Unit 1: Vectors and Differential Equations
        
        Vector Algebra
        
        Dot and Cross product and their properties
        Scalar (Dot) Product: A product of two vectors that results in a scalar.
        
            A · B = |A| |B| cos(θ)
        
        
            - Properties:
                
                    - Commutative: A · B = B · A
- Distributive: A · (B + C) = A · B + A · C
- If A ⊥ B (perpendicular), then θ = 90°, and A · B = 0.
- If A || B (parallel), then θ = 0°, and A · B = |A||B|.
- In components: A · B = AxBx + AyBy + AzBz
 
- Application: Work done, W = F · d.
Vector (Cross) Product: A product of two vectors that results in a new vector.
        
            A × B = |A| |B| sin(θ) n̂
        
        Where n̂ is a unit vector perpendicular to the plane of A and B, given by the Right-Hand Rule.
        
            - Properties:
                
            
- Application: Torque, τ = r × F. Angular Momentum, L = r × p.
Scalar and Vector triple products
        Scalar Triple Product (Box Product):
        
            A · (B × C)
        
        
        
        Vector Triple Product:
        
            A × (B × C)
        
        
            - Result: A vector.
- Significance: The resulting vector lies in the plane formed by B and C.
- Expansion (BAC-CAB Rule): This is a crucial identity.
                
                    A × (B × C) = B(A · C) - C(A · B)
                 
- Property: Not associative: A × (B × C) ≠ (A × B) × C.
            Common Mistake: Confusing the two triple products. Remember `A · (B × C)` gives a scalar (volume) and can be calculated with a 3x3 determinant. `A × (B × C)` gives a vector and is solved with the BAC-CAB rule.
        
        
        
        Vector Calculus: Grad, Div, Curl
        This section introduces the vector differential operator del (∇).
        
            In Cartesian coordinates: ∇ = î (∂/∂x) + ĵ (∂/∂y) + k̂ (∂/∂z)
        
        Gradient (grad) and its significance
        The gradient acts on a scalar field (f) and produces a vector field.
        
            grad(f) = ∇f = (∂f/∂x)î + (∂f/∂y)ĵ + (∂f/∂z)k̂
        
        Physical Significance:
        
            - The vector ∇f points in the direction of the steepest increase of the scalar field f.
- The magnitude |∇f| is the value of this maximum rate of change.
- ∇f is always perpendicular (normal) to the level surfaces of f (surfaces where f = constant).
- Application: Relates a potential field (scalar) to its force field (vector).
                
 Electric Field: E = -∇V
 Conservative Force: F = -∇U
Divergence (div) and its significance
        The divergence acts on a vector field (V) and produces a scalar field.
        
            div(V) = ∇ · V = (∂Vx/∂x) + (∂Vy/∂y) + (∂Vz/∂z)
        
        Physical Significance:
        
            - Divergence measures the "outflow" or "source strength" of a vector field at a point.
- ∇ · V > 0: The point is a source (field lines diverge). (e.g., a positive charge).
- ∇ · V < 0: The point is a sink (field lines converge). (e.g., a negative charge).
- ∇ · V = 0: The field is solenoidal (no sources or sinks). (e.g., a magnetic field, ∇ · B = 0).
Curl and its significance
        The curl acts on a vector field (V) and produces another vector field.
        
            
        |  î     ĵ     k̂   |
curl(V) = ∇ × V = | ∂/∂x  ∂/∂y  ∂/∂z |
        |  Vx    Vy    Vz  |
            
        
        Physical Significance:
        
            - Curl measures the microscopic "circulation" or "vorticity" of a vector field at a point.
- Imagine a tiny paddlewheel in the field. If it spins, the field has a curl.
- ∇ × V ≠ 0: The field is rotational. (e.g., water in a whirlpool, magnetic field around a wire).
- ∇ × V = 0: The field is irrotational.
                
 Key Property: All conservative fields are irrotational (∇ × F = 0).
        Vector Integration
        The syllabus specifies "statement only," so we will define the integrals and their primary meaning.
        Line Integral
        
            I = ∫C V · dr
        
        This represents the integral of the component of a vector field V that is tangent to a path C.
        
Application: Calculating the Work done by a force field F along a path: W = ∫ F · dr.
        
        Surface Integral
        
            Φ = ∫∫S V · dS
        
        This represents the integral of the component of a vector field V that is normal to a surface S.
        
Application: Calculating the Flux of a field (e.g., electric flux, magnetic flux) through a surface.
        
        Volume Integral
        
            Q = ∫∫∫V f dV
        
        This represents the integral of a scalar field f (like density) over a 3D volume V.
        
Application: Calculating the total quantity (e.g., total mass, total charge) inside a volume.
        
        
        1st Order Homogeneous Differential Equations
        This refers to linear homogeneous equations of the form:
        
            dy/dx + P(x)y = 0
        
        This type of equation is fundamental in physics, describing processes like radioactive decay or the discharge of a capacitor.
        
        Solution Method (Separation of Variables)
        
            - Rearrange the equation:
                
 dy/dx = -P(x)y
- Separate the variables (bring all 'y' terms to one side, 'x' terms to the other):
                
 dy / y = -P(x) dx
- Integrate both sides:
                
 ∫ (1/y) dy = -∫ P(x) dx + C'
 ln(y) = -∫ P(x) dx + C'
- Solve for y by taking the exponential of both sides:
                
 y = e(-∫P(x)dx + C') = e-∫P(x)dx * eC'
- Let the constant `e^C'` be `C`. The general solution is:
                
                    y(x) = C e-∫P(x)dx
                 
            Example (Radioactive Decay):
            
The rate of decay (dN/dt) is proportional to the number of atoms (N).
            
dN/dt = -λN  (where λ is the decay constant)
            
This is a 1st order homogeneous ODE.
            
dN/N = -λ dt
            
∫ dN/N = ∫ -λ dt
            
ln(N) = -λt + C'
            
N(t) = e-λt + C' = (eC') e-λt
            
Let N₀ be the number at t=0.  N₀ = (eC')e⁰ → eC' = N₀.
            
Solution: N(t) = N₀ e-λt
        
        
        
        2nd Order Homogeneous Differential Equations with Constant Coefficients
        This is one of the most important equations in physics, as it describes all types of oscillations (Simple Harmonic Motion, Damped Oscillations).
        The general form is:
        
            a (d²y/dx²) + b (dy/dx) + c y = 0
        
        where `a`, `b`, and `c` are constants.
        Solution Method (Auxiliary Equation)
        
            - Assume a trial solution of the form y = emx.
- Substitute this into the ODE:
                
 y' = memx
 y'' = m²emx
 a(m²emx) + b(memx) + c(emx) = 0
- Factor out emx (which is never zero):
                
 emx (am² + bm + c) = 0
- This leaves the Auxiliary (or Characteristic) Equation:
                
                    am² + bm + c = 0
                 
- Solve this quadratic equation for its roots, m₁ and m₂. The form of the general solution `y(x)` depends on the nature of these roots.
Three Cases for the Solution
        
        
            
                
                    | Case (based on discriminant b²-4ac) | Roots (m₁, m₂) | General Solution y(x) | Physical Example (in time) | 
            
            
                
                    | Case 1: Real and Distinct (b²-4ac > 0)
 | m₁ and m₂ are real, m₁ ≠ m₂ | y = C₁em₁x + C₂em₂x | Overdamped motion | 
                
                    | Case 2: Real and Equal (b²-4ac = 0)
 | m₁ = m₂ = m (real) | y = (C₁ + C₂x)emx | Critically Damped motion | 
                
                    | Case 3: Complex Conjugate (b²-4ac < 0)
 | m = α ± iβ | y = eαx (C₁cos(βx) + C₂sin(βx)) | Underdamped (Oscillatory) motion | 
            
        
        
        
            Example (Simple Harmonic Motion):
            
d²x/dt² + (k/m)x = 0  or  x'' + ω₀²x = 0
            
Here, `a=1`, `b=0`, `c=ω₀²`.
            
Auxiliary Equation: m² + ω₀² = 0
            
Roots: m² = -ω₀²  →  m = ±√(-ω₀²) = ± iω₀.
            
This is Case 3, with α = 0 and β = ω₀.
            
Solution: x(t) = e0t (C₁cos(ω₀t) + C₂sin(ω₀t))
            
x(t) = C₁cos(ω₀t) + C₂sin(ω₀t), which is the standard SHM solution.