1. **Problem Statement:** We are given height data of a road at distances 0, 2, 4, and 6 km and need to: - a) Estimate the height at 3 km using Lagrange interpolation. - b) Analyze the slope at 2 km using forward-backward numerical differentiation. - c) Estimate the area under the height curve from 0 to 6 km using Simpson's rule. - d) Discuss how these calculations assist in road construction planning. 2. **Lagrange Interpolation Formula:** $$L(x) = \sum_{i=0}^n y_i \prod_{j=0,j\neq i}^n \frac{x - x_j}{x_i - x_j}$$ This formula constructs a polynomial passing through all given points. 3. **Data Points:** $(x_0,y_0)=(0,100), (x_1,y_1)=(2,150), (x_2,y_2)=(4,210), (x_3,y_3)=(6,280)$ 4. **Step a: Interpolation at $x=3$ km** Calculate each Lagrange basis polynomial: $$L_0(3) = \frac{3-2}{0-2} \cdot \frac{3-4}{0-4} \cdot \frac{3-6}{0-6} = \frac{1}{-2} \cdot \frac{-1}{-4} \cdot \frac{-3}{-6} = -\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} = -\frac{1}{16}$$ $$L_1(3) = \frac{3-0}{2-0} \cdot \frac{3-4}{2-4} \cdot \frac{3-6}{2-6} = \frac{3}{2} \cdot \frac{-1}{-2} \cdot \frac{-3}{-4} = \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} = \frac{9}{16}$$ $$L_2(3) = \frac{3-0}{4-0} \cdot \frac{3-2}{4-2} \cdot \frac{3-6}{4-6} = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{-3}{-2} = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{9}{16}$$ $$L_3(3) = \frac{3-0}{6-0} \cdot \frac{3-2}{6-2} \cdot \frac{3-4}{6-4} = \frac{3}{6} \cdot \frac{1}{4} \cdot \frac{-1}{2} = \frac{1}{2} \cdot \frac{1}{4} \cdot -\frac{1}{2} = -\frac{1}{16}$$ Calculate interpolated height: $$P(3) = 100 \cdot (-\frac{1}{16}) + 150 \cdot \frac{9}{16} + 210 \cdot \frac{9}{16} + 280 \cdot (-\frac{1}{16})$$ $$= -6.25 + 84.375 + 118.125 - 17.5 = 178.75$$ 5. **Step b: Numerical Differentiation at $x=2$ km** Using forward difference: $$f'(2) \approx \frac{f(4) - f(2)}{4-2} = \frac{210 - 150}{2} = 30$$ Using backward difference: $$f'(2) \approx \frac{f(2) - f(0)}{2-0} = \frac{150 - 100}{2} = 25$$ Average slope: $$\frac{30 + 25}{2} = 27.5$$ 6. **Step c: Simpson's Rule for area from 0 to 6 km** Simpson's rule formula for 4 points (3 intervals): $$\int_a^b f(x) dx \approx \frac{h}{3} [y_0 + 4y_1 + 2y_2 + 4y_3 + y_4]$$ Here, $h=2$ km, points are $y_0=100, y_1=150, y_2=210, y_3=280$ (only 4 points, so use 3 intervals): $$Area \approx \frac{2}{3} [100 + 4(150) + 2(210) + 4(280)]$$ Calculate inside bracket: $$100 + 600 + 420 + 1120 = 2240$$ Area: $$\frac{2}{3} \times 2240 = 1493.33$$ 7. **Step d: Application in Road Construction** - Interpolated height helps estimate elevation at unmeasured points for design. - Slope analysis aids in assessing road gradient for safety and drainage. - Area estimation informs earthwork volume for excavation planning. **Final answers:** - a) Height at 3 km: $178.75$ m - b) Slope at 2 km: $27.5$ m/km - c) Excavation area: $1493.33$ m·km