2.3 Analytic Geometry & Differential Calculus

PRC Electronics Engineering Licensure Examination Study Guide Exam

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1 Chapter 1: Introduction & Board Exam Strategy

1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy

2 Chapter 2: Mathematics Foundations

2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications

3 Chapter 3: Advanced Mathematics for ECE

3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability

4 Chapter 4: GEAS - Engineering Sciences & Economics

4.1 Engineering Physics & Materials Chemistry 4.2 Engineering Mechanics & Strength of Materials 4.3 Engineering Economics & Financial Analysis 4.4 Computer Fundamentals & C/C++ Programming

5 Chapter 5: GEAS - Management, Materials, Environment & CAD

5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics

6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts

6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)

7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes

7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators

8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits

8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design

9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems

9.1 Feedback Topologies & Oscillator Circuits 9.2 Boolean Algebra, Logic Gates & Combinational Circuits 9.3 Sequential Logic, Flip-Flops & Microprocessor Architecture 9.4 Industrial Power Electronics & Control System Theory

10 Chapter 10: EST - Communications Fundamentals & Transmission Lines

10.1 Signal Fundamentals, Decibel Calculations & Noise Analysis 10.2 Amplitude Modulation (AM) Principles & Receivers 10.3 Frequency Modulation (FM), Phase Modulation & Discriminators 10.4 Transmission Line Parameters, SWR & Smith Chart

11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems

11.1 Pulse & Digital Modulation Systems 11.2 Data Communications, OSI Model & Protocols 11.3 Antenna Principles & Electromagnetic Wave Propagation 11.4 Optical Fiber, Microwave, Satellite & Philippine NTC Regulations

2.3 Analytic Geometry & Differential Calculus

Key Takeaways Explain This Topic AI

Identify and analyze equations of conic sections (parabolas, ellipses, hyperbolas) applied to antenna reflectors and navigational positioning.

Apply L'Hôpital's Rule to resolve indeterminate limits in signal processing and transfer function analysis.

Master derivative rules for polynomial, exponential, logarithmic, and trigonometric functions.

Perform optimization using first and second derivatives to solve physical engineering problems, including the Maximum Power Transfer Theorem.

Last updated: July 2026

2.3 Analytic Geometry & Differential Calculus

Analytic geometry provides geometric interpretations of algebraic equations, while differential calculus allows electronics engineers to calculate instantaneous rates of change, optimize system performance, and analyze dynamic waveforms.

1. Conic Sections in ECE Applications

Conic sections are curves formed by intersecting a cone with a plane. Their reflective and geometric properties make them central to antenna design and radio navigation.

Summary of Conic Equations & ECE Uses

Conic Section Standard Equation Key Parameters ECE Application

Circle ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2 + (y-k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 Center ( h , k ) (h,k) ( h , k ) , Radius r r r Isotropic antenna radiation patterns

Parabola ( y − k ) 2 = 4 p ( x − h ) (y-k)^2 = 4p(x-h) ( y − k ) 2 = 4 p ( x − h ) Focal distance p p p , Vertex ( h , k ) (h,k) ( h , k ) Parabolic reflector dish antennas

Ellipse ( x − h ) 2 a 2 + ( y − k ) 2 b 2 = 1 \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 a 2 ( x − h ) 2 ​ + b 2 ( y − k ) 2 ​ = 1 Semi-major a a a , Semi-minor b b b , e = c a < 1 e = \frac{c}{a} < 1 e = a c ​ < 1 Satellite elliptical orbits (Keplerian)

Hyperbola ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 a 2 ( x − h ) 2 ​ − b 2 ( y − k ) 2 ​ = 1 Foci c = a 2 + b 2 c = \sqrt{a^2 + b^2} c = a 2 + b 2 ​ , e = c a > 1 e = \frac{c}{a} > 1 e = a c ​ > 1 LORAN hyperbolic navigation lines

Worked ECE Board Example 1: Parabolic Reflector Focal Point

Problem: A satellite dish antenna has a parabolic cross-section. The opening diameter of the dish is 2 meters 2\text{ meters} 2 meters wide, and its depth at the center vertex ( 0 , 0 ) (0,0) ( 0 , 0 ) is 0.5 meters 0.5\text{ meters} 0.5 meters . Find the parabolic equation y 2 = 4 p x y^2 = 4px y 2 = 4 p x and determine the exact position of the feed horn receiver at the focus ( p , 0 ) (p,0) ( p , 0 ) .

Solution:

Step 1: Set vertex at ( 0 , 0 ) (0,0) ( 0 , 0 ) . The outer rim edge point lies at x = 0.5 m x = 0.5\text{ m} x = 0.5 m and y = 2 m 2 = 1.0 m y = \frac{2\text{ m}}{2} = 1.0\text{ m} y = 2 2 m ​ = 1.0 m .

Step 2: Substitute point ( 0.5 , 1.0 ) (0.5, 1.0) ( 0.5 , 1.0 ) into y 2 = 4 p x y^2 = 4px y 2 = 4 p x : ( 1.0 ) 2 = 4 p ( 0.5 ) ⟹ 1.0 = 2 p ⟹ p = 0.5 meters (1.0)^2 = 4p(0.5) \implies 1.0 = 2p \implies p = 0.5\text{ meters} ( 1.0 ) 2 = 4 p ( 0.5 ) ⟹ 1.0 = 2 p ⟹ p = 0.5 meters

Step 3: Parabola equation is y 2 = 2 x y^2 = 2x y 2 = 2 x . The feed horn receiver must be placed at the focal point ( 0.5 m , 0 ) (0.5\text{ m}, 0) ( 0.5 m , 0 ) directly inside the dish.

2. Limits, Indeterminate Forms & L'Hôpital's Rule

When evaluating limits lim ⁡ x → a f ( x ) g ( x ) \lim_{x \to a} \frac{f(x)}{g(x)} lim x → a ​ g ( x ) f ( x ) ​ that result in indeterminate forms [ 0 0 ] \left[\frac{0}{0}\right] [ 0 0 ​ ] or [ ∞ ∞ ] \left[\frac{\infty}{\infty}\right] [ ∞ ∞ ​ ] , L'Hôpital's Rule applies: lim ⁡ x → a f ( x ) g ( x ) = lim ⁡ x → a f ′ ( x ) g ′ ( x ) \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} x → a lim ​ g ( x ) f ( x ) ​ = x → a lim ​ g ′ ( x ) f ′ ( x ) ​

provided the limit of the derivatives exists.

Worked ECE Board Example 2: Sinc Function Limit in Signal Processing

Problem: Evaluate the zero-frequency spectral magnitude limit L = lim ⁡ t → 0 1 − cos ⁡ ( 3 t ) t 2 L = \lim_{t \to 0} \frac{1 - \cos(3t)}{t^2} L = lim t → 0 ​ t 2 1 − c o s ( 3 t ) ​ .

Solution:

Step 1: Direct substitution yields 1 − cos ⁡ ( 0 ) 0 = 0 0 \frac{1 - \cos(0)}{0} = \frac{0}{0} 0 1 − c o s ( 0 ) ​ = 0 0 ​ (Indeterminate).

Step 2: Apply L'Hôpital's Rule (First Derivative): L = lim ⁡ t → 0 d d t ( 1 − cos ⁡ ( 3 t ) ) d d t ( t 2 ) = lim ⁡ t → 0 3 sin ⁡ ( 3 t ) 2 t L = \lim_{t \to 0} \frac{\frac{d}{dt}(1 - \cos(3t))}{\frac{d}{dt}(t^2)} = \lim_{t \to 0} \frac{3\sin(3t)}{2t} L = t → 0 lim ​ d t d ​ ( t 2 ) d t d ​ ( 1 − cos ( 3 t )) ​ = t → 0 lim ​ 2 t 3 sin ( 3 t ) ​

Step 3: Direct substitution still yields 0 0 \frac{0}{0} 0 0 ​ . Apply L'Hôpital's Rule a second time: L = lim ⁡ t → 0 d d t ( 3 sin ⁡ ( 3 t ) ) d d t ( 2 t ) = lim ⁡ t → 0 9 cos ⁡ ( 3 t ) 2 = 9 cos ⁡ ( 0 ) 2 = 9 2 = 4.5 L = \lim_{t \to 0} \frac{\frac{d}{dt}(3\sin(3t))}{\frac{d}{dt}(2t)} = \lim_{t \to 0} \frac{9\cos(3t)}{2} = \frac{9\cos(0)}{2} = \frac{9}{2} = 4.5 L = t → 0 lim ​ d t d ​ ( 2 t ) d t d ​ ( 3 sin ( 3 t )) ​ = t → 0 lim ​ 2 9 cos ( 3 t ) ​ = 2 9 cos ( 0 ) ​ = 2 9 ​ = 4.5

• 3. Derivative Rules & Transcendental Derivatives
• Differentiation Rules Summary Table

• Function Type Function f ( x ) f(x) f ( x ) Derivative f ′ ( x ) f'(x) f ′ ( x )
• Power Rule x n x^n x n n x n − 1 n x^{n-1} n x n − 1
• Exponential e u ( x ) e^{u(x)} e u ( x ) e u ⋅ d u d x e^{u} \cdot \frac{du}{dx} e u ⋅ d x d u ​

Natural Logarithm ln ⁡ ( u ( x ) ) \ln(u(x)) ln ( u ( x )) 1 u ⋅ d u d x \frac{1}{u} \cdot \frac{du}{dx} u 1 ​ ⋅ d x d u ​

Sine / Cosine sin ⁡ ( u ) / cos ⁡ ( u ) \sin(u) \quad / \quad \cos(u) sin ( u ) / cos ( u ) cos ⁡ ( u ) d u d x / − sin ⁡ ( u ) d u d x \cos(u)\frac{du}{dx} \quad / \quad -\sin(u)\frac{du}{dx} cos ( u ) d x d u ​ / − sin ( u ) d x d u ​

Product Rule u ( x ) ⋅ v ( x ) u(x) \cdot v(x) u ( x ) ⋅ v ( x ) u ′ v + u v ′ u'v + uv' u ′ v + u v ′

Quotient Rule u ( x ) v ( x ) \frac{u(x)}{v(x)} v ( x ) u ( x ) ​ u ′ v − u v ′ v 2 \frac{u'v - uv'}{v^2} v 2 u ′ v − u v ′ ​

Chain Rule f ( g ( x ) ) f(g(x)) f ( g ( x )) f ′ ( g ( x ) ) ⋅ g ′ ( x ) f'(g(x)) \cdot g'(x) f ′ ( g ( x )) ⋅ g ′ ( x )

4. Optimization & Maximum Power Transfer Theorem Proof

Differential calculus is used to find maximum or minimum values of physical variables by setting the first derivative to zero ( f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 ) and confirming concavity using the second derivative test ( f ′ ′ ( x ) < 0 f''(x) < 0 f ′′ ( x ) < 0 for maximum, f ′ ′ ( x ) > 0 f''(x) > 0 f ′′ ( x ) > 0 for minimum).

Worked ECE Board Example 3: Maximum Power Transfer Proof

Problem: A DC source has a Thevenin equivalent voltage V th V_{\text{th}} V th ​ and internal resistance R th R_{\text{th}} R th ​ . When connected to a variable load resistor R L R_L R L ​ , the power delivered to the load is: P ( R L ) = I 2 R L = ( V th R th + R L ) 2 R L = V th 2 R L ( R th + R L ) 2 P(R_L) = I^2 R_L = \left(\frac{V_{\text{th}}}{R_{\text{th}} + R_L}\right)^2 R_L = \frac{V_{\text{th}}^2 R_L}{(R_{\text{th}} + R_L)^2} P ( R L ​ ) = I 2 R L ​ = ( R th ​ + R L ​ V th ​ ​ ) 2 R L ​ = ( R th ​ + R L ​ ) 2 V th 2 ​ R L ​ ​

Use calculus to prove that maximum power transfer occurs when R L = R th R_L = R_{\text{th}} R L ​ = R th ​ , and compute P max P_{\text{max}} P max ​ .

Solution:

Step 1: Take the derivative of P P P with respect to R L R_L R L ​ using the quotient rule, treating V th V_{\text{th}} V th ​ and R th R_{\text{th}} R th ​ as constants: d P d R L = V th 2 [ ( 1 ) ( R th + R L ) 2 − R L ⋅ 2 ( R th + R L ) ( 1 ) ( R th + R L ) 4 ] \frac{dP}{dR_L} = V_{\text{th}}^2 \left[ \frac{(1)(R_{\text{th}} + R_L)^2 - R_L \cdot 2(R_{\text{th}} + R_L)(1)}{(R_{\text{th}} + R_L)^4} \right] d R L ​ d P ​ = V th 2 ​ [ ( R th ​ + R L ​ ) 4 ( 1 ) ( R th ​ + R L ​ ) 2 − R L ​ ⋅ 2 ( R th ​ + R L ​ ) ( 1 ) ​ ]

Step 2: Simplify the numerator: d P d R L = V th 2 [ ( R th + R L ) − 2 R L ( R th + R L ) 3 ] = V th 2 [ R th − R L ( R th + R L ) 3 ] \frac{dP}{dR_L} = V_{\text{th}}^2 \left[ \frac{(R_{\text{th}} + R_L) - 2R_L}{(R_{\text{th}} + R_L)^3} \right] = V_{\text{th}}^2 \left[ \frac{R_{\text{th}} - R_L}{(R_{\text{th}} + R_L)^3} \right] d R L ​ d P ​ = V th 2 ​ [ ( R th ​ + R L ​ ) 3 ( R th ​ + R L ​ ) − 2 R L ​ ​ ] = V th 2 ​ [ ( R th ​ + R L ​ ) 3 R th ​ − R L ​ ​ ]

Step 3: Set d P d R L = 0 \frac{dP}{dR_L} = 0 d R L ​ d P ​ = 0 : R th − R L = 0 ⟹ R L = R th R_{\text{th}} - R_L = 0 \implies R_L = R_{\text{th}} R th ​ − R L ​ = 0 ⟹ R L ​ = R th ​

Step 4: Substitute R L = R th R_L = R_{\text{th}} R L ​ = R th ​ back into the power equation: P max = V th 2 ( R th ) ( R th + R th ) 2 = V th 2 R th 4 R th 2 = V th 2 4 R th P_{\text{max}} = \frac{V_{\text{th}}^2 (R_{\text{th}})}{(R_{\text{th}} + R_{\text{th}})^2} = \frac{V_{\text{th}}^2 R_{\text{th}}}{4 R_{\text{th}}^2} = \frac{V_{\text{th}}^2}{4 R_{\text{th}}} P max ​ = ( R th ​ + R th ​ ) 2 V th 2 ​ ( R th ​ ) ​ = 4 R th 2 ​ V th 2 ​ R th ​ ​ = 4 R th ​ V th 2 ​ ​

5. Tangents & Normals to Curves

For a curve y = f ( x ) y = f(x) y = f ( x ) at point ( x 0 , y 0 ) (x_0, y_0) ( x 0 ​ , y 0 ​ ) :

• Slope of Tangent Line : m t = f ′ ( x 0 ) m_t = f'(x_0) m t ​ = f ′ ( x 0 ​ )
• Equation of Tangent Line : y − y 0 = m t ( x − x 0 ) y - y_0 = m_t(x - x_0) y − y 0 ​ = m t ​ ( x − x 0 ​ )

Slope of Normal Line : m n = − 1 m t = − 1 f ′ ( x 0 ) m_n = -\frac{1}{m_t} = -\frac{1}{f'(x_0)} m n ​ = − m t ​ 1 ​ = − f ′ ( x 0 ​ ) 1 ​

• Equation of Normal Line : y − y 0 = m n ( x − x 0 ) y - y_0 = m_n(x - x_0) y − y 0 ​ = m n ​ ( x − x 0 ​ )
• 6. Board Exam Summary Table

Calculus Concept Mathematical Condition ECE Board Exam Application

Parabola Focus y 2 = 4 p x ⟹ Focus at ( p , 0 ) y^2 = 4px \implies \text{Focus at }(p,0) y 2 = 4 p x ⟹ Focus at ( p , 0 ) Locating feed horn in parabolic dish antennas

L'Hôpital's Rule lim ⁡ f g = lim ⁡ f ′ g ′ \lim \frac{f}{g} = \lim \frac{f'}{g'} lim g f ​ = lim g ′ f ′ ​ for 0 0 , ∞ ∞ \frac{0}{0}, \frac{\infty}{\infty} 0 0 ​ , ∞ ∞ ​ Resolving indeterminate transfer function limits

Optimization d y d x = 0 , d 2 y d x 2 < 0 \frac{dy}{dx} = 0, \frac{d^2y}{dx^2} < 0 d x d y ​ = 0 , d x 2 d 2 y ​ < 0 Proving Maximum Power Transfer ( R L = R th R_L = R_{\text{th}} R L ​ = R th ​ )

Tangent Slope m t = f ′ ( x 0 ) m_t = f'(x_0) m t ​ = f ′ ( x 0 ​ ) Calculating beam launch angles in fiber optics