Long Question 10

Question 10 (10 marks):
Diagram shows a curve y = 2x² – 18 and the straight line PQ which is a tangent to the curve at point K.

It is given that the gradient of the straight line PQ is 4.
(a) Find the coordinates of point K
(b) Calculate the area of the shaded region.

(c) When the region bounded by the curve, the x-axis and the straight line y = h is rotated through 180^o about the y-axis, the volume generated is 65π unit³.
Find the value of h.

Solution: 
(a)
$\begin{array}{l}y=2x^2-18\\ \frac{dy}{dx}=4x\\ \\ \text{Gradient of straight line }PQ=4\\ 4x=4\\ x=1\\ \\ \text{When }x=1,\\ y=2{\left(1\right)}^2-18=-16\\ \\ \text{Coordinates of }K=\left(1,-16\right).\end{array}$


(b)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \therefore 2x^2-18=0\\ 2x^2=18\\ x^2=9\\ x=\pm 3\\ \text{The curve cuts the }x\text{-axis at }\left(-3,0\right)\text{ and }\left(3,0\right)\text{.}\\ \\ \text{Area of shaded region}\\ \text{= Area of triangle}-\text{Area under the curve}\\ =\frac{1}{2}\left(5-1\right)\left(16\right)-{\displaystyle {\int }_1^{3}ydx}\\ =32-{\displaystyle {\int }_1^3\left(2x^2-18\right)dx}\\ =32-\left|{\left[\frac{2x^3}{3}-18x\right]}_1^3\right|\\ =32-\left|\left(\frac{2{\left(3\right)}^3}{3}-18\left(3\right)\right)-\left(\frac{2{\left(1\right)}^3}{3}-18\left(1\right)\right)\right|\\ =32-\left|\left(18-54-\frac{2}{3}+18\right)\right|\\ =32-\left|-18\frac{2}{3}\right|\\ =32-18\frac{2}{3}\\ =13\frac{1}{3}{\text{ units}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume generated}=65\pi \\ \pi {\displaystyle {\int }_h^0{x}^{2}dy}=65\pi \\ \pi {\displaystyle {\int }_h^0\left(\frac{y}{2}+9\right)dx}=65\pi \leftarrow \boxed{\begin{array}{l}y=2x^2-18\\ x^2=\frac{y}{2}+9\end{array}}\\ {\left[\frac{y^2}{4}+9y\right]}_h^0=65\\ 0-\left(\frac{h^2}{4}+9h\right)=65\\ -\frac{h^2}{4}-9h=65\\ h^2+36h+260=0\\ \left(h+10\right)\left(h+26\right)=0\\ h=-10\text{ or }h=-26\left(\text{rejected}\right)\\ \text{Thus, }h=-10\end{array}$