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State Board · Class 11 · Physics

Lesson 2: Mathematical Methods

7 concepts · 56 practice questions with answers · Free to read

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Scalars and vectors

From daily life

'The shop is 2 km away' makes you ask 'in which direction?' — that is where vectors begin. Link to earlier classes : the algebra, trigonometry and Pythagoras theorem of classes 9-10 are used here for vectors and calculus.

Simple definition

A vector has magnitude and direction; a scalar has magnitude only.

Displacement versus distance4 m3 mdisplacement 5 mdistance = 7 m (scalar)AB

In the figure : Orange path = distance; green arrow = displacement.

Rule 1: Scalar and vector — Scalar : magnitude only (mass, time, distance, speed, energy, work). Vector : magnitude + direction (displacement, velocity, acceleration, force, momentum).

  • Easy : 50 kg mass : scalar; 50 N force northwards : vector.
  • Medium : Speed 20 m/s : scalar; velocity 20 m/s eastwards : vector.
  • Exam (CET) : Which is a vector? (1) force (2) energy (3) speed (4) time → answer 1) Trick : if it has a direction, it is a vector.

Rule 2: Displacement and distance — Distance = actual path length (scalar); displacement = straight-line gap from start to end with direction (vector). For perpendicular legs |displacement| = √(a² + b²).

  • Easy : 4 m east + 3 m north ⇒ distance 7 m, displacement √(16 + 9) = 5 m.
  • Medium : After one full circle displacement = 0, distance = 2π r.
  • Exam (CET) : Distance equals |displacement| when? (1) in a straight line (2) in a circle (3) always (4) never → answer 1)

Rule 3: Properties of vectors — Equal vectors : same magnitude and direction. Negative −A : same magnitude, opposite direction. k A : magnitude |k| A (reversed if k < 0).

  • Easy : A = 5 m east ⇒ −A = 5 m west.
  • Medium : 3A = 15 m east; −2A = 10 m west.
  • Exam (CET) : A + (−A) = ? (1) null vector (2) 2A (3) A² (4) 1 → answer 1)

Derivation : Magnitude of displacement

  • Step 1 : A → B : 4 m east, then 3 m north.
  • Step 2 : The two legs are perpendicular ⇒ a right triangle.
  • Step 3 : Pythagoras : AB² = 4² + 3² = 25.
  • Step 4 : |displacement| = 5 m, direction : tanθ = 3/4.
Remember

|displacement| ≤ distance; vector = magnitude + direction.

Common mistake / CET trap

Speed is scalar, velocity is vector; current has a direction yet is a scalar.

Shortcut

Memory clue : "Direction = vector".

Spot the difference

PropertyScalarVector
Magnitudeyesyes
Directionnoyes
E.g.speed, energyvelocity, force
Additionordinarytriangle law

Practice questions

Easy
  1. 1What is a scalar?
  2. 2Difference between speed and velocity?
  3. 3What is −A?
Medium
  1. 46 m east + 8 m north : displacement?
  2. 5Its distance?
  3. 6A = 4 m : 2.5 A?
Challenging
  1. 7Semicircle r = 7 m : distance and displacement?
  2. 8A vector changes but its magnitude stays same : how?
Show answers (try them yourself first)
  1. 1A quantity with magnitude only
  2. 2Velocity has direction
  3. 3Same magnitude, opposite direction
  4. 410 m
  5. 514 m
  6. 610 m
  7. 722 m; 14 m
  8. 8By changing direction

Addition and subtraction of vectors

From daily life

A boat crossing a river is pushed forward and sideways — both together fix the real direction.

Simple definition

The resultant is the single vector that has the same effect as the two vectors together.

Parallelogram lawABRθR = √(A² + B² + 2AB cosθ)tanα = B sinθ ÷ (A + B cosθ)θ = 0° : R = A + Bθ = 180° : R = A − B

In the figure : The diagonal of A and B is R.

Live figure

Addition and subtraction of vectors

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Rule 1: Triangle and parallelogram laws — Add two vectors by joining head to tail (triangle) or by the diagonal from a common origin (parallelogram). R = √(A² + B² + 2AB cosθ); tanα = B sinθ ÷ (A + B cosθ) (α = angle between R and A).

  • Easy : A = 3, B = 4, θ = 90° ⇒ R = √(9 + 16) = 5.
  • Medium : A = B = 10, θ = 60° ⇒ R = 17.32 (= 10√3).
  • Exam (CET) : θ = 0° ⇒ R? (1) A + B (2) A − B (3) √(A²+B²) (4) 0 → answer 1) Trick : same direction = maximum.

Rule 2: Maximum and minimum resultant — θ = 0° ⇒ R_max = A + B; θ = 180° ⇒ R_min = |A − B|. So |A − B| ≤ R ≤ A + B.

  • Easy : The resultant of 7 N and 4 N lies between 3 N and 11 N.
  • Medium : Two equal vectors at 120° ⇒ R equals the magnitude (e.g. 5, 5 ⇒ 5).
  • Exam (CET) : Can the resultant of 5 N and 5 N be 12 N? (1) no (2) yes (3) sometimes (4) unknown → answer 1)

Rule 3: Subtraction — A − B = A + (−B). R = √(A² + B² − 2AB cosθ).

  • Easy : A = B = 10, θ = 60° ⇒ |A − B| = 10.
  • Medium : A = 6, B = 8, θ = 90° ⇒ |A − B| = 10.
  • Exam (CET) : Sign in |A − B| formula? (1) − 2AB cosθ (2) + 2AB cosθ (3) 2AB (4) AB → answer 1)

Derivation : Formula for R

  • Step 1 : Put B's tail on A's head; B has component B cosθ along A and B sinθ perpendicular.
  • Step 2 : R² = (A + B cosθ)² + (B sinθ)².
  • Step 3 : = A² + 2AB cosθ + B²(cos²θ + sin²θ).
  • Step 4 : R = √(A² + B² + 2AB cosθ); tanα = B sinθ ÷ (A + B cosθ).
Remember

R = √(A² + B² + 2AB cosθ); |A − B| ≤ R ≤ A + B.

Common mistake / CET trap

θ is the angle between the tails; the sign is − for subtraction.

Shortcut

Memory clue : "Together = big; opposite = small".

Spot the difference

θR
0°A + B (max)
90°√(A² + B²)
180°A − B(min)

Practice questions

Easy
  1. 1State the triangle law.
  2. 2When is R maximum?
  3. 3Formula for A − B?
Medium
  1. 4A = 6, B = 8, θ = 90° : R?
  2. 5A = B = 5, θ = 120° : R?
  3. 612 N and 5 N : maximum and minimum?
Challenging
  1. 7A = 3, B = 4, θ = 90° : α?
  2. 8A = 2, B = 2, θ = 60° : |A − B|?
Show answers (try them yourself first)
  1. 1Tail of B on head of A; R from first tail to last head
  2. 2θ = 0°
  3. 3A + (−B)
  4. 4R = 10
  5. 5R = 5
  6. 617 N; 7 N
  7. 7α = 53.1°
  8. 82

Resolution of vectors and unit vectors

From daily life

Climbing a slope you move forward and upward — both parts can be measured separately.

Simple definition

A component is the projection of a vector on an axis; a unit vector has magnitude 1 and shows direction.

Components of a vectorAA_x = A cosθA_y = A sinθθA = A_x î + A_y ĵ|A| = √(A_x² + A_y²)Â = A ÷ |A|

In the figure : Red A has blue A_x and green A_y.

Rule 1: Resolution into components — A_x = A cosθ, A_y = A sinθ; A = A_x î + A_y ĵ; |A| = √(A_x² + A_y²); tanθ = A_y ÷ A_x.

  • Easy : A = 10 N, θ = 30° ⇒ A_x = 8.66, A_y = 5.00.
  • Medium : A = 3î + 4ĵ ⇒ |A| = 5; tanθ = 4/3 ⇒ θ = 53°.
  • Exam (CET) : At θ = 45° the components are? (1) equal (2) A_x > A_y (3) A_y > A_x (4) zero → answer 1)

Rule 2: Unit vectors — Â = A ÷ |A|; magnitude 1, direction of A. î, ĵ, k̂ are unit vectors along x, y, z.

  • Easy : A = 3î + 4ĵ ⇒ Â = 0.6î + 0.8ĵ.
  • Medium : |0.6î + 0.8ĵ| = √(0.36 + 0.64) = 1 ✓.
  • Exam (CET) : Unit of a unit vector? (1) none (2) m (3) N (4) s → answer 1)

Rule 3: Adding by components — A + B : add x components and y components separately. R_x = A_x + B_x, R_y = A_y + B_y.

  • Easy : A = 2î + 3ĵ, B = 4î − ĵ ⇒ A + B = 6î + 2ĵ; |A + B| = √40 = 6.32.
  • Medium : A − B = −2î + 4ĵ; |A − B| = √20 = 4.47.
  • Exam (CET) : (î + ĵ) + (î − ĵ) = ? (1) 2î (2) 2ĵ (3) 0 (4) î → answer 1)

Derivation : A_x = A cosθ, A_y = A sinθ

  • Step 1 : Take A as hypotenuse and drop a perpendicular to the x-axis ⇒ a right triangle.
  • Step 2 : cosθ = adjacent ÷ hypotenuse = A_x ÷ A.
  • Step 3 : sinθ = opposite ÷ hypotenuse = A_y ÷ A.
  • Step 4 : A_x = A cosθ; A_y = A sinθ; A² = A_x² + A_y².
Remember

A_x = A cosθ; A_y = A sinθ; Â = A/|A|; add by components.

Common mistake / CET trap

Check which axis θ is measured from : from the y-axis, sin and cos swap.

Shortcut

Memory clue : "x = cos; y = sin".

Spot the difference

QuantityFormulaUnit
A_xA cosθsame as A
A_yA sinθsame as A
A√(A_x²+A_y²)same as A
ÂA/Anone

Practice questions

Easy
  1. 1Formula for A_x?
  2. 2What is a unit vector?
  3. 3What are î, ĵ, k̂?
Medium
  1. 4A = 20, θ = 60° : A_x, A_y?
  2. 5A = 6î + 8ĵ : |A| and Â?
  3. 6A = 3î + 2ĵ, B = î − 5ĵ : A + B?
Challenging
  1. 7A = 5î − 12ĵ : |A| and angle?
  2. 8A = 4î + 3ĵ, B = 2î + 6ĵ : |A − B|?
Show answers (try them yourself first)
  1. 1A cosθ
  2. 2Vector of magnitude 1 showing direction
  3. 3Unit vectors along x, y, z
  4. 4A_x = 10, A_y = 17.3
  5. 510; 0.6î + 0.8ĵ
  6. 64î − 3ĵ
  7. 713; −67.4°
  8. 8√13 = 3.61

Scalar (dot) product

From daily life

Even with a slanting push, only the part along the motion does work — a projection.

Simple definition

The dot product multiplies two vectors to give a scalar.

Dot product : projectionB cosθABθA·B = AB cosθ= A_xB_x + A_yB_yW = F·dî·î = 1 ; î·ĵ = 0

In the figure : The projection of B on A is B cosθ.

Rule 1: Scalar (dot) product — A·B = A B cosθ = A_xB_x + A_yB_y + A_zB_z. The result is a scalar. A·B = B·A. î·î = 1; î·ĵ = 0.

  • Easy : A = 3î + 4ĵ, B = 2î − ĵ ⇒ A·B = 6 − 4 = 2.
  • Medium : F = 10 N, d = 5 m, θ = 60° ⇒ W = 50 cos60° = 25 J.
  • Exam (CET) : Dot product of perpendicular vectors? (1) 0 (2) AB (3) 1 (4) −1 → answer 1) Trick : 90° ⇒ cos = 0.

Rule 2: Angle and projection — cosθ = A·B ÷ (A B). Projection of B on A = (A·B) ÷ A.

  • Easy : A = î + ĵ, B = î ⇒ cosθ = 1/√2 ⇒ θ = 45°.
  • Medium : A = 3î + 4ĵ, B = 2î + ĵ ⇒ A·B = 10; projection of B on A = 10 ÷ 5 = 2.
  • Exam (CET) : A·A = ? (1) A² (2) 0 (3) 2A (4) 1 → answer 1)

Rule 3: Work and power — W = F·d = F d cosθ; P = F·v. A perpendicular force does no work (e.g. in circular motion).

  • Easy : F = 20 N, v = 3 m/s in the same direction ⇒ P = 60 W.
  • Medium : F = 5î + 2ĵ, d = 3î ⇒ W = 15 J.
  • Exam (CET) : Work is zero when? (1) F ⟂ d (2) F ∥ d (3) θ = 0 (4) never → answer 1)

Derivation : A·B = A_xB_x + A_yB_y

  • Step 1 : A = A_x î + A_y ĵ, B = B_x î + B_y ĵ.
  • Step 2 : A·B = A_xB_x(î·î) + A_xB_y(î·ĵ) + A_yB_x(ĵ·î) + A_yB_y(ĵ·ĵ).
  • Step 3 : î·î = ĵ·ĵ = 1; î·ĵ = ĵ·î = 0.
  • Step 4 : A·B = A_xB_x + A_yB_y.
Remember

A·B = AB cosθ = A_xB_x + A_yB_y; W = F·d.

Common mistake / CET trap

The dot result is a scalar; use cosθ, not sinθ; perpendicular ⇒ 0.

Shortcut

Memory clue : "Dot = cos = scalar".

Spot the difference

PropertyDot A·BCross A×B
Resultscalarvector
FormulaAB cosθAB sinθ n̂
θ = 0°maximumzero
θ = 90°zeromaximum

Practice questions

Easy
  1. 1What is A·B?
  2. 2î·ĵ = ?
  3. 3Vector formula for work?
Medium
  1. 4A = 2î + 3ĵ, B = 4î + ĵ : A·B?
  2. 5F = 8 N, d = 4 m, θ = 30° : W?
  3. 6A = 3î + 4ĵ : A·A?
Challenging
  1. 7A = 2î + 2ĵ, B = 3î : angle?
  2. 8F = 4î − 2ĵ, v = 3î + 6ĵ : P?
Show answers (try them yourself first)
  1. 1AB cosθ (a scalar)
  2. 20
  3. 3W = F·d
  4. 411
  5. 527.7 J
  6. 625
  7. 745°
  8. 80 W

Vector (cross) product

From daily life

A spanner turns easiest when pushed perpendicular to its handle — the cross product at work.

Simple definition

The cross product of two vectors is a new vector perpendicular to both.

Cross product : A × BABA × Bθarea|A×B| = AB sinθτ = r × Fî × ĵ = k̂A×B = −B×A

In the figure : The blue parallelogram is |A × B|; the red arrow is its direction.

Rule 1: Vector (cross) product — A × B = A B sinθ n̂; n̂ is perpendicular to A and B by the right-hand rule. Magnitude = area of the parallelogram on A and B. A × B = −(B × A); A × A = 0.

  • Easy : A = 3, B = 4, θ = 30° ⇒ |A × B| = 12 sin30° = 6.
  • Medium : Parallelogram sides 5, 8 at 30° ⇒ area = 20.
  • Exam (CET) : Cross product of parallel vectors? (1) 0 (2) AB (3) 1 (4) −AB → answer 1) Trick : 0° ⇒ sin = 0.

Rule 2: Unit vectors and components — î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (reverse order gives −). In the x-y plane : (A × B)_z = A_xB_y − A_yB_x.

  • Easy : A = î + 2ĵ, B = 3î + ĵ ⇒ z = 1·1 − 2·3 = −5 ⇒ A × B = −5k̂; |A × B| = 5.
  • Medium : ĵ × î = −k̂.
  • Exam (CET) : k̂ × î = ? (1) ĵ (2) −ĵ (3) î (4) 0 → answer 1) Trick : cycle x→y→z→x.

Rule 3: Torque and angular momentum — τ = r × F ⇒ τ = r F sinθ; L = r × p.

  • Easy : r = 0.5 m, F = 20 N perpendicular ⇒ τ = 10 N m.
  • Medium : r = 2 m, F = 10 N, θ = 30° ⇒ τ = 10 N m.
  • Exam (CET) : τ is maximum when? (1) F ⟂ r (2) F ∥ r (3) θ = 0 (4) θ = 180° → answer 1)

Derivation : (A × B)_z = A_xB_y − A_yB_x

  • Step 1 : A = A_x î + A_y ĵ, B = B_x î + B_y ĵ.
  • Step 2 : A × B = A_xB_x(î×î) + A_xB_y(î×ĵ) + A_yB_x(ĵ×î) + A_yB_y(ĵ×ĵ).
  • Step 3 : î×î = ĵ×ĵ = 0; î×ĵ = k̂; ĵ×î = −k̂.
  • Step 4 : A × B = (A_xB_y − A_yB_x) k̂.
Remember

A × B = AB sinθ n̂; (A×B)_z = A_xB_y − A_yB_x; τ = r × F.

Common mistake / CET trap

Order matters : A × B = −B × A; use sinθ, not cosθ.

Shortcut

Memory clue : "Cross = sin = vector".

Spot the difference

PropertyDotCross
Resultscalarvector
Swap ordersamesign flips
ParallelAB0
Perpendicular0AB

Practice questions

Easy
  1. 1How is the direction of a cross product found?
  2. 2î × î = ?
  3. 3Vector formula for τ?
Medium
  1. 4A = 4, B = 5, θ = 90° : |A × B|?
  2. 5A = 2î + ĵ, B = î + 3ĵ : A × B?
  3. 6r = 0.4 m, F = 25 N, θ = 90° : τ?
Challenging
  1. 7A = 3, B = 4, A·B = 6 : |A × B|?
  2. 8Relation between A × B and B × A?
Show answers (try them yourself first)
  1. 1By the right-hand rule
  2. 20
  3. 3τ = r × F
  4. 420
  5. 55k̂
  6. 610 N m
  7. 7cosθ = 0.5 ⇒ θ = 60° ⇒ 10.39
  8. 8A × B = −B × A

Differentiation

From daily life

A speedometer shows the speed at each instant — the rate of change of position, i.e. differentiation.

Simple definition

The derivative dy/dx is the rate of change of y with x.

Differentiation : slope of the tangentx →ydy/dx = lim Δy ÷ Δxd(xⁿ)/dx = n xⁿ⁻¹d(sin x)/dx = cos xv = dx/dt ; a = dv/dt

In the figure : The slope of the red tangent is dy/dx.

Live figure

Differentiation

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Rule 1: Idea of a derivative — dy/dx = lim (Δx→0) Δy ÷ Δx = slope of the tangent. In physics : v = dx/dt, a = dv/dt.

  • Easy : x = 3t² + 2t ⇒ v = 6t + 2; t = 2 s ⇒ v = 14 m/s.
  • Medium : a = dv/dt = 6 m/s² (constant).
  • Exam (CET) : v = dx/dt means? (1) rate of change of position (2) x × t (3) only x/t (4) a → answer 1)

Rule 2: Rules of differentiation — d(c)/dx = 0; d(xⁿ)/dx = n xⁿ⁻¹; d(sin x) = cos x; d(cos x) = −sin x; d(eˣ) = eˣ; d(ln x) = 1/x; product (uv)' = u'v + uv'; chain d f(g)/dx = f'(g) g'.

  • Easy : d(5x⁴)/dx = 20x³.
  • Medium : d(sin 2x)/dx = 2 cos 2x; d(x² sin x)/dx = 2x sin x + x² cos x.
  • Exam (CET) : d(cos x)/dx = ? (1) −sin x (2) sin x (3) −cos x (4) 1 → answer 1) Trick : 'co' brings a minus.

Rule 3: Maxima and minima — dy/dx = 0 ⇒ turning point; d²y/dx² < 0 ⇒ maximum, > 0 ⇒ minimum.

  • Easy : y = −x² + 4x ⇒ dy/dx = −2x + 4 = 0 ⇒ x = 2; y = 4 (maximum, d²y/dx² = −2).
  • Medium : h = 20t − 5t² ⇒ dh/dt = 20 − 10t = 0 ⇒ t = 2 s; h_max = 20 m.
  • Exam (CET) : At a maximum dy/dx is? (1) 0 (2) 1 (3) infinite (4) negative → answer 1)

Derivation : d(x²)/dx = 2x (from first principles)

  • Step 1 : y = x²; (x + h)² − x² = 2xh + h².
  • Step 2 : Δy ÷ Δx = (2xh + h²) ÷ h = 2x + h.
  • Step 3 : As h → 0 the h vanishes.
  • Step 4 : dy/dx = 2x; in general n xⁿ⁻¹.
Remember

d(xⁿ) = n xⁿ⁻¹; v = dx/dt; a = dv/dt; maximum : dy/dx = 0.

Common mistake / CET trap

The derivative of a constant is 0; reduce the power by 1; d(cos) is negative.

Shortcut

Memory clue : "Bring the power down, power −1".

Spot the difference

FunctionDerivative
xⁿn xⁿ⁻¹
sin xcos x
cos x−sin x
eˣeˣ
ln x1/x

Practice questions

Easy
  1. 1d(x³)/dx = ?
  2. 2v and a from differentiation?
  3. 3d(c)/dx = ?
Medium
  1. 4x = 4t³ − t : v at t = 1?
  2. 5y = sin 3x : dy/dx?
  3. 6y = x² + 6x : minimum point?
Challenging
  1. 7x = 2t³ − 3t² : a = 0 when?
  2. 8h = 30t − 5t² : maximum height?
Show answers (try them yourself first)
  1. 13x²
  2. 2v = dx/dt; a = dv/dt
  3. 30
  4. 411
  5. 53 cos 3x
  6. 6x = −3; y = −9
  7. 7t = 0.5 s
  8. 845 m

Integration

From daily life

How much water filled a bucket is found by adding the flow of each instant — integration.

Simple definition

Integration adds many tiny parts to get a total; the area under a curve.

Integration : area under a curvex = 0 → 3 mFW = ∫F dx∫xⁿ dx = xⁿ⁺¹ ÷ (n+1) + C∫sin x dx = −cos x∫cos x dx = sin xx = ∫v dt ; v = ∫a dt

In the figure : The area under an F – x graph is the work done.

Rule 1: Indefinite integral — Integration reverses differentiation. ∫xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + C; ∫sin x = −cos x + C; ∫cos x = sin x + C; ∫eˣ = eˣ + C; ∫dx/x = ln x + C.

  • Easy : ∫3x² dx = x³ + C.
  • Medium : v = 6t + 2 ⇒ x = ∫v dt = 3t² + 2t + C.
  • Exam (CET) : ∫x dx = ? (1) x²/2 + C (2) x² + C (3) 2x (4) 1 → answer 1)

Rule 2: Definite integral — ∫ₐᵇ f dx = [F]ₐᵇ = F(b) − F(a) = area under the curve. C cancels.

  • Easy : ∫₀² 3t² dt = [t³]₀² = 8 .
  • Medium : ∫₀^π sin x dx = [−cos x]₀^π = 2.
  • Exam (CET) : ∫₁² dx = ? (1) 1 (2) 2 (3) 0 (4) 3 → answer 1)

Rule 3: Uses in physics — x = ∫v dt, v = ∫a dt, W = ∫F dx. Constant a ⇒ v = u + at, x = ut + ½at².

  • Easy : F = 4x N, 0 to 3 m ⇒ W = ∫4x dx = 2x² = 18 J.
  • Medium : a = 6t, u = 0 ⇒ v = 3t²; t = 2 s ⇒ v = 12 m/s.
  • Exam (CET) : W = ? (1) ∫F dx (2) dF/dx (3) F/x (4) F + x → answer 1)

Derivation : x = ut + ½ a t² (by integration)

  • Step 1 : a = dv/dt constant ⇒ v = ∫a dt = a t + C₁.
  • Step 2 : At t = 0, v = u ⇒ C₁ = u ⇒ v = u + a t.
  • Step 3 : x = ∫v dt = ∫(u + a t) dt = u t + ½ a t² + C₂.
  • Step 4 : At t = 0, x = 0 ⇒ C₂ = 0 ⇒ x = u t + ½ a t².
Remember

∫xⁿ dx = xⁿ⁺¹/(n+1) + C; x = ∫v dt; W = ∫F dx.

Common mistake / CET trap

Do not forget + C in an indefinite integral; in a definite one use upper − lower limit.

Shortcut

Memory clue : "Power + 1, divide by the new power".

Spot the difference

FunctionIntegral
xⁿxⁿ⁺¹/(n+1)
sin x−cos x
cos xsin x
eˣeˣ
1/xln x

Practice questions

Easy
  1. 1∫x² dx?
  2. 2What is integration?
  3. 3Integral formula for W?
Medium
  1. 4∫₀³ 2x dx?
  2. 5a = 4 (constant), u = 0 : v at t = 3 s?
  3. 6F = 6x, 0 to 2 m : W?
Challenging
  1. 7v = 3t² : displacement from t = 0 to 2 s?
  2. 8∫₀^(π/2) cos x dx?
Show answers (try them yourself first)
  1. 1x³/3 + C
  2. 2Reverse of differentiation
  3. 3W = ∫F dx
  4. 49
  5. 512 m/s
  6. 612 J
  7. 78 m
  8. 81

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
ResultantR = √(A² + B² + 2AB cosθ)same as A
ComponentsA_x = A cosθ; A_y = A sinθsame as A
Unit vector = A/Anone
DotA·B = AB cosθscalar
CrossA×B= AB sinθvector
Derivatived(xⁿ) = n xⁿ⁻¹—
Motionv = dx/dt; a = dv/dtm/s; m/s²
Integral∫xⁿ dx = xⁿ⁺¹/(n+1)—

🎯 Most important points for the exam

  • |A − B| ≤ R ≤ A + B.
  • Adding by components is easy.
  • Perpendicular vectors : dot = 0.
  • Parallel vectors : cross = 0.
  • Cross depends on order : A × B = −B × A.
  • Max/min : dy/dx = 0.
  • Integral = area; W = ∫F dx.
Study mantra

Vectors : resolve → add; dot = cos; cross = sin; calculus : power down / power + 1.