An engineering team is designing a section of a new roller coaster track. The vertical profile of…
An engineering team is designing a section of a new roller coaster track. The vertical profile of the track for a specific horizontal stretch is modelled by the function:
$f(x) = \frac{1}{3}x^3 - 2x^2 + 3x + 5$
where $x$ represents the horizontal distance from the start of the section (in meters) and $f(x)$ represents the height of the track (in meters). To ensure safety and a thrilling experience, the engineering team must analyse the steepness and the peaks of this track.
(a)[1.0]
Identify the intervals of $x$ where the roller coaster is climbing (increasing height) and where it is descending (decreasing height).
(b)[1.0]
A support beam must be placed at the point where the track's slope is exactly zero. Find the coordinates of the points where support beam must be placed.
(c)[2.0]
Determine the maximum and minimum heights reached by the roller coaster in the interval $x \in [0, 4]$.
(d)[1.0]
At the point $x = 1$, a maintenance ladder must be placed perpendicular to the track, (along the normal). Find the equation of the maintenance ladder at $x = 1$.
Answer
Answer (a)
AIWritten by AI - it can contain mistakes.
Given $f(x) = \frac{1}{3}x^3 - 2x^2 + 3x + 5$.
$f'(x) = x^2 - 4x + 3 = (x - 1)(x - 3)$.
Critical points are $x = 1$ and $x = 3$.
- For $x \in [0, 1) \cup (3, \infty)$, $f'(x) > 0$, so the roller coaster is climbing (increasing height).
- For $x \in (1, 3)$, $f'(x) < 0$, so the roller coaster is descending (decreasing height).
Answer (b)
AIWritten by AI - it can contain mistakes.
The track slope is zero where $f'(x) = 0 \Rightarrow x = 1$ and $x = 3$.
- At $x = 1$: $f(1) = \frac{1}{3}(1)^3 - 2(1)^2 + 3(1) + 5 = \frac{1}{3} - 2 + 3 + 5 = 6\frac{1}{3} \approx 6.33$.
- At $x = 3$: $f(3) = \frac{1}{3}(27) - 2(9) + 3(3) + 5 = 9 - 18 + 9 + 5 = 5$.
Therefore, support beams must be placed at $(1, 6\frac{1}{3})$ and $(3, 5)$.
Answer (c)
AIWritten by AI - it can contain mistakes.
We evaluate $f(x)$ at the critical points and endpoints of $[0, 4]$:
- At $x = 0$: $f(0) = 5$
- At $x = 1$: $f(1) = 6\frac{1}{3} \approx 6.33$
- At $x = 3$: $f(3) = 5$
- At $x = 4$: $f(4) = \frac{1}{3}(64) - 2(16) + 3(4) + 5 = \frac{64}{3} - 32 + 12 + 5 = \frac{64}{3} - 15 = \frac{19}{3} = 6\frac{1}{3} \approx 6.33$.
Therefore, the maximum height reached is $6\frac{1}{3}$ m (or $6.33$ m) at $x = 1$ and $x = 4$, and the minimum height reached is $5$ m at $x = 0$ and $x = 3$.
Answer (d)
AIWritten by AI - it can contain mistakes.
At $x = 1$, the slope of the tangent is $f'(1) = 0$, which means the tangent line is horizontal ($y = 6\frac{1}{3}$).
The normal (maintenance ladder) is perpendicular to the tangent, so it is a vertical line passing through $x = 1$.
Therefore, the equation of the maintenance ladder is $x = 1$.
Final answer: x = 1
From ISC 2027 Specimen Mathematics Paper 1, question 17(i).