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Highway Exit Ramp Grade Design (Construction Engineering) Solution

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Problem Overview

We evaluate the instantaneous slope and tangent grade equation for the highway off-ramp elevation profile H(x)=24x+2H(x) = \frac{24}{x+2} at station x=2x = 2.

Step-by-Step Mathematical Solution

Step 1: Set up Difference Quotient Limit for H′(x)H'(x)

H′(x)=lim⁡h→0H(x+h)−H(x)h=lim⁡h→024(x+h)+2−24x+2hH'(x) = \lim_{h \to 0} \frac{H(x+h) - H(x)}{h} = \lim_{h \to 0} \frac{\frac{24}{(x+h)+2} - \frac{24}{x+2}}{h}

Step 2: Combine Numerator Terms with Common Denominator (x+h+2)(x+2)(x+h+2)(x+2)

H′(x)=lim⁡h→024(x+2)−24(x+h+2)(x+h+2)(x+2)hH'(x) = \lim_{h \to 0} \frac{\frac{24(x+2) - 24(x+h+2)}{(x+h+2)(x+2)}}{h}
=lim⁡h→024x+48−24x−24h−48h(x+h+2)(x+2)=lim⁡h→0−24hh(x+h+2)(x+2)= \lim_{h \to 0} \frac{24x + 48 - 24x - 24h - 48}{h(x+h+2)(x+2)} = \lim_{h \to 0} \frac{-24h}{h(x+h+2)(x+2)}

Step 3: Divide out hh and Substitute h=0h = 0

H′(x)=lim⁡h→0−24(x+h+2)(x+2)=−24(x+0+2)(x+2)=−24(x+2)2H'(x) = \lim_{h \to 0} \frac{-24}{(x+h+2)(x+2)} = \frac{-24}{(x+0+2)(x+2)} = -\frac{24}{(x+2)^2}

Step 4: Evaluate Road Slope at Station x=2x = 2

H′(2)=−24(2+2)2=−2416=−1.50 m per stationH'(2) = -\frac{24}{(2+2)^2} = -\frac{24}{16} = -1.50\text{ m per station}
The instantaneous slope matches the required specification of −1.50-1.50 (a −1.5%-1.5\% grade).

Step 5: Construct the Tangent Grade Equation at x=2x = 2

  • Elevation at x=2x = 2: H(2)=242+2=6.00 m  ⟹  (2,6)H(2) = \frac{24}{2+2} = 6.00\text{ m} \implies (2, 6)
  • Point-slope equation:
    y−6=−1.5(x−2)  ⟹  y−6=−1.5x+3  ⟹  y=−1.5x+9y - 6 = -1.5(x - 2) \implies y - 6 = -1.5x + 3 \implies y = -1.5x + 9

Engineering Conclusion

At station x=2x = 2, the off-ramp reaches an elevation of 6.00 m6.00\text{ m} with an instantaneous grade of −1.50-1.50, satisfying the safety specification. Surveyors can utilize the tangent line formula y=−1.5x+9y = -1.5x + 9 to align grade stakes along the embankment, guaranteeing a smooth transition into the collector roadway without hazardous vertical slope kinks.