Long Question 6

Question 6:
Diagram below shows part of the curve   $y=\frac{2}{{\left(3x-2\right)}^2}$  which passes through B (1, 2).


(a) Find the equation of the tangent to the curve at the point B.
(b) A region is bounded by the curve, the x-axis and the straight lines x = 2 and x = 3.
(i) Find the area of the region.
(ii) The region is revolved through 360° about the x–axis. Find the volume generated, in terms of p.
Solution:
(a)
$\begin{array}{l}y=\frac{2}{{\left(3x-2\right)}^2}=2{\left(3x-2\right)}^{-2}\\ \frac{dy}{dx}=-4{\left(3x-2\right)}^{-3}\left(3\right)\\ \frac{dy}{dx}=\frac{-12}{{\left(3x-2\right)}^3}\\ \frac{dy}{dx}=\frac{-12}{{\left(3\left(1\right)-2\right)}^3}\text{,}x=1\\ \frac{dy}{dx}=-12\\ y-2=-12\left(x-1\right)\\ y-2=-12x+12\\ y=-12x+14\end{array}$


(b)(i)
$\begin{array}{l}\text{Area =}{\displaystyle {\int }_2^{3}ydx}\\ ={\displaystyle {\int }_2^3\frac{2}{{\left(3x-2\right)}^2}dx=}{\displaystyle {\int }_2^{3}2}{\left(3x-2\right)}^{-2}dx\\ ={\left[\frac{2{\left(3x-2\right)}^{-1}}{-1\left(3\right)}\right]}_2^3\\ ={\left[-\frac{2}{3\left(3x-2\right)}\right]}_2^3\\ =\left[-\frac{2}{3\left[3\left(3\right)-2\right]}\right]-\left[-\frac{2}{3\left[3\left(2\right)-2\right]}\right]\\ =-\frac{2}{21}+\frac{1}{6}\\ =\frac{1}{14}{\text{unit}}^2\end{array}$


(b)(ii)
$\begin{array}{l}\text{Volume generated}\\ \text{=}\pi {\displaystyle \int y^{2}dx}\\ =\pi {\displaystyle {\int }_2^3\frac{4}{{\left(3x-2\right)}^4}}dx=\pi {\displaystyle {\int }_2^3\text{4}{\left(3x-2\right)}^{-4}}dx\\ =\pi {\left[\frac{\text{4}{\left(3x-2\right)}^{-3}}{-3\left(3\right)}\right]}_2^3=\pi {\left[-\frac{\text{4}}{9{\left(3x-2\right)}^3}\right]}_2^3\\ =\pi \left[-\frac{\text{4}}{9{\left[3\left(3\right)-2\right]}^3}\right]-\left[-\frac{\text{4}}{9{\left[3\left(2\right)-2\right]}^3}\right]\\ =\pi \left(-\frac{4}{3087}+\frac{4}{576}\right)\\ =\frac{31}{5488}\pi {\text{unit}}^3\end{array}$