Critical Points in Multivariable Functions

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Concept Explanation

Geometric Meaning: The Flat Tangent Plane

In single-variable calculus, critical points occur where the tangent line is horizontal ($f'(x) = 0$) or sharp/undefined. In multivariable calculus, for a surface $z = f(x, y)$, the tangent plane equation at $(a, b, f(a, b))$ is given by:

When $(a, b)$ is a critical point where $f_x(a, b) = 0$ and $f_y(a, b) = 0$, the tangent plane simplifies directly to:

This means the tangent plane is completely horizontal (parallel to the $xy$-plane). At this point, the surface has zero slope along the $x$-axis, zero slope along the $y$-axis, and in fact zero directional derivative in every direction $\vec{u}$:

               z ^
                 |        Flat Tangent Plane: z = f(a, b)
                 |        +-------------------------+
                 |       /                         /
                 |      /            * (a, b, f)  /
                 |     +-------------------------+
                 |                .-'   '-.
                 |              .'         '.
                 |             /   Surface   \
                 |
                 +----------------------------------> y
                /
               /
            x v

Critical Points vs. Extrema

Locating a critical point is a necessary condition for an interior local extremum of a differentiable function (Fermat's Theorem for Multivariable Functions):

• If $f$ has a local maximum or local minimum at an interior point $(a, b)$, and the first partial derivatives exist, then $\nabla f(a, b) = \mathbf{0}$.
• However, having $\nabla f(a, b) = \mathbf{0}$ is not sufficient to guarantee a maximum or minimum. The point could be a saddle point (where the surface curves upward in one direction and downward in another), or an inflection trough. Classifying the nature of a critical point requires The Second Derivative Test.

Step-by-Step Procedure to Locate Critical Points

1. Compute First Partial Derivatives: Calculate the symbolic expressions for $f_x(x, y)$ and $f_y(x, y)$ across the domain.
2. Formulate the Simultaneous System: Set both partial derivatives equal to zero:
   $$\begin{cases} f_x(x, y) = 0 \\ f_y(x, y) = 0 \end{cases}$$
3. Solve the System Algebraically:
   • Look for single-variable equations first to isolate numerical values.
   • If equations are coupled or non-linear, factor where possible, or substitute expressions (e.g., $y = g(x)$) into the other equation.
   • Caution: Avoid dividing both sides of an equation by a variable expression without first checking if that variable could equal zero.
4. Identify Singular Points: Check the original domain of $f$ for points where $f_x$ or $f_y$ ceases to exist (e.g., roots in denominators, cone vertices like $f(x, y) = \sqrt{x^2 + y^2}$).

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Engineering Applications

• Aerospace & Mechanical Engineering: Equilibrium configurations of mechanical mechanisms occur at stationary points of the potential energy surface $V(q_1, q_2)$. Setting $\nabla V = \mathbf{0}$ yields the equilibrium states of linkages and robotic trusses.
• Electrical Engineering: Static bias points in multi-transistor analog circuits correspond to critical points of node-voltage equations, ensuring amplifiers operate in their linear active regions without clipping.
• Chemical Engineering: In thermodynamic phase equilibrium, stationary points of the Gibbs free energy function determine the coexistence of vapor and liquid phases in distillation columns.

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Example Problems

Example 1: Decoupled Polynomial Critical Points

Find all critical points of the function:

Solution

1. Compute first partial derivatives:

   $$f_x(x, y) = \frac{\partial}{\partial x}(x^3 - 3x + y^2 - 4y + 7) = 3x^2 - 3$$
   $$f_y(x, y) = \frac{\partial}{\partial y}(x^3 - 3x + y^2 - 4y + 7) = 2y - 4$$

2. Set each partial derivative to zero:

   $$3x^2 - 3 = 0 \implies 3(x^2 - 1) = 0 \implies x = 1 \quad \text{or} \quad x = -1$$
   $$2y - 4 = 0 \implies 2y = 4 \implies y = 2$$

3. Assemble coordinate pairs: Combining the $x$-solutions with the unique $y$-solution yields two distinct critical points:

   $$(1, 2) \quad \text{and} \quad (-1, 2)$$

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Example 2: Coupled Non-Linear Critical Points

Find all critical points of the function:

Solution

1. Compute first partial derivatives:

   $$f_x(x, y) = \frac{\partial}{\partial x}(y^3 - 3xy + 12x) = -3y + 12$$
   $$f_y(x, y) = \frac{\partial}{\partial y}(y^3 - 3xy + 12x) = 3y^2 - 3x$$

2. Solve the simultaneous system: Set $f_x = 0$:

   $$-3y + 12 = 0 \implies 3y = 12 \implies y = 4$$
   Substitute $y = 4$ into $f_y = 0$:
   $$3(4)^2 - 3x = 0 \implies 3(16) - 3x = 0 \implies 48 = 3x \implies x = 16$$

3. Conclusion: There is exactly one critical point:

   $$(16, 4)$$

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Connections & References

• Prerequisites:
  • Partial Derivatives — Calculating $f_x$ and $f_y$.
  • The Gradient Vector — Vector definition $\nabla f = \langle f_x, f_y \rangle = \mathbf{0}$.
• Direct Successors:
  • Second Derivative Test for Multivariable Functions — Classifying critical points as local maxima, minima, or saddle points.
  • Extrema on Closed and Bounded Regions — Finding candidate points for global optimization.
  • Lagrange Multipliers — Finding constrained critical points along restricted boundaries.
• Course Sessions:
  • Session 13 - Optimization with Constraints