High-Speed Railway Transition (Civil Engineering) Solution

Problem Overview

We analyze the continuity and differentiability of the railway track alignment profile at the junction coordinate $x = 3$:

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Step-by-Step Mathematical Solution

Step 1: Check for Continuity at $x = 3$

Evaluate the left-hand limit, right-hand limit, and function value at $x = 3$:

• Left-hand limit:
  $$\lim_{x \to 3^-} T(x) = \lim_{x \to 3^-} (2x) = 2(3) = 6$$
• Right-hand limit:
  $$\lim_{x \to 3^+} T(x) = \lim_{x \to 3^+} (x^2 - 4x + 9) = (3)^2 - 4(3) + 9 = 9 - 12 + 9 = 6$$
• Function value:
  $$T(3) = 2(3) = 6$$

Since $\lim_{x \to 3^-} T(x) = \lim_{x \to 3^+} T(x) = T(3) = 6$, the two track segments meet continuously at $(3, 6)$ without any physical gap.

Step 2: Check for Differentiability at $x = 3$

Compute the one-sided derivative limits approaching $x = 3$:

• Left-hand derivative (straight track):
  $$\lim_{h \to 0^-} \frac{T(3+h) - T(3)}{h} = \lim_{h \to 0^-} \frac{2(3+h) - 6}{h} = \lim_{h \to 0^-} \frac{2h}{h} = 2$$
• Right-hand derivative (curved bypass):
  $$\lim_{h \to 0^+} \frac{T(3+h) - T(3)}{h} = \lim_{h \to 0^+} \frac{[(3+h)^2 - 4(3+h) + 9] - 6}{h}$$
  $$= \lim_{h \to 0^+} \frac{9 + 6h + h^2 - 12 - 4h + 9 - 6}{h} = \lim_{h \to 0^+} \frac{2h + h^2}{h} = \lim_{h \to 0^+} (2 + h) = 2$$

Since the incoming tangent slope ($2$) matches the outgoing tangent slope ($2$), the function is differentiable at $x = 3$ with $T'(3) = 2$.

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Engineering Conclusion

Because the track profile is differentiable at $x = 3$, the alignment forms a smooth curve without a sharp geometric corner ($C^1$ continuity). This layout is safe for high-speed operation. If the one-sided slopes did not match (creating a non-differentiable corner), high-speed trains crossing the junction would experience an instantaneous impulse step in lateral acceleration and jerk, creating severe derailment risks and structural wheel-flange damage.