Linear Approximation and Applications

Foundations of Linear Approximation

• Differentiability and local linearity: If a function $f$ is differentiable at a point $x = a$, the tangent line to the graph of $y = f(x)$ at $x = a$ provides an accurate approximation for the function $f(x)$ when $x$ is close to $a$.

• Definition of input and output changes: Let $\Delta x$ represent a small change in the input variable. The corresponding change in the output values of $f$, denoted as $\Delta f$, is defined by:   $\Delta f = f(a + \Delta x) - f(a)$

• Approximation of change in output: Using the symbol $\approx$ to denote "approximately equal to", the change in the function outputs is approximated by:   $\Delta f \approx f'(a)\Delta x$

• Geometric and physical interpretation of the derivative: The derivative $f'(a)$ indicates the rate of change and specifies how much the function $f$ "stretches" points located near $a$.

• Derivation from the limit definition of the derivative: The relationship $\Delta f \approx f'(a)\Delta x$ is derived directly from the limit definition of the derivative by taking $a$ as the base point and $\Delta x$ as the perturbation:   $f'(a) = \lim_{\Delta x \to 0} \frac{f(a + \Delta x) - f(a)}{\Delta x}$$f'(a) = \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}$$\frac{\Delta f}{\Delta x} \approx f'(a) \implies \Delta f \approx f'(a)\Delta x \quad \text{when } \Delta x \to 0$

Linearization and Tangent Line Approximation

• Transition from differential change to function estimation: By adding $f(a)$ to both sides of the output change approximation and substituting $x$ for $a + \Delta x$ (such that $\Delta x = x - a$), the approximation for function values near $a$ is obtained:   $f(x) \approx f'(a)(x - a) + f(a)$

• Equivalence to the tangent line: The right-hand side of the expression $f'(a)(x - a) + f(a)$ matches the point-slope formula for the tangent line to the curve $y = f(x)$ at the point $(a, f(a))$.

• Definition 4.1.1 (Linearization): If $f$ is differentiable at $a$, the linearization of $f$ centered at $x = a$ is defined as the function $L$ given by:   $L(x) = f'(a)(x - a) + f(a)$

• Characteristics of the linearization:

  • The graph of the linearization function $L$ is the tangent line to $y = f(x)$.
  • For inputs $x$ near $a$, the approximation holds:     $L(x) \approx f(x)$

• Computational purpose: The formulas for $\Delta f$ and $L(x)$ enable the approximation of changes in function values and absolute function values without requiring direct computation of complex expressions.

Worked Examples and Practical Applications

• Example 1 — Radical Function Approximation:

  • Problem statement: Determine the linearization of $f(x) = \sqrt{x + 1}$ centered at $x = 0$, and use this linearization to estimate $\sqrt{1.2}$.
  • Differentiation step: Express the radical function in power form and compute its first derivative:     $f(x) = (x + 1)^{1/2}$$f'(x) = \frac{d}{dx}\left((x + 1)^{1/2}\right) = \frac{1}{2}(x + 1)^{-1/2} = \frac{1}{2\sqrt{x + 1}}$
  • Center evaluation: Evaluate both the original function $f(x)$ and its derivative $f'(x)$ at the center point $x = 0$:     $f(0) = \sqrt{0 + 1} = \sqrt{1} = 1$$f'(0) = \frac{1}{2\sqrt{0 + 1}} = \frac{1}{2(1)} = \frac{1}{2}$
  • Constructing the linearization: Substitute the evaluated values into the linearization equation:     $L(x) = \frac{1}{2}(x - 0) + 1 = \frac{1}{2}x + 1$
  • Numerical estimation: Since $f(0.2) = \sqrt{0.2 + 1} = \sqrt{1.2}$, evaluate the linearization at $x = 0.2$:     $\sqrt{1.2} \approx L(0.2) = \frac{1}{2}(0.2) + 1 = 0.1 + 1 = 1.1$
  • Comparison with true value: The actual value of $\sqrt{1.2}$ to four decimal places is approximately $1.0954$, showing that the linear approximation $1.1$ provides a close estimate.

• Example 2 — Travel Time Estimation with Speeding:

  • Problem statement: A driver is traveling a distance of $25\,\text{miles}$ along a highway. Considering driving $5\,\text{mph}$ above the posted speed limit of $75\,\text{mph}$, estimate the amount of time saved using derivatives.
  • Mathematical model: Let $t$ represent travel time in hours and $v$ represent speed in miles per hour ($\text{mph}$). The relationship between distance, speed, and time gives:     $v = \frac{25}{t} \implies t = \frac{25}{v}$
  • Derivative computation: Differentiate $t$ with respect to $v$:     $\frac{dt}{dv} = -\frac{25}{v^2}$
  • Differential approximation: Apply the change approximation formula with initial speed $a = 75\,\text{mph}$ and change in speed $\Delta v = 5\,\text{mph}$:     $\Delta t \approx -\frac{25}{75^2}(5)$$\Delta t \approx -\frac{25}{5625}(5) = -\frac{125}{5625} = -\frac{1}{45}\,\text{hours}$
  • Conversion to minutes: Convert the change in time to minutes by multiplying by $60\,\text{minutes/hour}$:     $\Delta t \approx -\frac{1}{45}(60) = -\frac{60}{45} = -\frac{4}{3}\,\text{minutes} \approx -1.3\,\text{minutes}$
  • Interpretation and real-world implications: Driving $5\,\text{mph}$ over the speed limit reduces travel time by approximately $1.3\,\text{minutes}$. That amount of time could be saved simply by walking more quickly across the parking lot upon arrival. Furthermore, the linear approximation provides an overestimate of the actual time saved.

Summary of Core Approximation Formulas

• Change in function output:   $\Delta f \approx f'(a)\Delta x$   where $\Delta f = f(a + \Delta x) - f(a)$ represents the true change in output for a small change $\Delta x$.

• Tangent line function estimation:   $f(x) \approx L(x)$   where the linearization $L(x)$ centered at $x = a$ is given by:   $L(x) = f'(a)(x - a) + f(a)$