Highway Grading & Haul Road Stability (Civil Engineering) Solution

Problem Overview

We evaluate the ground elevation function $h(x, y) = 80 - 0.02x^2 - 0.01y^3$ at bulldozer position $P(10, 4)$ along direction $\vec{v} = \langle 12, 5 \rangle$ . We normalize the vector, evaluate the directional derivative $D_{\vec{u}} h(10, 4)$ , and compare the resulting road grade against geotechnical vehicle stability thresholds.

Step-by-Step Mathematical Solution

1. Compute and Evaluate the Terrain Gradient

State the Gradient Vector Formula (defined in The Gradient Vector):

\nabla h(x, y) = \left\langle \frac{\partial h}{\partial x}, \frac{\partial h}{\partial y} \right\rangle

Compute the partial derivatives:

h_x = \frac{\partial}{\partial x}[80 - 0.02x^2 - 0.01y^3] = -0.04x

h_y = \frac{\partial}{\partial y}[80 - 0.02x^2 - 0.01y^3] = -0.03y^2

Evaluate at $P(10, 4)$ :

h_x(10, 4) = -0.04(10) = -0.40\text{ m/m}

h_y(10, 4) = -0.03(4^2) = -0.03(16) = -0.48\text{ m/m}

\nabla h(10, 4) = \langle -0.40, -0.48 \rangle

2. Normalize the Proposed Direction Vector

State the Vector Normalization Formula:

\vec{u} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{\langle v_1, v_2 \rangle}{\sqrt{v_1^2 + v_2^2}}

Compute the magnitude:

\|\vec{v}\| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13

\vec{u} = \frac{\langle 12, 5 \rangle}{13} = \left\langle \frac{12}{13}, \frac{5}{13} \right\rangle

3. Compute the Directional Derivative

State the Directional Derivative Dot Product Formula (defined in Directional Derivatives):

D_{\vec{u}} h(P) = \nabla h(P) \cdot \vec{u}

Evaluate the dot product:

D_{\vec{u}} h(10, 4) = \langle -0.40, -0.48 \rangle \cdot \left\langle \frac{12}{13}, \frac{5}{13} \right\rangle = \frac{(-0.40)(12) + (-0.48)(5)}{13}

D_{\vec{u}} h(10, 4) = \frac{-4.80 - 2.40}{13} = -\frac{7.20}{13} \approx -0.5538\text{ m/m}

4. Safety & Design Analysis

Physical Interpretation: Traveling along heading $\vec{v} = \langle 12, 5 \rangle$ causes an instantaneous vertical descent of approximately $0.554\text{ meters}$ per 1 horizontal meter traveled (a downward grade of $55.4\%$ ).
Engineering Decision: The calculated road slope ( $55.4\%$ ) vastly exceeds the maximum allowable safe operational grade ( $30\%$ or $0.30$ ) for heavy tracked bulldozers. Operating on this slope creates a high probability of track slippage, loss of steering control, or catastrophic rollover. The civil engineer must reject this haul route and realign the path closer to an elevation contour line (where $D_{\vec{u}} h \approx 0$ ) to keep the slope safely below $30\%$ .