Roller Coaster Track G-Force Curvature Transitions (Entertainment Engineering) Solution

1. Initial Track Elevation ( $y$ -Intercept)

Evaluate $y(0)$ :

y(0) = \frac{1}{4}(0)^4 - (0)^3 + 8 = 8\text{ meters}

The drop begins at an initial elevation of

8.0\text{ meters}

(above track datum) at horizontal station

x = 0

2. Track Grade & Critical Points (First Derivative Analysis)

Name the Governing Formula: The Power Rule for Polynomial Differentiation and the First Derivative Test for Local Extrema: (Defined in First Derivative Test.)
Compute $y'(x)$ : $y'(x) = \frac{d}{dx}\left[\frac{1}{4}x^4 - x^3 + 8\right] = x^3 - 3x^2 = x^2(x - 3)$
Locate Critical Numbers on $[0, 4]$ : $x^2(x - 3) = 0 \implies x = 0 \quad \text{and} \quad x = 3$
Sign Chart and Classification:
- On $(0, 3)$ : For test point $x = 1 \implies y'(1) = (1)^2(1 - 3) = -2 < 0 \implies$ Track is descending ( $\searrow$ ).
- On $(3, 4]$ : For test point $x = 3.5 \implies y'(3.5) = (3.5)^2(0.5) > 0 \implies$ Track is ascending ( $\nearrow$ ).
- Classification at $x = 0$ : $y'(0) = 0$ , but the track is descending immediately after; this is a horizontal plateau/start bench.
- Classification at $x = 3$ : $y'$ changes from negative to positive ( $- \to +$ ) $\implies$ Local (and Absolute) Minimum Valley: $y(3) = \frac{1}{4}(3)^4 - (3)^3 + 8 = \frac{81}{4} - 27 + 8 = 20.25 - 19 = 1.25\text{ meters}$ The coaster bottoms out in a high-G pullout valley at station $(3, 1.25)$ .

3. Curvature & Inflection Points (Second Derivative Analysis)

Name the Governing Formula: The Second Derivative Inflection Criterion: $y''(x) = 0 \quad \text{with concavity sign transition}$ (Formalized in Concavity and the Second Derivative Test.)
Compute $y''(x)$ : $y''(x) = \frac{d}{dx}[x^3 - 3x^2] = 3x^2 - 6x = 3x(x - 2)$
Locate Inflection Candidates on $[0, 4]$ : $3x(x - 2) = 0 \implies x = 0 \quad \text{and} \quad x = 2$
Concavity Sign Analysis:
- For $x \in (0, 2)$ : Test $x = 1 \implies y''(1) = 3(1)(-1) = -3 < 0 \implies$ Concave Downward ( $\cap$ ).
- For $x \in (2, 4]$ : Test $x = 3 \implies y''(3) = 3(3)(1) = 9 > 0 \implies$ Concave Upward ( $\cup$ ). Across $x = 2$ , $y''$ flips sign from negative to positive. $y(2) = \frac{1}{4}(2)^4 - (2)^3 + 8 = \frac{16}{4} - 8 + 8 = 4.0\text{ meters}$ An inflection point occurs at $(2, 4.0)$ .

4. Information Consolidation Table

Station $x$	Elevation $y(x)$	Slope $y'(x)$	Concavity $y''(x)$	Track Feature / Passenger G-Force State
$x = 0$	$8.0\text{ m}$	$0$	$0$	Horizontal launch bench; zero vertical grade.
$x \in (0, 2)$	Descending	Negative	Negative ( $\cap$ )	Crest roll-off; tangent lies above track; airtime sensation.
$x = 2$	$4.0\text{ m}$	$-4.0$	$0$ (Inflection)	Steepest descent; inflection point; pure $1.0\text{ G}$ transition.
$x \in (2, 3)$	Descending	Negative	Positive ( $\cup$ )	Pullout transition begins; increasing seat compression.
$x = 3$	$1.25\text{ m}$	$0$	Positive ( $\cup$ )	Bottom of valley; maximum seat compression G-forces.
$x \in (3, 4]$	Ascending	Positive	Positive ( $\cup$ )	Camelback climb into subsequent element.

5. Entertainment Safety & Trim Brake Engineering

At the inflection point $(2, 4.0)$ , the vertical geometric curvature is zero ( $\kappa = \frac{|y''|}{[1 + (y')^2]^{3/2}} = 0$ ), so the normal centripetal acceleration vanishes ( $a_n = \kappa v^2 = 0$ ). Passengers experience neither airtime floating nor seat compression, but pure $1.0\text{ G}$ gravitational force. Placing eddy-current magnetic trim brakes exactly around this inflection window allows engineers to modulate vehicle train speed without subjecting riders to combined braking jerk and high vertical G-loads simultaneously.