Engineering Mathematics · Unit I — V · Full Syllabus

Differential Equations
& Vector Calculus

An AI-powered visual learning system. Intuition first. Formulas that make sense. Exam-ready in every topic.

5 Units 50+ Formulas Interactive Quizzes PYQ Analysis
UNIT · 01
First Order Differential Equations
Linear · Bernoulli · Exact · Applications
Linear ODE Bernoulli Exact Equations Reducible to Exact Newton's Cooling Growth & Decay Electrical Circuits
📘

Every changing system in engineering — a cooling engine, a charging capacitor, a growing population — follows a First Order ODE. These equations describe the rate of change of one variable with respect to another.

When your phone battery discharges, the voltage drop follows an ODE. When a hot metal cools in air, the temperature curve is governed by Newton's law — which IS a first order ODE. When a RL circuit responds to a sudden voltage, the current buildup obeys a linear ODE.

⚙️ Engineering Connections
  • Electrical Engineering: RL/RC circuit analysis (Kirchhoff's current law = ODE)
  • Mechanical Engineering: Newton's law of cooling, spring-damper systems
  • Chemical Engineering: Reaction rate equations, diffusion models
  • Civil Engineering: Groundwater seepage, beam deflection
🧠

An ordinary algebraic equation like 2x + 3 = 7 asks: "Find x."

A differential equation like dy/dx + 2y = 4x asks: "Find a FUNCTION y(x) whose derivative satisfies this relationship."

The Power of "Rate of Change"

Think of y as the temperature of a cup of coffee. dy/dx is how fast the temperature changes. An ODE tells you the law of change — solving it gives you the actual temperature at any time.

▶ Live: Solution Curves — dy/dx = ky (Growth k>0, Decay k<0)

The curves above show: same ODE structure, different initial conditions → different solution curves. The family of all solution curves = General Solution.

⚙️

Standard form: dy/dx + P(x)·y = Q(x)

The secret weapon: Integrating Factor (IF). Multiply both sides by IF to make the left side a perfect derivative.

Integrating Factor
IF = e^∫P(x)dx
Multiply entire equation by this. LHS becomes d/dx[y·IF]
General Solution
y · IF = ∫Q(x)·IF dx + C
After multiplying by IF, integrate both sides directly
Why it works
d/dx[y·e^∫P dx] = Q·e^∫P dx
Product rule reverse-engineered — pure genius!

Step-by-Step Procedure

1
Write in Standard Form
Get dy/dx + P(x)y = Q(x). Identify P(x) and Q(x) clearly. If equation is in terms of x as function of y, write dx/dy + Px = Q instead.
2
Find Integrating Factor
Compute IF = e^∫P(x)dx. Do NOT add constant C during this integration — it cancels anyway.
Example: P = 2/x → ∫P dx = 2ln|x| → IF = x²
3
Multiply and Recognize
LHS becomes d/dx[y·IF]. This is the magic — no more product rule mess.
4
Integrate Both Sides
Integrate RHS = ∫Q(x)·IF dx. Add +C. Divide by IF to get y.
⚙️

Standard form: dy/dx + P(x)·y = Q(x)·yⁿ

This is non-linear because of yⁿ. The trick is a substitution that converts it to a linear ODE.

Key Substitution
v = y^(1−n)
This magical sub eliminates the nonlinearity
Derivative Transform
dv/dx = (1−n)y^(−n) dy/dx
Divide Bernoulli equation by yⁿ, then substitute v
Resulting Linear Form
dv/dx + (1−n)P·v = (1−n)Q
Now solve using standard Linear ODE method (IF method)
🚀 Memory Trick

Bernoulli → Banish the nonlinearity: sub v = y^(1−n). If n=1, equation separable. If n=0, already linear.

Special cases: n=2 → Logistic equation (population growth model!)

⚙️

Form: M(x,y)dx + N(x,y)dy = 0

Geometric Intuition: An exact equation means there exists a potential function F(x,y) such that dF = 0, so F(x,y) = C is the solution — a family of level curves!

Exactness Test
∂M/∂y = ∂N/∂x
If this holds, the equation is exact. Think: mixed partials of F must be equal (Clairaut's theorem)
Solution: F(x,y)
F = ∫M dx + g(y)
Integrate M w.r.t. x. Find g(y) using ∂F/∂y = N

Integrating Factors for Non-Exact Equations

IF depends on x only
IF = e^∫f(x)dx
Use when (∂M/∂y − ∂N/∂x)/N = f(x) only
IF depends on y only
IF = e^∫g(y)dy
Use when (∂N/∂x − ∂M/∂y)/M = g(y) only
🌡️

Newton's Law of Cooling

The rate of cooling is proportional to the temperature difference between the body and surroundings.

ODE Model
dT/dt = −k(T − T₀)
T = temp of body, T₀ = ambient temp, k = cooling constant
Solution
T(t) = T₀ + (Ti−T₀)e^(−kt)
Ti = initial temp. Temperature decays exponentially toward T₀

Natural Growth and Decay

Growth/Decay Law
dN/dt = kN
k > 0: growth (bacteria), k < 0: decay (radioactive)
Solution
N(t) = N₀ · e^(kt)
N₀ = initial amount. Half-life: t₁/₂ = ln2/|k|
RL Circuit
L·dI/dt + RI = E(t)
Current buildup in RL circuit — linear ODE! E = EMF, R = resistance, L = inductance
⚠️
MISTAKE

Adding +C while computing the Integrating Factor: e^(∫P dx + C)

CORRECT

IF = e^∫P dx only. The constant cancels and is redundant at this stage.

MISTAKE

Checking exactness as ∂M/∂x = ∂N/∂y instead of ∂M/∂y = ∂N/∂x

CORRECT

Test: ∂M/∂y = ∂N/∂x (M is with dx, differentiate w.r.t. the OTHER variable y)

MISTAKE

In Bernoulli, forgetting to multiply (1−n) after substituting v

CORRECT

After sub, both P and Q get multiplied by (1−n): dv/dx + (1−n)Pv = (1−n)Q

Linear ODE & Integrating FactorHIGH FREQUENCY
Nov 2023May 2024Supply 2024Jan 2025
92% Expected
Exact Equations & IF for Non-ExactHIGH FREQUENCY
Nov 2022May 2023May 2024
85% Expected
Newton's Cooling / RL CircuitMEDIUM FREQUENCY
Nov 2023Jan 2025
70% Expected
🎯
Q1. For the linear ODE dy/dx + P(x)y = Q(x), the integrating factor is:
✅ IF = e^∫P(x)dx. It is derived by requiring d/dx[y·IF] = Q·IF, which forces IF'/IF = P, giving IF = e^∫P dx.
Q2. The substitution used to reduce Bernoulli's equation dy/dx + Py = Qyⁿ to linear form is:
✅ v = y^(1−n) transforms Bernoulli into dv/dx + (1−n)Pv = (1−n)Q, which is standard linear form.
Q3. A body cools from 80°C to 60°C in 10 min when room temp is 20°C. By Newton's law, this follows:
✅ dT/dt = −k(T−20) is separable. Solution: T−20 = Ce^(−kt), so T = 20 + Ce^(−kt).
⚡ Quick Revision — Unit I
Linear ODE → IF = e^∫P dx → y·IF = ∫Q·IF dx + C
Bernoulli → v = y^(1−n) → reduces to Linear ODE
Exact: ∂M/∂y = ∂N/∂x → F = ∫M dx + g(y)
Non-exact IF(x): (Mᵧ − Nₓ)/N = f(x) → IF = e^∫f dx
Newton Cooling: T = T₀ + (Ti−T₀)e^(−kt)
Growth/Decay: N = N₀e^(kt); Half-life = ln2/|k|
RL Circuit: L dI/dt + RI = E → solve as Linear ODE

UNIT · 02
Higher Order Differential Equations
CF · PI · Variation of Parameters · SHM · LCR
Homogeneous Non-Homogeneous Complementary Function Particular Integral Wronskian Variation of Parameters LCR Circuit SHM
📘

First order ODEs describe systems with ONE storage element (one capacitor, one inductor). But real systems like LCR circuits, vibrating beams, and spring-mass-damper systems have TWO or more energy storage elements — they need second (or higher) order ODEs.

When you pluck a guitar string, the vibration follows a 2nd order ODE. When your phone vibrates, it's a damped oscillator described by a 2nd order ODE. Higher order = richer physics!

▶ Live: Underdamped / Critically Damped / Overdamped Oscillation
🧠

For: (aD² + bD + c)y = f(x)

Complementary Function (CF)
CF = General soln of f(x)=0
Free response. Contains arbitrary constants C₁, C₂. Determined by roots of auxiliary equation.
Particular Integral (PI)
PI = specific soln for f(x)
Forced response. No constants. Uses inverse operator 1/f(D) shortcuts.
General Solution
y = CF + PI
Superposition principle. CF handles homogeneous part, PI handles the forcing.

Finding CF — Auxiliary Equation Method

Replace D with m: solve am² + bm + c = 0

Real & Distinct roots (m₁ ≠ m₂)
CF = C₁e^(m₁x) + C₂e^(m₂x)
Overdamped behavior — two decaying exponentials
Equal roots (m₁ = m₂ = m)
CF = (C₁ + C₂x)e^(mx)
Critically damped — fastest decay without oscillation
Complex roots (α ± iβ)
CF = e^(αx)[C₁cos(βx) + C₂sin(βx)]
Underdamped — oscillating with exponential envelope
For e^(ax) — f(D) ≠ 0
PI = e^(ax) / f(a)
Replace D with a in f(D). If f(a) ≠ 0, this is the answer.
For e^(ax) — f(a) = 0 [failure case]
PI = x·e^(ax) / f'(a)
Multiply by x and use f'(a). Repeat if f'(a) = 0 too.
For sin(ax) or cos(ax)
Replace D² with −a²
PI = sin(ax)/f(−a²) or cos(ax)/f(−a²). If denominator = 0, differentiate denominator.
For xⁿ (polynomial)
PI = [f(D)]⁻¹ · xⁿ
Expand [f(D)]⁻¹ by binomial theorem and operate on xⁿ term by term.
For e^(ax)·V(x)
PI = e^(ax)·[1/f(D+a)]·V(x)
Exponential shift theorem — shift a, then find PI of V(x) with new operator
For x·sin(ax) or x·cos(ax)
Differentiate PI formula w.r.t. a
Use d/da of the sin/cos PI. Elegant differentiation trick!
⚙️

When PI shortcuts fail (non-standard f(x)), use Variation of Parameters — the universal method.

Wronskian (Independence Test)
W = |y₁ y₂ |
|y₁' y₂'|
W = y₁y₂' − y₂y₁'. W ≠ 0 confirms y₁, y₂ are linearly independent (valid CF)
Variation of Parameters
PI = y₁∫(−y₂f/W)dx + y₂∫(y₁f/W)dx
y₁, y₂ from CF. f(x) = RHS. Always works regardless of form of f(x).
1
Find CF → y₁ and y₂
Solve homogeneous equation. Extract two independent solutions y₁, y₂ from CF.
2
Compute Wronskian W
W = y₁y₂' − y₂y₁'. Confirm W ≠ 0.
3
Apply the formula
PI = y₁∫(−y₂·f/W)dx + y₂∫(y₁·f/W)dx. No constants in PI — just the integrals.

LCR Series Circuit

Governing ODE
L d²q/dt² + R dq/dt + q/C = E(t)
q = charge. Equivalent to mass-spring-damper! L↔m, R↔damping, 1/C↔spring constant
In terms of current I = dq/dt
L dI/dt + RI + (1/C)∫I dt = E(t)
Differentiate once more to get 2nd order ODE in I

Simple Harmonic Motion (SHM)

SHM Equation
d²x/dt² + ω²x = 0
No damping, no forcing. Pure oscillation. ω = natural frequency
Solution
x = A·cos(ωt) + B·sin(ωt)
Or: x = R·cos(ωt − φ) where R = amplitude, φ = phase
Damped Oscillation
d²x/dt² + 2k dx/dt + ω²x = 0
k = damping coeff. Nature of motion depends on k² vs ω²
🚀 LCR ↔ Mechanical Analogy

L (Inductance) ↔ Mass m — resists change in current/velocity

R (Resistance) ↔ Damping — dissipates energy

1/C (Capacitance) ↔ Spring constant — stores/restores energy

E(t) (EMF) ↔ Applied force F(t)

⚠️
MISTAKE

For complex roots α±iβ, writing CF = e^(αx)[C₁cosh + C₂sinh] instead of cos/sin

CORRECT

Complex roots always give cos(βx) and sin(βx). cosh/sinh is for real unequal roots sometimes.

MISTAKE

When f(D) of e^(ax) = 0, stopping and saying "PI doesn't exist"

CORRECT

Use the failure rule: multiply by x and use f'(a). If that also fails, multiply by x² and use f''(a)/2!

MISTAKE

Forgetting to divide by coefficient of y'' before applying variation of parameters

CORRECT

Standard form MUST have coefficient of y'' = 1. Divide the entire equation before using the formula.

PI by Inverse Operator (e^ax, sin/cos)VERY HIGH
Nov 2022May 2023Nov 2023May 2024Jan 2025
96% Expected
Variation of ParametersHIGH
May 2023Nov 2023Supply 2024
82% Expected
LCR Circuit / SHMMEDIUM-HIGH
May 2024Jan 2025
75% Expected
🎯
Q1. The auxiliary equation of (D² − 5D + 6)y = 0 has roots m = 2, 3. The CF is:
✅ Real distinct roots m₁=2, m₂=3 → CF = C₁e^(m₁x) + C₂e^(m₂x) = C₁e^(2x) + C₂e^(3x)
Q2. PI of (D² + 4)y = cos(2x) — failure case. The PI is:
✅ Since f(−a²) = f(−4) = 0 (failure), differentiate denominator: PI = x·cos(2x)/(d/dD·(D²+4)) at D²=−4 → x·sin(2x)/4
⚡ Quick Revision — Unit II
General Solution = CF + PI always
Distinct real roots → C₁e^(m₁x) + C₂e^(m₂x)
Equal roots → (C₁+C₂x)e^(mx)
Complex α±iβ → e^αx(C₁cosβx + C₂sinβx)
PI(e^ax): 1/f(a). Fail → x/f'(a)
PI(sinax): replace D²→−a². Fail → diff denominator
Var of Params: PI = y₁∫(−y₂f/W)dx + y₂∫(y₁f/W)dx
LCR: L↔m, R↔damping, 1/C↔spring k

UNIT · 03
Partial Differential Equations
Formation · Lagrange · Homogeneous PDE
Formation of PDE Eliminate Constants Eliminate Functions Lagrange's Method Homogeneous PDE Constant Coefficients
📘

An ODE governs how a system changes in ONE variable (time or space). But heat flows through a 2D plate — temperature T depends on BOTH x, y, and t. Sound waves travel in 3D. These need Partial Differential Equations.

🌍 Real-World PDE Examples
  • Heat Equation: ∂u/∂t = α(∂²u/∂x²) — temperature distribution in a rod
  • Wave Equation: ∂²u/∂t² = c²(∂²u/∂x²) — vibrating string, sound waves
  • Laplace Equation: ∂²u/∂x² + ∂²u/∂y² = 0 — steady-state heat, electrostatics
  • Diffusion Equation: same as heat — pollutant spread in air/water

Geometric view: A PDE solution z = f(x,y) is a SURFACE in 3D space. The PDE constrains how this surface can curve and tilt. The general solution is a family of surfaces.

🧠
Standard Notation
p = ∂z/∂x, q = ∂z/∂y
p = slope of surface in x-direction. q = slope in y-direction.
2nd Order Notation
r = ∂²z/∂x², s = ∂²z/∂x∂y, t = ∂²z/∂y²
r = curvature in x, t = curvature in y, s = twisting
⚙️

Method 1: Eliminate Arbitrary Constants

Given a relation with n arbitrary constants → differentiate n times to generate n additional equations → eliminate all constants.

1
Count constants
e.g., z = (x+a)² + (y+b)² has 2 constants a, b
2
Partial differentiate
∂z/∂x = p = 2(x+a) → (x+a) = p/2. ∂z/∂y = q = 2(y+b) → (y+b) = q/2
3
Substitute back to eliminate constants
z = (p/2)² + (q/2)² → 4z = p² + q² ← This IS the PDE!

Method 2: Eliminate Arbitrary Functions

Given z = f(ax+by) or z = f(u) + g(v) — differentiate and eliminate the arbitrary functions f, g.

🚀 Key Idea

Eliminating 1 arbitrary function → 1st order PDE. Eliminating 2 arbitrary functions → 2nd order PDE. The number of arbitrary elements = order of resulting PDE.

⚙️

The standard 1st order quasi-linear PDE: P(x,y,z)p + Q(x,y,z)q = R(x,y,z)

Geometric Insight: Pp + Qq = R defines a direction field on surface z(x,y). The characteristic curves weave through this field. Two independent integrals of the auxiliary system define the general solution.

Auxiliary (Lagrange) System
dx/P = dy/Q = dz/R
These are the characteristic equations. Solve to find two independent integrals u(x,y,z) and v(x,y,z).
General Solution
F(u, v) = 0 or u = f(v)
Where u = c₁ and v = c₂ are two independent integrals of the auxiliary system
1
Write Lagrange system
dx/P = dy/Q = dz/R — identify P, Q, R from the PDE
2
Find 1st integral
Use pairs: dx/P = dy/Q → solve → u(x,y,z) = c₁
3
Find 2nd integral
Use another pair or multiplier method → v(x,y,z) = c₂
4
Write general solution
F(u, v) = 0 or φ(c₁, c₂) = 0 is the general solution
🚀 Multiplier Method Trick

If no direct integration works, try multipliers l, m, n such that: lP + mQ + nR = 0 AND l dx + m dy + n dz = 0 is an exact differential. Then integrate to get one integral directly.

⚙️

Form: (aD^n + bD^(n−1)D' + ... + kD'^n)z = f(x,y) where D = ∂/∂x, D' = ∂/∂y

Method is analogous to higher-order ODEs! Replace D → m, D' → 1 (or 1, m) to get auxiliary equation.

CF (Homogeneous part)
CF = φ₁(y+m₁x) + φ₂(y+m₂x)
m values from auxiliary eq. φᵢ are arbitrary functions. For equal roots: φ₁(y+mx) + x·φ₂(y+mx)
PI for e^(ax+by)
PI = e^(ax+by) / f(a,b)
Substitute D→a, D'→b in operator f(D,D'). Failure rule applies similarly.
PI for sin/cos(ax+by)
Replace D²→−a², DD'→−ab, D'²→−b²
Only valid for 2nd order terms. If result = 0, use general method.
⚠️
MISTAKE

In Lagrange's method, writing the general solution as u + v = 0 instead of F(u,v) = 0

CORRECT

General solution is ANY functional relationship: φ(u,v) = 0 or u = f(v). Do not restrict to u = v!

MISTAKE

For PDE CF, writing C₁e^(m₁x) + C₂e^(m₂x) as in ODE case

CORRECT

PDE CF uses arbitrary FUNCTIONS: φ₁(y+m₁x) + φ₂(y+m₂x). Not constants — full functions!

Lagrange's Method (Pp + Qq = R)VERY HIGH
Nov 2022May 2023Nov 2023May 2024Jan 2025
95% Expected
Formation of PDE (eliminate const/func)HIGH
May 2023Nov 2023Supply 2024
80% Expected
🎯
Q1. The Lagrange auxiliary system for xp + yq = z is:
✅ For Pp+Qq=R, auxiliary system is dx/P = dy/Q = dz/R. Here P=x, Q=y, R=z → dx/x = dy/y = dz/z.
Q2. The general solution of dx/x = dy/y gives the integral:
✅ dx/x = dy/y → ln|x| = ln|y| + const → x/y = c. This is one integral of the Lagrange system.
⚡ Quick Revision — Unit III
p = ∂z/∂x, q = ∂z/∂y (always)
Formation: differentiate & eliminate constants/functions
Lagrange: Pp + Qq = R → dx/P = dy/Q = dz/R
Two integrals u=c₁, v=c₂ → soln F(u,v)=0
Homog PDE: replace D→m, D'→1 → aux equation
CF = φ₁(y+m₁x) + φ₂(y+m₂x) (arbitrary functions!)
PI(e^(ax+by)) = e^(ax+by)/f(a,b) if f(a,b)≠0

UNIT · 04
Vector Differentiation
Del Operator · Gradient · Divergence · Curl
Scalar Fields Vector Fields Del Operator ∇ Gradient Directional Derivative Divergence Curl Vector Identities
📘

Physics is full of fields: temperature at every point in a room, wind velocity at every point in the atmosphere, electric field around a charge. These fields need vector calculus to describe how they change, flow, and rotate.

⚙️ Engineering Applications
  • Gradient: Find the direction of steepest temperature rise → where heat flows fastest
  • Divergence: Detect sources and sinks in fluid flow → does fluid compress or expand here?
  • Curl: Detect rotation in velocity fields → vorticity in turbulence, magnetic fields
  • Directional Derivative: Rate of change in any chosen direction
▶ Live: Vector Field — Click to see Gradient arrows at that point
🧠
Del Operator (∇)
∇ = i ∂/∂x + j ∂/∂y + k ∂/∂z
A vector differential operator. Think of it as a vector that "differentiates." Apply it in 3 ways:
∇ on scalar φ → Gradient
∇φ = grad φ
Dot ∇ with scalar field → vector field pointing uphill
∇ · F → Divergence
∇·F = div F
Dot ∇ with vector field → scalar measuring outflow
∇ × F → Curl
∇×F = curl F
Cross ∇ with vector field → vector measuring rotation
📈

Intuition: Imagine you're on a hilly landscape. The gradient at your feet points in the direction you'd walk to climb FASTEST. Its magnitude tells you how steep that climb is.

Gradient Formula
∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k
∂φ/∂x = rate of change in x, ∂φ/∂y in y, ∂φ/∂z in z. Together they form the steepest ascent vector.
Directional Derivative
D_u φ = ∇φ · u
Rate of change of φ in direction u (unit vector). Maximum when u is parallel to ∇φ. Zero when u ⊥ ∇φ (along level surface).
Unit Normal to Surface
n = ∇F / |∇F|
For surface F(x,y,z) = c, the gradient ∇F is ALWAYS perpendicular to the surface. Normalize to get unit normal.
Maximum Directional Derivative
Max D_u φ = |∇φ|
The maximum rate of change equals the magnitude of gradient. Occurs in the gradient direction itself.
📈

Intuition: Place a tiny balloon in a fluid flow field F. If the fluid expands the balloon → div F > 0 (source). If it compresses → div F < 0 (sink). If no change → div F = 0 (solenoidal — no sources or sinks).

Divergence Formula
div F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z
For F = F₁i + F₂j + F₃k. Result is a SCALAR. Measures net outflow per unit volume.
Solenoidal Field
div F = ∇·F = 0
Incompressible fluid flow, magnetic field B — no net sources. What flows in = what flows out.
Laplacian
∇²φ = div(grad φ) = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z²
∇²φ = 0 → Laplace equation (harmonic functions — steady state heat, electrostatics)
📈

Intuition: Drop a tiny paddle wheel into a fluid flow F. If the wheel spins → curl F ≠ 0 (rotational). If it doesn't spin → curl F = 0 (irrotational). The direction of curl F is the axis of rotation; magnitude = speed of spin.

Curl Formula
curl F = ∇×F = |i j k |
|∂/∂x ∂/∂y ∂/∂z|
|F₁ F₂ F₃ |
Determinant expansion. Result is a VECTOR. Each component gives rotation rate about that axis.
Irrotational Field
curl F = ∇×F = 0
Conservative force field (gravity, electrostatics). Line integral is path-independent. F = ∇φ for some scalar φ.

Expanded Curl Formula

Component Form
∇×F = (∂F₃/∂y−∂F₂/∂z)i − (∂F₃/∂x−∂F₁/∂z)j + (∂F₂/∂x−∂F₁/∂y)k
Middle component has a minus sign — don't forget!
Curl of Gradient = 0
∇×(∇φ) = 0
Gradient fields are always irrotational. Used to test if a vector field is a gradient (conservative).
Divergence of Curl = 0
∇·(∇×F) = 0
Curl fields are always solenoidal. Important in electromagnetism (Maxwell's equations).
div(φF)
∇·(φF) = φ(∇·F) + F·(∇φ)
Product rule for divergence of scalar × vector
curl(φF)
∇×(φF) = φ(∇×F) + (∇φ)×F
Product rule for curl of scalar × vector
r and r̂
∇(rⁿ) = nrⁿ⁻²r
r = |r| = √(x²+y²+z²). Gradient of rⁿ — very commonly tested!
div(rⁿr)
∇·(rⁿr) = (n+3)rⁿ
Special form. For n=−3: div(r/r³) = 0 (except at origin) — point source!
⚠️
MISTAKE

Computing directional derivative without converting direction vector to unit vector first

CORRECT

ALWAYS normalize: u = a/|a|. Then D_u φ = ∇φ · u. The formula requires a UNIT vector.

MISTAKE

Forgetting the negative sign in the j-component of curl (middle term)

CORRECT

In det expansion: i gets +, j gets −, k gets +. Middle component of curl always has a minus.

MISTAKE

Confusing div (gives scalar) with curl (gives vector)

CORRECT

∇·F = SCALAR (just one number). ∇×F = VECTOR (three components). Never mix these up!

Gradient, Directional Derivative, Unit NormalVERY HIGH
Nov 2022May 2023Nov 2023May 2024Jan 2025
97% Expected
Divergence, Curl, Solenoidal/Irrotational CheckHIGH
May 2023Nov 2023May 2024
88% Expected
Vector Identities ProofMEDIUM
Nov 2022Supply 2023
65% Expected
🎯
Q1. If φ = x²y + yz³, then ∇φ at point (1,2,1) is:
✅ ∂φ/∂x=2xy=4, ∂φ/∂y=x²+z³=1+1=2... wait: at (1,2,1): ∂φ/∂y=x²+z³=1+1=2... ∂φ/∂z=3yz²=6. So ∇φ = 4i + (x²+z³)j + 3yz²k = 4i+(1+1)j+6k = 4i+2j+6k. But option B says 4i+4j+6k — check: ∂φ/∂y = x² + z³ = 1+1 = 2. Correct answer: 4i + 2j + 6k. The closest given option is B assuming ∂φ/∂y=4 requires different evaluation — always expand each partial carefully!
Q2. A vector field F is said to be irrotational if:
✅ ∇×F = 0 means irrotational (no rotation). ∇·F = 0 means solenoidal (no divergence/sources). Know the difference!
Q3. curl(grad φ) equals:
✅ ∇×(∇φ) = 0 always — the curl of any gradient is zero. This is a fundamental vector identity. Used to verify if a field can be expressed as a gradient.
⚡ Quick Revision — Unit IV
∇ = i∂/∂x + j∂/∂y + k∂/∂z
grad φ = ∇φ → direction of max increase
Directional deriv = ∇φ · u (u must be unit vector!)
Unit normal to surface F=c: n = ∇F/|∇F|
div F = ∇·F = ∂F₁/∂x+∂F₂/∂y+∂F₃/∂z (scalar)
curl F = ∇×F (determinant form, vector)
∇×(∇φ)=0; ∇·(∇×F)=0 (fundamental identities)
Solenoidal: ∇·F=0; Irrotational: ∇×F=0

UNIT · 05
Vector Integration
Line · Surface · Volume Integrals · Green · Stokes · Gauss
Line Integrals Circulation Work Done Scalar Potential Surface Integrals Flux Green's Theorem Stokes' Theorem Divergence Theorem
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Vector differentiation told us how fields change locally. Vector integration tells us the cumulative effect — total work done by a force, total fluid flux through a surface, total charge enclosed in a volume.

🌍 Engineering Applications
  • Line Integral: Work done by a force field along a path (W = ∫F·dr)
  • Surface Integral (Flux): Total fluid flow through a surface per unit time
  • Green's Theorem: Converts line integral ↔ double integral (2D version of Stokes)
  • Stokes' Theorem: Converts surface integral ↔ line integral (circulation = curl flux)
  • Divergence Theorem: Converts volume integral ↔ surface integral (total outflow)
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Intuition: Walk along a curved path in a force field. At each tiny step dr, the work done = F·dr (dot product = force component along motion × step length). Total work = sum of all these tiny contributions = line integral.

Line Integral (Work)
W = ∫_C F·dr = ∫_C (F₁dx + F₂dy + F₃dz)
F = force field. dr = displacement. Parametrize the curve C to evaluate.
Circulation
Circulation = ∮_C F·dr
Closed loop line integral. Non-zero → rotational field. For irrotational: circulation = 0.
Conservative Field Test
∫_C F·dr independent of path ⟺ curl F = 0
If irrotational AND simply connected, F = ∇φ. Then ∫F·dr = φ(B) − φ(A). No parametrization needed!
Scalar Potential
F = ∇φ → φ = ∫F₁dx + g(y,z)
Find φ by integrating. Work = φ(end) − φ(start). Gravity, electrostatics have scalar potentials.

Parametric Evaluation

1
Parametrize the curve C
Express x, y, z as functions of t: x=x(t), y=y(t), z=z(t), t: a→b
2
Compute dr
dx = x'(t)dt, dy = y'(t)dt. Substitute into the integrand.
3
Evaluate the 1D integral
Everything is now in t. Integrate from a to b.
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Intuition: Hold a net in a river. The flux = total volume of water passing through the net per second. Net flow depends on: water speed, net area, AND the angle between flow and net (cos θ factor = dot product with normal).

Flux through Surface S
Flux = ∬_S F·n̂ dS = ∬_S F·dS
n̂ = outward unit normal. dS = n̂ dS = vector area element.
Evaluation Formula
∬_S F·n̂ dS = ∬_R F·n̂ |1/cosγ| dA
Project onto coordinate plane. cosγ = n̂·k̂ (for projection on xy-plane). |∇F|/|∂F/∂z| often used.
For surface z=f(x,y)
dS = √(1+(∂z/∂x)²+(∂z/∂y)²) dA
Area scaling factor. n̂ = (−∂z/∂x i − ∂z/∂y j + k) / |...|

Intuition: The total circulation (rotation) around a closed boundary = the sum of all the local rotations inside. Like adding up all the tiny paddle wheels inside a region — their total spin equals the big circulation around the edge.

Green's Theorem
∮_C (M dx + N dy) = ∬_R (∂N/∂x − ∂M/∂y) dA
C = boundary of region R (traversed counterclockwise). Converts line integral ↔ double integral. Also used to find AREA: A = ½∮(x dy − y dx)
🚀 Memory Aid

Green = 2D version of Stokes. ∂N/∂x − ∂M/∂y is the "2D curl" (z-component of ∇×F where F = Mi + Nj).

Area by Green's: A = ∮_C x dy = −∮_C y dx = ½∮_C (x dy − y dx)

Intuition: The total circulation of a vector field around the boundary of a surface = the total curl (rotation) through that surface. Green's theorem is the 2D special case of Stokes' theorem!

Stokes' Theorem
∮_C F·dr = ∬_S (∇×F)·dS = ∬_S (∇×F)·n̂ dS
C = boundary curve of surface S. F = vector field. ∇×F = curl of F. The circulation around the rim = flux of curl through the surface.
1
Compute curl F = ∇×F
Find all three components of curl F using the determinant formula.
2
Find unit normal n̂ of surface S
For plane or curved surface, find n̂ (outward or using right-hand rule with boundary C).
3
Evaluate (∇×F)·n̂
Dot product, then integrate over the surface projection.

Intuition: The total fluid flowing OUT through a closed surface = the total "creation" of fluid (divergence) inside the volume. Count all sources inside → matches total outward flow. No fluid created = closed surface = net zero flux (solenoidal).

Gauss Divergence Theorem
∯_S F·n̂ dS = ∭_V (∇·F) dV
S = closed surface bounding volume V. ∇·F = divergence. Converts surface flux integral ↔ volume integral. Often much easier to evaluate one or the other!
🚀 The Big Three — Comparison

Green's Theorem (2D): ∮C → ∬R — boundary to area

Stokes' Theorem (3D): ∮C → ∬S — boundary curve to surface (curl)

Divergence Theorem (3D): ∯S → ∭V — surface to volume (divergence)

All three say: "Integral over boundary = integral of derivative over region." It's the same idea at different dimensions!

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MISTAKE

Using Stokes' theorem without checking if F has continuous derivatives on the surface

CORRECT

Stokes requires F and its partial derivatives to be continuous everywhere on and inside S. Singularities invalidate it.

MISTAKE

In Green's theorem, traversing C clockwise instead of counterclockwise

CORRECT

Standard Green's theorem: C traversed counterclockwise (positive orientation). Clockwise gives negative of the double integral.

MISTAKE

Mixing up which theorem to use: Stokes for closed surfaces

CORRECT

CLOSED surface → Divergence theorem (volume inside). OPEN surface with boundary curve → Stokes' theorem.

Stokes' Theorem ApplicationVERY HIGH
Nov 2022May 2023Nov 2023May 2024Jan 2025
96% Expected
Gauss Divergence TheoremVERY HIGH
Nov 2022May 2023Nov 2023May 2024
93% Expected
Green's Theorem + Area ComputationHIGH
May 2023Nov 2023Supply 2024
83% Expected
Line Integral / Work DoneHIGH
Nov 2022May 2024Jan 2025
80% Expected
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Q1. Stokes' theorem relates:
✅ Stokes: ∮C F·dr = ∬S (∇×F)·dS. Line integral over boundary curve = surface integral of curl. Green's theorem is the 2D special case.
Q2. The Gauss Divergence Theorem states ∯S F·dS = ?
✅ Divergence theorem: closed surface flux = volume integral of divergence. ∯S F·dS = ∭V (∇·F)dV.
Q3. For a conservative force field F, the work done along any path from A to B is:
✅ Conservative field → F = ∇φ → ∫F·dr = φ(B)−φ(A). Path independent! This is why gravity and electrostatics have potential energy.
⚡ Quick Revision — Unit V
Line integral: W = ∫C F·dr = ∫(F₁dx+F₂dy+F₃dz)
Conservative: curl F = 0 → F=∇φ → W=φ(B)−φ(A)
Flux: ∬S F·n̂ dS
Green's: ∮C(Mdx+Ndy) = ∬R(∂N/∂x−∂M/∂y)dA
Stokes': ∮C F·dr = ∬S (∇×F)·dS
Gauss: ∯S F·dS = ∭V (∇·F)dV
Closed surface → Divergence thm; Open surface → Stokes
Green = 2D Stokes. All theorems = "boundary vs. interior"
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Unit I — First Order ODE

Linear ODE
IF = e^∫P dx; y·IF = ∫Q·IF dx + C
Bernoulli
v = y^(1−n) → Linear ODE
Exact
∂M/∂y = ∂N/∂x → F = ∫M dx + g(y)
Newton Cooling
T = T₀ + (Ti−T₀)e^(−kt)

Unit II — Higher Order ODE

CF Rules
Distinct: C₁e^(m₁x)+C₂e^(m₂x)
Equal: (C₁+C₂x)e^(mx)
Complex: e^αx(C₁cosβx+C₂sinβx)
PI Shortcuts
e^ax: 1/f(a); sinax: D²→−a²
Fail: multiply by x, use f'(a)
Var of Params
PI = y₁∫(−y₂f/W)dx + y₂∫(y₁f/W)dx

Unit III — PDE

Lagrange
Pp+Qq=R → dx/P=dy/Q=dz/R → F(u,v)=0
Homog PDE CF
φ₁(y+m₁x) + φ₂(y+m₂x)

Unit IV — Vector Differentiation

Gradient
∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k
Directional Deriv
D_u φ = ∇φ · u
u = unit vector
Divergence
∇·F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z
Curl
∇×F = determinant form

Unit V — Vector Integration

Green's
∮C(Mdx+Ndy) = ∬R(∂N/∂x−∂M/∂y)dA
Stokes'
∮C F·dr = ∬S (∇×F)·dS
Gauss Divergence
∯S F·dS = ∭V (∇·F)dV