Linear Approximations

Concept Explanation

The Principle of Local Linearity

In single-variable calculus, zooming in very closely to any smooth, differentiable curve causes the curve to look indistinguishable from its tangent line ($L(x) = f(x_0) + f'(x_0)(x - x_0)$). In multivariable calculus, this geometric property extends directly to higher dimensions: zooming in on any smooth surface $z = f(x, y)$ causes the curved topography to appear virtually flat.

In engineering applications, real-world physical equations often involve complex non-linear functions (radicals, logarithms, trigonometric terms, or polynomials of high degree). Evaluating these equations at messy decimal coordinates can be computationally expensive or algebraically intractable. By constructing a flat tangent plane at a nearby "clean" integer base point $(x_0, y_0)$, engineers can substitute the complex function with its linearization $L(x, y)$, converting non-linear calculations into simple multiplications and additions.

       ^ z
       |                      . True function surface z = f(x, y)
       |                     /
       |       L(x, y)      /  <--- Tangent plane approximation
       |       ------+-----+------------------
       |            /|     |
       |           / |     |  Error = |f(x, y) - L(x, y)|
       |          /  |     |
       |         /   |     . f(x, y)
       |        /    . P_0(x_0, y_0, z_0)
       +-------+------------------------------> y
              /
             v x

Anatomy of the Linearization Formula

The linearization formula consists of three intuitive components:

1. Baseline Value ($z_0 = f(x_0, y_0)$): The exact, easily computed starting output of the system at the chosen reference coordinate.
2. First-Order $x$-Correction ($f_x(x_0, y_0)\Delta x$): The rate of change in the $x$-direction multiplied by the distance traveled away from $x_0$.
3. First-Order $y$-Correction ($f_y(x_0, y_0)\Delta y$): The rate of change in the $y$-direction multiplied by the distance traveled away from $y_0$.

Strategic Selection of the Base Point $(x_0, y_0)$

To produce an accurate estimate, the base point $(x_0, y_0)$ must satisfy two criteria:

1. Clean Evaluation: The function value $f(x_0, y_0)$ and partial derivatives $f_x(x_0, y_0), f_y(x_0, y_0)$ must evaluate to simple integers or clean rational numbers without requiring a calculator.
2. Proximity: The displacements $\Delta x = x - x_0$ and $\Delta y = y - y_0$ must be small (typically $|\Delta x|, |\Delta y| \ll 1$). Because a tangent plane only models first-order derivatives, surface curvature causes the approximation error to grow quadratically as distance from the base point increases.

Extension to Three or More Variables

The principle of linearization extends seamlessly to functions of three or more variables. For a scalar field $w = f(x, y, z)$ at base point $(x_0, y_0, z_0)$:

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Engineering Applications

• Embedded Real-Time Control Systems: Microcontrollers in autonomous drones, automotive anti-lock braking systems (ABS), and robotic manipulators use linearized state-space matrices to compute real-time motor commands in microseconds.
• Finite Element Analysis (FEA): Structural and geotechnical simulation software linearizes non-linear stress-strain relationships across small spatial mesh elements to solve large systems of linear algebraic equations ($[K]\{u\} = \{F\}$).
• Aerospace Stability & Control: Aircraft flight dynamics linearize non-linear aerodynamic forces around steady cruise trim conditions to evaluate longitudinal and lateral stability modes.
• Electrical Circuit Small-Signal Modeling: Modeling non-linear semiconductor diodes and transistors as linear equivalent circuits (transconductance and dynamic resistance) around a fixed DC operating point (Q-point).

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Example Problems

Example 1: Radical and Quadratic Linearization

Find the linear approximation $L(x, y)$ of $f(x, y) = \sqrt{x^2 + y}$ at the base point $(2, 1)$, and use it to estimate $f(2.1, 0.9)$.

Solution

1. Evaluate Baseline Value at Clean Base Point $(2, 1)$:
   $$f(2, 1) = \sqrt{2^2 + 1} = \sqrt{4 + 1} = \sqrt{5}$$
2. Compute General Partial Derivatives:
   $$f_x(x, y) = \frac{\partial}{\partial x}[(x^2 + y)^{1/2}] = \frac{1}{2}(x^2 + y)^{-1/2}(2x) = \frac{x}{\sqrt{x^2 + y}}$$
   $$f_y(x, y) = \frac{\partial}{\partial y}[(x^2 + y)^{1/2}] = \frac{1}{2}(x^2 + y)^{-1/2}(1) = \frac{1}{2\sqrt{x^2 + y}}$$
3. Evaluate Partial Derivatives at $(2, 1)$:
   $$f_x(2, 1) = \frac{2}{\sqrt{2^2 + 1}} = \frac{2}{\sqrt{5}}$$
   $$f_y(2, 1) = \frac{1}{2\sqrt{2^2 + 1}} = \frac{1}{2\sqrt{5}}$$
4. Formulate Linearization Function $L(x, y)$:
   $$L(x, y) = f(2, 1) + f_x(2, 1)(x - 2) + f_y(2, 1)(y - 1)$$
   $$L(x, y) = \sqrt{5} + \frac{2}{\sqrt{5}}(x - 2) + \frac{1}{2\sqrt{5}}(y - 1)$$
5. Evaluate at Target Point $(2.1, 0.9)$:
   • Offsets: $\Delta x = 2.1 - 2 = 0.1$, $\Delta y = 0.9 - 1 = -0.1$.
     $$L(2.1, 0.9) = \sqrt{5} + \frac{2}{\sqrt{5}}(0.1) + \frac{1}{2\sqrt{5}}(-0.1) = \sqrt{5} + \frac{0.2}{\sqrt{5}} - \frac{0.05}{\sqrt{5}} = \sqrt{5} + \frac{0.15}{\sqrt{5}}$$
     Using $\sqrt{5} \approx 2.23607$:
     $$L(2.1, 0.9) \approx 2.23607 + \frac{0.15}{2.23607} \approx 2.23607 + 0.06708 = 2.30315$$
     (Exact value: $\sqrt{2.1^2 + 0.9} = \sqrt{4.41 + 0.9} = \sqrt{5.31} \approx 2.30434$; error is only $0.00119$ or $0.05\%$.)

Example 2: Polynomial Linearization and Curvature Error

Construct the linearization of $f(x, y) = x^2 + y^2$ at $(1, 1)$, estimate $f(1.1, 0.9)$, and calculate the approximation error.

Solution

1. Baseline Value:
   $$f(1, 1) = 1^2 + 1^2 = 2$$
2. Partial Derivatives at $(1, 1)$:
   $$f_x(x, y) = 2x \implies f_x(1, 1) = 2(1) = 2$$
   $$f_y(x, y) = 2y \implies f_y(1, 1) = 2(1) = 2$$
3. Linearization Equation:
   $$L(x, y) = 2 + 2(x - 1) + 2(y - 1)$$
4. Estimate at $(1.1, 0.9)$:
   $$\Delta x = 1.1 - 1 = 0.1, \quad \Delta y = 0.9 - 1 = -0.1$$
   $$L(1.1, 0.9) = 2 + 2(0.1) + 2(-0.1) = 2 + 0.2 - 0.2 = 2.0$$
5. Compare with Exact Value:
   $$f(1.1, 0.9) = (1.1)^2 + (0.9)^2 = 1.21 + 0.81 = 2.02$$
   $$\text{Approximation Error} = |f(1.1, 0.9) - L(1.1, 0.9)| = |2.02 - 2.00| = 0.02$$
   Insight: The linear terms exactly canceled ($0.2 - 0.2 = 0$), so the remaining discrepancy $0.02$ represents second-order quadratic curvature ($(\Delta x)^2 + (\Delta y)^2 = 0.01 + 0.01 = 0.02$).

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Connections & References

• Parent Concepts: Tangent Planes | Partial Derivatives
• Related Notes: Differentials and Error Propagation | Equations of Planes in 3D
• Course Sessions: Calc-3 Session 10