2.11.2 Differentiation SPM Practice (Paper 2 Question 6 – 10)

Question 6 (SPM 2018 – 6 marks):
Diagram 4 shows the front view of a part of a roller coaster track in a miniature park.

The curve part of the track of the roller coaster is represented by an equation $y=\frac{1}{64}{x}^3-\frac{3}{16}{x}^2$ , with point A as the region.
Find the shortest vertical distance, in m, from the track to ground level.


Solution:
$\begin{array}{l}y=\frac{1}{64}{x}^3-\frac{3}{16}{x}^2\mathrm{\dots \dots \dots \dots \dots }\left(1\right)\\ \frac{dy}{dx}=3\left(\frac{1}{64}\right)x^2-2\left(\frac{3}{16}\right)x^1\\ =\frac{3}{64}{x}^2-\frac{3}{8}x\\ \\ \text{At turning point, }\frac{dy}{dx}=0\\ \frac{3}{64}{x}^2-\frac{3}{8}x=0\\ x\left(\frac{3}{64}x-\frac{3}{8}\right)=0\\ x=0\text{ or}\\ \frac{3}{64}x-\frac{3}{8}=0\\ \frac{3}{64}x=\frac{3}{8}\\ x=\frac{3}{8}\times \frac{64}{3}\\ x=8\\ \\ \text{Substitute values of }x\text{ into equation (1):}\\ \text{When }x=0,\\ y=\frac{1}{64}{\left(0\right)}^3-\frac{3}{16}{\left(0\right)}^2\\ y=0\\ \\ \text{When }x=8,\\ y=\frac{1}{64}{\left(8\right)}^3-\frac{3}{16}{\left(8\right)}^2\\ y=-4\\ \\ \text{Thus, turning points : }\left(0,\text{ 0}\right)\text{ and }\left(8,-4\right)\end{array}$

$\begin{array}{l}\frac{dy}{dx}=\frac{3}{64}{x}^2-\frac{3}{8}x\\ \frac{d^{2}y}{d{x}^2}=2\left(\frac{3}{64}\right)x-\frac{3}{8}\\ =\frac{3}{32}x-\frac{3}{8}\\ \\ \text{When }x=0,\\ \frac{d^{2}y}{d{x}^2}=\frac{3}{32}\left(0\right)-\frac{3}{8}\\ =-\frac{3}{8}\left(<0\right)\\ \left(0,0\right)\text{ is maximum point}\text{.}\\ \\ \text{When }x=8,\\ \frac{d^{2}y}{d{x}^2}=\frac{3}{32}\left(8\right)-\frac{3}{8}\\ =\frac{3}{8}\left(>0\right)\\ \left(8,-4\right)\text{ is minimum point}\text{.}\\ \\ \text{Shortest vertical distance between track }\\ \text{and ground level is at the minimum point}\text{.}\\ \text{Shortest vertical distance}\\ =5-4\\ =1\text{ m}\end{array}$

Question 7 (SPM 2019):
Diagram 1 shows a sector POA with centre O.

It is given that the length of arc PB is 2.56 cm. [Use π = 3.142]
Calculate
(a) ∠ POB in radians, [2 marks]
(b) the area, in cm², of the shaded region. [4 marks]


Solution:
(a)
$\begin{array}{l}s=r\theta \\ 2.56=10\theta \\ \theta =0.256\text{ radian}\end{array}$

(b)
$\begin{array}{l}\text{Area of }OBP\\ =\frac{1}{2}{r}^2\theta \\ =\frac{1}{2}\times {10}^2\times 0.256\\ =12.8{\text{ cm}}^2\\ \\ \frac{AC}{10}=\tan {60}^o\\ AC=17.32\\ \\ \text{Area of shaded region }ABC\\ =\left(\frac{1}{2}\times 10\times 17.32\right)-\left(\frac{1}{2}\times {10}^2\times \frac{\pi }{3}\right)\\ =34.233{\text{ cm}}^2\\ \\ \text{Total area of shaded regions}\\ =12.8+34.233\\ =47.033{\text{ cm}}^2\end{array}$