3.1 Differential Calculus, Limits, Derivatives & Applications

3.1 Differential Calculus, Limits, Derivatives & Applications

Differential calculus is the branch of engineering mathematics concerned with instantaneous rates of change, slopes of curves, and optimization of physical systems. For electrical engineers preparing for the PRC Licensure Examination, mastery of differential calculus is foundational—it underpins signal analysis, circuit transient response, electromagnetic field theory, and power distribution optimization.

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1. Limits and Continuity

Mathematical Definition of a Limit

The limit of a function $f(x)$ as $x$ approaches a real number $c$ is $L$, written as: $\lim_{x \to c} f(x) = L$ This signifies that $f(x)$ can be made arbitrarily close to $L$ by choosing $x$ sufficiently close to $c$, but not equal to $c$.

Continuity Criteria

A function $f(x)$ is continuous at a point $x = c$ if and only if three conditions are satisfied simultaneously:

1. $f(c)$ is defined (i.e., $c$ is in the domain of $f$).
2. $\lim_{x \to c} f(x)$ exists (meaning the left-hand limit $\lim_{x \to c^-} f(x)$ equals the right-hand limit $\lim_{x \to c^+} f(x)$).
3. $\lim_{x \to c} f(x) = f(c)$.

If any of these conditions fail, the function exhibits a discontinuity (removable, jump, or infinite/essential).

Indeterminate Forms & L'Hôpital's Rule

When evaluating a limit $\lim_{x \to c} \frac{f(x)}{g(x)}$ results in an indeterminate quotient of the form $\left[\frac{0}{0}\right]$ or $\left[\frac{\infty}{\infty}\right]$, L'Hôpital's Rule applies: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$ provided the limit on the right exists or is infinite.

For other indeterminate forms ($0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, $\infty^0$), algebraic transformation into a quotient form is required prior to applying L'Hôpital's Rule.

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2. Fundamental Differentiation Rules

The derivative of $y = f(x)$ with respect to $x$ is defined as the limit of the difference quotient: $f'(x) = \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

Core Operational Rules

Standard Derivatives Table

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3. Implicit Differentiation

When a relation between $x$ and $y$ is expressed implicitly as $F(x, y) = 0$ rather than explicitly as $y = f(x)$, we differentiate both sides of the equation with respect to $x$, applying the chain rule to terms containing $y$ (treating $y$ as an implicit function of $x$), and then solve algebraically for $\frac{dy}{dx}$.

Alternatively, using partial derivatives: $\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y}$

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4. Maxima, Minima & Optimization Problems

Critical Points and Extremities

• A critical point of $f(x)$ occurs where $f'(x) = 0$ or where $f'(x)$ is undefined.
• First Derivative Test: If $f'(x)$ changes sign from positive to negative at $x_0$, $f(x_0)$ is a local maximum. If $f'(x)$ changes from negative to positive, $f(x_0)$ is a local minimum.
• Second Derivative Test: If $f'(x_0) = 0$:
  • If $f''(x_0) < 0$, $f(x)$ has a local maximum at $x_0$ (concave down).
  • If $f''(x_0) > 0$, $f(x)$ has a local minimum at $x_0$ (concave up).
  • If $f''(x_0) = 0$, the test is inconclusive.
• An inflection point occurs where concavity changes, requiring $f''(x) = 0$ (or undefined) with a sign change in $f''(x)$.

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5. Related Rates & Curvature

Related Rates

In related rates problems, physical quantities are functions of time $t$. Differentiating governing geometric or physical equations implicitly with respect to $t$ establishes relationships between rates of change (e.g., $\frac{dV}{dt}, \frac{dI}{dt}, \frac{dx}{dt}$).

Curvature & Radius of Curvature

Curvature ($\kappa$) measures how sharply a curve bends per unit arc length. For a curve $y = f(x)$: $\kappa = \frac{|y''|}{[1 + (y')^2]^{3/2}}$

The radius of curvature ($\rho$) is the reciprocal of curvature: $\rho = \frac{1}{\kappa} = \frac{[1 + (y')^2]^{3/2}}{|y''|}$

For parametric equations $x = x(t)$ and $y = y(t)$: $\kappa = \frac{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}}$

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6. Partial Derivatives

For a function of multiple variables $z = f(x, y)$:

• The partial derivative with respect to $x$ ($\frac{\partial f}{\partial x}$ or $f_x$) is calculated by differentiating $f(x,y)$ with respect to $x$ while holding $y$ constant.
• The partial derivative with respect to $y$ ($\frac{\partial f}{\partial y}$ or $f_y$) treats $x$ as constant.

Clairaut's Theorem (Mixed Partials)

If $f(x,y)$ and its partial derivatives are continuous, mixed second-order partials are equal: $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$

Total Differential & Error Propagation

$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$ For electrical power $P = V \cdot I$ or $P = I^2 R$, total differentials estimate measurement error propagation: $\frac{dP}{P} = \frac{dV}{V} + \frac{dI}{I}$.

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Step-by-Step Worked Numerical Examples

Example 3.1.1: Indeterminate Limit via L'Hôpital's Rule

Problem: Evaluate the limit $\lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x^2}$.

Solution:

1. Evaluate direct substitution: At $x = 0$, $\frac{e^0 - 1 - 0}{0^2} = \frac{1 - 1 - 0}{0} = \frac{0}{0}$ (indeterminate form).
2. Apply L'Hôpital's Rule (1st differentiation): $\lim_{x \to 0} \frac{\frac{d}{dx}(e^{2x} - 1 - 2x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{2e^{2x} - 2}{2x}$
3. Re-evaluate substitution: At $x = 0$, $\frac{2(1) - 2}{0} = \frac{0}{0}$ (still indeterminate).
4. Apply L'Hôpital's Rule (2nd differentiation): $\lim_{x \to 0} \frac{\frac{d}{dx}(2e^{2x} - 2)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{4e^{2x}}{2}$
5. Evaluate final limit: $\frac{4 e^{2(0)}}{2} = \frac{4(1)}{2} = 2$

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Example 3.1.2: Electrical Conductor Insulation Optimization

Problem: An insulated high-voltage cylindrical power cable has an inner conductor radius $r$ and outer insulation radius $R_0 = 10\text{ mm}$ (fixed). The breakdown voltage capability of the cable insulation is given by $V(r) = E_{\text{crit}} \cdot r \ln\left(\frac{R_0}{r}\right)$, where $E_{\text{crit}}$ is a dielectric constant. Find the conductor radius $r$ that maximizes the breakdown voltage.

Solution:

1. Express breakdown voltage function: $V(r) = E_{\text{crit}} \cdot r [\ln(10) - \ln(r)]$
2. Take the first derivative with respect to $r$: $\frac{dV}{dr} = E_{\text{crit}} \left[ 1 \cdot (\ln(10) - \ln(r)) + r \cdot \left(-\frac{1}{r}\right) \right]$ $\frac{dV}{dr} = E_{\text{crit}} [\ln(10) - \ln(r) - 1] = E_{\text{crit}} \left[ \ln\left(\frac{10}{r}\right) - 1 \right]$
3. Set derivative to zero to find critical points: $E_{\text{crit}} \left[ \ln\left(\frac{10}{r}\right) - 1 \right] = 0 \implies \ln\left(\frac{10}{r}\right) = 1$ $\frac{10}{r} = e^1 = e \implies r = \frac{10}{e}$
4. Calculate numerical value: $r = \frac{10}{2.71828} \approx 3.6788\text{ mm}$
5. Verify maximum via second derivative: $\frac{d^2V}{dr^2} = E_{\text{crit}} \cdot \left(-\frac{1}{r}\right) < 0$ Since the second derivative is negative for all $r > 0$, $r = 3.68\text{ mm}$ maximizes breakdown voltage capability.

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Example 3.1.3: Calculation of Radius of Curvature

Problem: Calculate the radius of curvature $\rho$ for the curve $y = \frac{1}{3} x^3$ at the point $P\left(1, \frac{1}{3}\right)$.

Solution:

1. Find the first derivative $y'$: $y' = \frac{d}{dx}\left(\frac{1}{3}x^3\right) = x^2$ At $x = 1$, $y'(1) = 1^2 = 1$.
2. Find the second derivative $y''$: $y'' = \frac{d}{dx}(x^2) = 2x$ At $x = 1$, $y''(1) = 2(1) = 2$.
3. Apply the radius of curvature formula: $\rho = \frac{[1 + (y')^2]^{3/2}}{|y''|} = \frac{[1 + (1)^2]^{3/2}}{|2|} = \frac{(2)^{3/2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$
4. Numerical evaluation: $\rho = \sqrt{2} \approx 1.4142\text{ units}$