Long Questions (Question 4)

Question 4 (6 marks):
Diagram 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:
$\begin{array}{l}y=\frac{1}{64}{x}^3-\frac{3}{16}{x}^2\mathrm{...............}\left(1\right)\\ \frac{dy}{dx}=3\left(\frac{1}{64}\right)x^2-2\left(\frac{3}{16}\right)x^1\\ =\frac{3}{64}{x}^2-\frac{3}{8}x\\ \\ \text{At turning point, }\frac{dy}{dx}=0\\ \frac{3}{64}{x}^2-\frac{3}{8}x=0\\ x\left(\frac{3}{64}x-\frac{3}{8}\right)=0\\ x=0\text{ or}\\ \frac{3}{64}x-\frac{3}{8}=0\\ \frac{3}{64}x=\frac{3}{8}\\ x=\frac{3}{8}\times \frac{64}{3}\\ x=8\\ \\ \text{Substitute values of }x\text{ into equation (1):}\\ \text{When }x=0,\\ y=\frac{1}{64}{\left(0\right)}^3-\frac{3}{16}{\left(0\right)}^2\\ y=0\\ \\ \text{When }x=8,\\ y=\frac{1}{64}{\left(8\right)}^3-\frac{3}{16}{\left(8\right)}^2\\ y=-4\\ \\ \text{Thus, turning points : }\left(0,\text{ 0}\right)\text{ and }\left(8,-4\right)\end{array}$

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