Important Derivations

Steps

1. 1

   Define acceleration: a = dv/dt → dv = a dt; integrate both sides

2. 2

   ∫ᵤᵛ dv = ∫₀ᵗ a dt → v − u = at → v = u + at ...(i)

3. 3

   Velocity v = ds/dt; substitute v = u + at: ds = (u + at) dt

4. 4

   Integrate: s = ut + ½at² ...(ii)

5. 5

   From (i): t = (v−u)/a; substitute into s = ut + ½at²

6. 6

   s = u(v−u)/a + ½a(v−u)²/a² → simplify → v² = u² + 2as ...(iii)

Steps

1. 1

   Resolve initial velocity: horizontal uₓ = u cosθ, vertical u_y = u sinθ

2. 2

   Vertical motion: y = u sinθ · t − ½gt²; at maximum height v_y = 0

3. 3

   Time to reach max height: t_H = u sinθ/g → Total time T = 2u sinθ/g

4. 4

   Maximum height H = u_y · t_H − ½g t_H² = u² sin²θ / 2g

5. 5

   Horizontal range R = uₓ × T = u cosθ × 2u sinθ/g = u² sin2θ/g

6. 6

   R is maximum when sin2θ = 1, i.e., θ = 45°

Steps

1. 1

   Consider particle moving in a circle of radius r with uniform speed v

2. 2

   Position vectors P₁ and P₂ are separated by angle Δθ; |Δv| = v Δθ (since |v₁|=|v₂|=v)

3. 3

   Centripetal acceleration a = Δv/Δt = v (Δθ/Δt) = v ω

4. 4

   Since v = rω, substitute ω = v/r: a = v · v/r = v²/r

5. 5

   Direction: Δv points towards centre as Δθ → 0

Steps

1. 1

   From Newton's second law: F = ma

2. 2

   Work done by net force: W = ∫F · ds = ∫m a ds

3. 3

   Use a ds = v dv (since a = dv/dt and ds = v dt): W = ∫ᵤᵛ mv dv

4. 4

   Integrate: W = ½mv² − ½mu²

5. 5

   Therefore W_net = ΔKE — the work-energy theorem

Steps

1. 1

   Consider a body of mass m falling from height h under gravity (conservative force)

2. 2

   At height h: KE = 0, PE = mgh → Total E = mgh

3. 3

   At height h': velocity v'; using v² = 2g(h−h'): KE = mg(h−h'), PE = mgh'

4. 4

   Total E = mg(h−h') + mgh' = mgh — same as initial

5. 5

   For any conservative force: W_conservative = −ΔPE; W = ΔKE → ΔKE + ΔPE = 0

6. 6

   Therefore KE + PE = constant

Steps

1. 1

   Consider a thin rod of mass M and length L; let linear mass density λ = M/L

2. 2

   Take a small element of length dx at distance x from centre (x: −L/2 to +L/2)

3. 3

   Mass of element: dm = λ dx = (M/L) dx

4. 4

   Contribution to MI: dI = x² dm = (M/L) x² dx

5. 5

   I = ∫₋L/2^(L/2) (M/L) x² dx = (M/L) [x³/3]₋L/2^(L/2)

6. 6

   I = (M/L) × 2(L/2)³/3 = (M/L) × L³/12 = ML²/12

Steps

1. 1

   Torque is defined as τ = dL/dt where L = angular momentum

2. 2

   If net external torque τ_net = 0, then dL/dt = 0

3. 3

   Therefore L = constant (angular momentum is conserved)

4. 4

   For a system: I₁ω₁ = I₂ω₂ when no external torque acts

5. 5

   Example: figure skater pulling arms in → I decreases → ω increases

Steps

1. 1

   For a body to escape, its kinetic energy must equal the magnitude of gravitational PE at surface

2. 2

   KE at surface: ½mv_e²; Gravitational PE: −GMm/R

3. 3

   Total energy must be ≥ 0 for escape: ½mv_e² − GMm/R = 0

4. 4

   Solve: v_e = √(2GM/R)

5. 5

   Since g = GM/R², substitute GM = gR²: v_e = √(2gR)

6. 6

   Numerically: v_e = √(2 × 9.8 × 6.4×10⁶) ≈ 11.2 km/s

Steps

1. 1

   For a satellite at height h, gravitational force provides centripetal force: GMm/(R+h)² = mv₀²/(R+h)

2. 2

   Solve for orbital velocity: v₀ = √(GM/(R+h))

3. 3

   For h << R (near-Earth orbit): v₀ ≈ √(gR) ≈ 7.9 km/s

4. 4

   Time period: T = circumference/speed = 2π(R+h)/v₀

5. 5

   Substitute v₀: T = 2π√((R+h)³/GM)

6. 6

   This gives Kepler's third law: T² ∝ r³ where r = R+h

Steps

1. 1

   Restoring force for spring: F = −kx (Hooke's Law)

2. 2

   Newton's second law: F = ma → ma = −kx → a = −(k/m)x

3. 3

   Comparing with SHM: a = −ω²x, we get ω² = k/m → ω = √(k/m)

4. 4

   Time period T = 2π/ω = 2π√(m/k)

Steps

1. 1

   For small angle θ, restoring torque τ = −mg sinθ · l ≈ −mglθ

2. 2

   Moment of inertia about pivot: I = ml²

3. 3

   Angular equation: I(d²θ/dt²) = −mglθ → d²θ/dt² = −(g/l)θ

4. 4

   Comparing with SHM: ω² = g/l → ω = √(g/l)

5. 5

   Time period T = 2π/ω = 2π√(l/g)

Steps

1. 1

   Consider a small element of string of length Δl forming an arc of radius R under tension T

2. 2

   Net upward (restoring) force = 2T sin(Δθ/2) ≈ T Δθ for small angle

3. 3

   Since Δl = R Δθ, the restoring force = TΔl/R

4. 4

   Mass of element: Δm = μ Δl; centripetal acceleration = v²/R

5. 5

   Apply Newton's second law: TΔl/R = μ Δl × v²/R

6. 6

   Simplify: T = μv² → v = √(T/μ)

Steps

1. 1

   Let ν₀ = frequency of source, v = speed of sound, v_s = speed of source towards observer

2. 2

   In time T₀ (one period), source moves a distance v_s T₀ towards observer

3. 3

   Wavelength ahead of source: λ' = (v − v_s)T₀ = (v − v_s)/ν₀

4. 4

   Observer receives waves with wavelength λ'; apparent frequency ν' = v/λ'

5. 5

   ν' = v/[(v − v_s)/ν₀] = ν₀ v/(v − v_s)

6. 6

   When source moves away: ν' = ν₀ v/(v + v_s) (frequency decreases)

Steps

1. 1

   Let GP be a, ar, ar², …, arⁿ⁻¹ with first term a and common ratio r

2. 2

   Write Sₙ = a + ar + ar² + … + arⁿ⁻¹ ...(i)

3. 3

   Multiply both sides by r: rSₙ = ar + ar² + … + arⁿ ...(ii)

4. 4

   Subtract (ii) from (i): Sₙ − rSₙ = a − arⁿ

5. 5

   Sₙ(1 − r) = a(1 − rⁿ) → Sₙ = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1) for r ≠ 1

6. 6

   For r = 1, each term equals a, so Sₙ = na

Steps

1. 1

   Base case (n=1): (a+b)¹ = a+b = ¹C₀ a + ¹C₁ b — holds true

2. 2

   Inductive hypothesis: assume (a+b)ᵏ = Σ_{r=0}^{k} ᵏCᵣ aᵏ⁻ʳ bʳ

3. 3

   Multiply both sides by (a+b): (a+b)ᵏ⁺¹ = (a+b) Σᵏ_{r=0} ᵏCᵣ aᵏ⁻ʳ bʳ

4. 4

   Separate: a·Σ + b·Σ and re-index the second sum (shift r → r−1)

5. 5

   Use Pascal's identity: ᵏCᵣ + ᵏCᵣ₋₁ = ᵏ⁺¹Cᵣ to combine terms

6. 6

   Obtain: (a+b)ᵏ⁺¹ = Σ_{r=0}^{k+1} ᵏ⁺¹Cᵣ aᵏ⁺¹⁻ʳ bʳ — proved by induction

7. 7

   General term: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ (r starts from 0)

Steps

1. 1

   Let line L: ax + by + c = 0 and point P(x₁, y₁)

2. 2

   Draw perpendicular PM from P to line L; let M = (x₀, y₀) on L

3. 3

   Slope of L: m₁ = −a/b; slope of PM: m₂ = b/a (perpendicular condition)

4. 4

   Write equations of PM; solve simultaneously with L to find M(x₀, y₀)

5. 5

   x₀ = x₁ − a(ax₁+by₁+c)/(a²+b²); y₀ = y₁ − b(ax₁+by₁+c)/(a²+b²)

6. 6

   d = PM = √[(x₁−x₀)² + (y₁−y₀)²] = |ax₁+by₁+c|/√(a²+b²)

Steps

1. 1

   Drop perpendiculars from each vertex to the x-axis; create three trapeziums

2. 2

   Area of triangle = (Area of trapezium 1 + Area of trapezium 2) − Area of trapezium 3

3. 3

   Trap 1: ½(y₁+y₂)(x₂−x₁); Trap 2: ½(y₂+y₃)(x₃−x₂); Trap 3: ½(y₁+y₃)(x₃−x₁)

4. 4

   Combine and expand all three expressions

5. 5

   After simplification: Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

6. 6

   Use absolute value to ensure non-negative area regardless of vertex order

Steps

1. 1

   By first principles: f'(x) = lim_{h→0} [f(x+h) − f(x)] / h

2. 2

   For f(x) = xⁿ: f'(x) = lim_{h→0} [(x+h)ⁿ − xⁿ] / h

3. 3

   Expand (x+h)ⁿ using Binomial theorem: xⁿ + nxⁿ⁻¹h + ⁿC₂xⁿ⁻²h² + … + hⁿ

4. 4

   Subtract xⁿ and divide by h: nxⁿ⁻¹ + ⁿC₂xⁿ⁻²h + … + hⁿ⁻¹

5. 5

   As h → 0, all terms with h vanish: f'(x) = nxⁿ⁻¹