Bab 14 Pengamiran

Posted on May 17, 2016 by user

3.8.5 Pengamiran, SPM Praktis (Kertas 2)

Soalan 6:

Rajah di bawah menunjukkan sebahagian daripada lengkung

$y = \frac{2}{{(3 x - 2)}^{2}}$ yang melalui B (1, 2).

(a) Carikan persamaan tangen kepada lengkung itu pada titik B.
(b) Suatu rantau dibatasi oleh lengkung itu, paksi-x, garis lurus x = 2 dan x = 3.
(i) Cari luas rantau yang berlorek.
(ii) Rantau itu dikisarkan melalui 360^o pada paksi-x.
Carikan isipadu janaan, dalam sebutan π.

Penyelesaian:

(a)

$\begin{array}{l} y = \frac{2}{{(3 x - 2)}^{2}} = 2 {(3 x - 2)}^{- 2} \\ \frac{d y}{d x} = - 4 {(3 x - 2)}^{- 3} (3) \\ \frac{d y}{d x} = \frac{- 12}{{(3 x - 2)}^{3}} \\ \frac{d y}{d x} = \frac{- 12}{{(3 (1) - 2)}^{3}}, x = 1 \end{array}$

y – 2 = –12 (x – 1)

y – 2 = –12x + 12

y = –12x + 14

(b)(i)

$\begin{array}{l} Area = \int_{2}^{3} y d x \\ = \int_{2}^{3} \frac{2}{{(3 x - 2)}^{2}} d x = \int_{2}^{3} 2 {(3 x - 2)}^{- 2} d x \\ = {[\frac{2 {(3 x - 2)}^{- 1}}{- 1 (3)}]}_{2}^{3} \\ = {[- \frac{2}{3 (3 x - 2)}]}_{2}^{3} \\ = [- \frac{2}{3 [3 (3) - 2]}] - [- \frac{2}{3 [3 (2) - 2]}] \\ = - \frac{2}{21} + \frac{1}{6} \\ = \frac{1}{14} {unit}^{2} \end{array}$

(b)(ii)

Isipadu janaan

$\begin{array}{l} = π \int y^{2} d x \\ = π \int_{2}^{3} \frac{4}{{(3 x - 2)}^{4}} d x = π \int_{2}^{3} 4 {(3 x - 2)}^{- 4} d x \\ = π {[\frac{4 {(3 x - 2)}^{- 3}}{- 3 (3)}]}_{2}^{3} = π {[- \frac{4}{9 {(3 x - 2)}^{3}}]}_{2}^{3} \\ = π [- \frac{4}{9 {[3 (3) - 2]}^{3}}] - [- \frac{4}{9 {[3 (2) - 2]}^{3}}] \\ = π (- \frac{4}{3087} + \frac{4}{576}) \\ = \frac{31}{5488} π {unit}^{3} \end{array}$