. [4 markah]
Jawapan:
(a) $$ \begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{2} \\ & y=\int \frac{x}{2} \mathrm{~d} x \\ & y=\frac{x^2}{(2)(2)}+c \\ & y=\frac{x^2}{4}+c, \quad \text { pada A(4, 6) } \\ & 6=\frac{(4)^2}{4}+c \\ & 6=4+c \\ & c=2 \\ & \therefore y=\frac{1}{4} x^2+2 \end{aligned} $$
(b)
$$ \begin{aligned} \text { Luas kawasan berlorek } & =\text { Luas trapezium }- \text { Luas di bawah suatu lengkung } \\ & =\frac{1}{2}(6+14)(4)-\int_0^4\left(\frac{1}{4} x^2+2\right) \mathrm{d} x \\ & =40-\left[\frac{x^3}{(3)(4)}+2 x\right]_0^4 \\ & =40-\left[\frac{x^3}{12}+2 x\right]_0^4 \\ & =40-\left[\left(\frac{4^3}{12}+2(4)\right)-\left(\frac{0^3}{12}+2(0)\right)\right] \\ & =40-\left(\frac{40}{3}-0\right) \\ & =\frac{80}{3} \text { unit }^2 \end{aligned} $$
(c) $$ \begin{aligned} y & =\frac{1}{4} x^2+2 \\ 4(y-2) & =x^2 \\ x^2 & =4 y-8 \end{aligned} $$
$$ \begin{aligned} &\text { Isi padu }\\ &\begin{aligned} & =V_{\text {kon }}+V_{y=2 \rightarrow 6} \\ & =\pi r^2 h+\pi \int_2^6 x^2 \mathrm{~d} y \\ & =\pi(4)^2(14-6)+\pi \int_2^6(4 y-8) \mathrm{d} y \\ & =128 \pi+\pi\left[\frac{4 y^2}{2}-8 y\right]_2^6 \\ & =128 \pi+\pi\left[2 y^2-8 y\right]_2^6 \\ & =128 \pi+\left[\left(2(6)^2-8(6)\right)-\left(2(2)^2-8(2)\right)\right] \pi \\ & =128 \pi+[24-(-8)] \pi \\ & =128 \pi+32 \pi \\ & =160 \pi \text { unit }^3 \end{aligned} \end{aligned} $$
Rajah 8 menunjukkan lengkung y = f(x) bersilang dengan garis lurus y = -2x + 14 pada titik A.
Diberi bahawa fungsi kecerunan lengkung itu ialah x/2, cari(a) persamaan lengkung itu, [3 markah]
(b) luas rantau yang dibatasi oleh lengkung, garis lurus x = 4, paksi-x dan paksi-y, [3 markah]
(c) isi padu janaan, dalam sebutan π, apabila rantau berlorek dikisarkan melalui 360° pada paksi-y. [4 markah]
Jawapan:
(a) $$ \begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{2} \\ & y=\int \frac{x}{2} \mathrm{~d} x \\ & y=\frac{x^2}{(2)(2)}+c \\ & y=\frac{x^2}{4}+c, \quad \text { pada A(4, 6) } \\ & 6=\frac{(4)^2}{4}+c \\ & 6=4+c \\ & c=2 \\ & \therefore y=\frac{1}{4} x^2+2 \end{aligned} $$
(b)
$$ \begin{aligned} \text { Luas kawasan berlorek } & =\text { Luas trapezium }- \text { Luas di bawah suatu lengkung } \\ & =\frac{1}{2}(6+14)(4)-\int_0^4\left(\frac{1}{4} x^2+2\right) \mathrm{d} x \\ & =40-\left[\frac{x^3}{(3)(4)}+2 x\right]_0^4 \\ & =40-\left[\frac{x^3}{12}+2 x\right]_0^4 \\ & =40-\left[\left(\frac{4^3}{12}+2(4)\right)-\left(\frac{0^3}{12}+2(0)\right)\right] \\ & =40-\left(\frac{40}{3}-0\right) \\ & =\frac{80}{3} \text { unit }^2 \end{aligned} $$
(c) $$ \begin{aligned} y & =\frac{1}{4} x^2+2 \\ 4(y-2) & =x^2 \\ x^2 & =4 y-8 \end{aligned} $$
$$ \begin{aligned} &\text { Isi padu }\\ &\begin{aligned} & =V_{\text {kon }}+V_{y=2 \rightarrow 6} \\ & =\pi r^2 h+\pi \int_2^6 x^2 \mathrm{~d} y \\ & =\pi(4)^2(14-6)+\pi \int_2^6(4 y-8) \mathrm{d} y \\ & =128 \pi+\pi\left[\frac{4 y^2}{2}-8 y\right]_2^6 \\ & =128 \pi+\pi\left[2 y^2-8 y\right]_2^6 \\ & =128 \pi+\left[\left(2(6)^2-8(6)\right)-\left(2(2)^2-8(2)\right)\right] \pi \\ & =128 \pi+[24-(-8)] \pi \\ & =128 \pi+32 \pi \\ & =160 \pi \text { unit }^3 \end{aligned} \end{aligned} $$