Aerospace Wing Profile Lift-to-Drag Optimization (Aerospace & Mechanical Engineering) Solution

An aerospace engineer at the UNLV Howard R. Hughes College of Engineering is prototyping a high-efficiency transonic wing section for an autonomous aerial survey drone. Aerodynamic wind-tunnel testing reveals that the non-dimensional lift-to-drag efficiency ratio $E(x, y)$ depends on the wing's angle of attack $x$ (in degrees, centered around nominal cruise trim) and the maximum camber percentage $y$ (in percent of chord length).

The empirical efficiency performance surface is modeled across the operational flight envelope by the unconstrained multivariable cubic function:

1. Determine the first-order partial derivatives $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ to formulate the gradient vector $\nabla E(x, y)$.
2. Locate all critical points $(x_0, y_0)$ where the aerodynamic performance gradient vanishes ($\nabla E = \mathbf{0}$).
3. Compute all second-order partial derivatives and formulate the Hessian discriminant $D(x, y) = E_{xx}E_{yy} - (E_{xy})^2$.
4. Apply the Second Derivative Test to classify each critical point, identifying which configuration represents a stable local maximum in cruise efficiency versus an unstable saddle state.
5. Evaluate the maximum aerodynamic efficiency ratio $E_{\max}$ at the optimal design point.

Step-by-Step Solution:

1. Formulate the Gradient Vector: State the Two-Variable Gradient Formula (defined in The Gradient Vector):

   $$\nabla E(x, y) = \left\langle \frac{\partial E}{\partial x}, \frac{\partial E}{\partial y} \right\rangle = \frac{\partial E}{\partial x}\hat{i} + \frac{\partial E}{\partial y}\hat{j}$$
   Compute the first-order partial derivatives treating other variables as constants:
   $$\frac{\partial E}{\partial x} = \frac{\partial}{\partial x}(-2x^3 + 6x - 4y^2 + 16y + 45) = -6x^2 + 6$$
   $$\frac{\partial E}{\partial y} = \frac{\partial}{\partial y}(-2x^3 + 6x - 4y^2 + 16y + 45) = -8y + 16$$
   Thus, the gradient vector is:
   $$\nabla E(x, y) = \langle -6x^2 + 6, \, -8y + 16 \rangle = (-6x^2 + 6)\hat{i} + (-8y + 16)\hat{j}$$

2. Locate Critical Points: State the Critical Point Condition (defined in Critical Points in Multivariable Functions):

   $$\nabla E(x, y) = \mathbf{0} \iff \frac{\partial E}{\partial x} = 0 \quad \text{and} \quad \frac{\partial E}{\partial y} = 0$$
   Set each component to zero:
   $$-6x^2 + 6 = 0 \implies -6(x^2 - 1) = 0 \implies x^2 = 1 \implies x = 1 \quad \text{or} \quad x = -1$$
   $$-8y + 16 = 0 \implies 8y = 16 \implies y = 2$$
   Pairing the solutions gives two candidate critical points:
   $$(1, 2) \quad \text{and} \quad (-1, 2)$$

3. Compute Second Partial Derivatives and Discriminant: State the Second-Order Partial Derivatives (defined in Second-Order Partial Derivatives):

   $$E_{xx} = \frac{\partial}{\partial x}(-6x^2 + 6) = -12x$$
   $$E_{yy} = \frac{\partial}{\partial y}(-8y + 16) = -8$$
   $$E_{xy} = \frac{\partial}{\partial y}(-6x^2 + 6) = 0$$
   State the Hessian Discriminant Formula (defined in Second Derivative Test for Multivariable Functions):
   $$D(x, y) = E_{xx}E_{yy} - (E_{xy})^2 = (-12x)(-8) - (0)^2 = 96x$$

4. Classify Critical Points via the Second Derivative Test:

   • At Point $(1, 2)$:
     $$D(1, 2) = 96(1) = 96 > 0$$
     $$E_{xx}(1, 2) = -12(1) = -12 < 0$$
     Because $D > 0$ and $E_{xx} < 0$, the point $(1, 2)$ is a local maximum. Aerodynamically, this represents a stable peak in the lift-to-drag envelope where small variations in angle of attack or camber do not cause rapid loss of flight efficiency.
   • At Point $(-1, 2)$:
     $$D(-1, 2) = 96(-1) = -96 < 0$$
     Because $D < 0$, the point $(-1, 2)$ is a saddle point. Aerodynamically, this represents an unstable saddle configuration where slight pitch adjustments induce severe aerodynamic divergence.

5. Evaluate Maximum Cruise Efficiency: Substitute the optimal coordinates $x = 1^\circ$ and $y = 2\%$ into $E(x, y)$:

   $$E(1, 2) = -2(1)^3 + 6(1) - 4(2)^2 + 16(2) + 45$$
   $$E(1, 2) = -2 + 6 - 4(4) + 32 + 45 = 4 - 16 + 32 + 45 = 65$$
   The maximum achievable lift-to-drag efficiency ratio is $E_{\max} = 65$, achieved at an angle of attack of $1^\circ$ and a maximum camber of $2\%$.