MTH 103 Vectors and Dynamics

Vectors and Dynamics

1. Course Introduction

This course develops two connected areas of applied mathematics: vectors and dynamics. Vectors provide the language for describing quantities that have both magnitude and direction, while dynamics applies vector ideas to the motion of particles and bodies under the action of forces.

The central aim is to help students solve problems involving vector operations, motion, force, momentum, work, projectile motion, impulse, impact, elastic strings, and simple harmonic systems.

Key Takeaways

• A scalar has magnitude only; a vector has magnitude and direction.
• Most physical quantities in dynamics, such as velocity, acceleration, force, and momentum, are vectors.
• Vectors are essential for resolving motion and forces into components.

2. Vectors and Their Component Form

A vector is a directed quantity represented geometrically by an arrowed line segment. The length of the arrow shows the magnitude, while the arrowhead shows the direction.

If a vector begins at point P(x₁, y₁, z₁) and ends at point Q(x₂, y₂, z₂), then its component form is:

PQ = ⟨x₂ - x₁, y₂ - y₁, z₂ - z₁⟩

In unit-vector notation, the same vector may be written as:

v = v₁i + v₂j + v₃k

Magnitude of a Vector

The magnitude or length of v = ⟨v₁, v₂, v₃⟩ is:

|v| = √(v₁² + v₂² + v₃²)

Example

Find the component form and magnitude of the vector from P(-3, 4, 1) to Q(-5, 2, 2).

Solution:

PQ = ⟨-5 - (-3), 2 - 4, 2 - 1⟩ = ⟨-2, -2, 1⟩

|PQ| = √[(-2)² + (-2)² + 1²] = √9 = 3

Exam Points

• Always subtract initial coordinates from terminal coordinates.
• Magnitude is always non-negative.
• Use i, j, k notation for three-dimensional vectors.

3. Direction Cosines and Unit Vectors

The direction cosines of a vector are the cosines of the angles it makes with the positive coordinate axes.

If:

v = xi + yj + zk

and α, β, γ are the angles made with the positive x, y, and z axes respectively, then:

cos α = ^x⁄_|v|, cos β = ^y⁄_|v|, cos γ = ^z⁄_|v|

Unit Vector

A unit vector is a vector of magnitude 1. The unit vector in the direction of a non-zero vector a is:

â = ^a⁄_|a|

Example

Find the unit vector in the direction of AB, where A(1, 2, 1) and B(2, 5, 4).

Solution:

AB = (2 - 1)i + (5 - 2)j + (4 - 1)k = i + 3j + 3k

|AB| = √(1² + 3² + 3²) = √19

Unit vector = ¹⁄_√19i + ³⁄_√19j + ³⁄_√19k

Key Takeaways

• Direction cosines describe orientation in three-dimensional space.
• A unit vector preserves direction but has magnitude 1.
• To find a unit vector, divide the vector by its magnitude.

4. Vector Addition, Subtraction, and Scalar Multiplication

Let:

u = ⟨u₁, u₂, u₃⟩ and v = ⟨v₁, v₂, v₃⟩

Then:

u + v = ⟨u₁ + v₁, u₂ + v₂, u₃ + v₃⟩

u - v = ⟨u₁ - v₁, u₂ - v₂, u₃ - v₃⟩

ku = ⟨ku₁, ku₂, ku₃⟩

Geometric Laws

• Triangle law: If two vectors are placed head-to-tail, their sum is the vector from the beginning of the first to the end of the second.
• Parallelogram law: If two vectors start from the same point, their resultant is the diagonal of the parallelogram formed.

Example

Let u = ⟨-1, 3, 1⟩ and v = ⟨4, 7, 0⟩. Find 2u + 3v.

Solution:

2u = ⟨-2, 6, 2⟩

3v = ⟨12, 21, 0⟩

2u + 3v = ⟨10, 27, 2⟩

Exam Points

• Add and subtract vectors component by component.
• Scalar multiplication changes magnitude and may reverse direction if the scalar is negative.
• The negative of a vector has the same magnitude but opposite direction.

5. Linear Dependence and Linear Independence

A set of vectors {a₁, a₂, ..., a_n} is linearly dependent if there exist scalars k₁, k₂, ..., k_n, not all zero, such that:

k₁a₁ + k₂a₂ + ... + k_na_n = 0

The vectors are linearly independent if the only solution to this equation is:

k₁ = k₂ = ... = k_n = 0

Example

Determine whether v₁ = ⟨12, -8⟩ and v₂ = ⟨-9, 6⟩ are dependent.

We solve:

k₁⟨12, -8⟩ + k₂⟨-9, 6⟩ = ⟨0, 0⟩

This gives:

12k₁ - 9k₂ = 0

-8k₁ + 6k₂ = 0

A non-zero solution exists, for example k₁ = 3 and k₂ = 4. Hence, the vectors are linearly dependent.

Key Takeaways

• Dependence means one vector can be expressed using others.
• Independence means no vector in the set is redundant.
• For two vectors in two dimensions, dependence usually means the vectors are parallel.

6. Scalar Product or Dot Product

The scalar product, also called the dot product, multiplies two vectors and produces a scalar.

If u and v are vectors, then:

u · v = |u||v|cos θ

In component form:

u · v = u₁v₁ + u₂v₂ + u₃v₃

Angle Between Two Vectors

cos θ = ^{u · v}⁄_|u||v|

Example

Find the dot product of ⟨1, -2, -1⟩ and ⟨-6, 2, -3⟩.

Solution:

u · v = (1)(-6) + (-2)(2) + (-1)(-3)

= -6 - 4 + 3 = -7

Vector Projection

The projection of u onto v is:

proj_vu = (^{u · v}⁄_|v|²)v

The scalar component of u in the direction of v is:

^{u · v}⁄_|v|

Work Done

If a constant force F moves an object through displacement d, then:

W = F · d

Exam Points

• The dot product gives a scalar.
• If u · v = 0, the vectors are perpendicular.
• Work done is a direct application of the dot product.
• Projection measures the component of one vector in the direction of another.

7. Vector Product or Cross Product

The cross product multiplies two vectors and produces a vector perpendicular to both.

If p and q are vectors, then:

|p × q| = |p||q|sin θ

The direction of p × q is determined by the right-hand rule.

Determinant Form

For p = p₁i + p₂j + p₃k and q = q₁i + q₂j + q₃k:

p × q = (p₂q₃ - p₃q₂)i - (p₁q₃ - p₃q₁)j + (p₁q₂ - p₂q₁)k

Example

Let u = 2i + j + k and v = -4i + 3j + k. Find u × v.

Solution:

u × v = (1 × 1 - 1 × 3)i - (2 × 1 - 1 × -4)j + (2 × 3 - 1 × -4)k

= -2i - 6j + 10k

Applications

• Area: The area of the parallelogram formed by p and q is |p × q|.
• Moment of force: M = r × F.
• Linear velocity in rotation: v = ω × r.

Exam Points

• The cross product gives a vector.
• p × q = -(q × p).
• The cross product is zero when the vectors are parallel.
• A vector perpendicular to a plane can be found by crossing two non-parallel vectors in that plane.

8. Differentiation and Integration of Vector Functions

A vector-valued function assigns a vector to each value of a scalar variable, usually time.

If:

r(t) = x(t)i + y(t)j + z(t)k

then the velocity and acceleration are:

v(t) = ^dr⁄_dt = ^dx⁄_dti + ^dy⁄_dtj + ^dz⁄_dtk

a(t) = ^dv⁄_dt = ^d2x⁄_dt²i + ^d2y⁄_dt²j + ^d2z⁄_dt²k

Vector Integration

If:

v(t) = v₁(t)i + v₂(t)j + v₃(t)k

then:

∫ v(t)dt = i∫ v₁(t)dt + j∫ v₂(t)dt + k∫ v₃(t)dt

Example

If r(t) = (40t² + 8t)i + 2cos(3t)j + 2sin(3t)k, then:

v(t) = (80t + 8)i - 6sin(3t)j + 6cos(3t)k

a(t) = 80i - 18cos(3t)j - 18sin(3t)k

Key Takeaways

• Differentiate or integrate each vector component separately.
• Velocity is the first derivative of position.
• Acceleration is the derivative of velocity and the second derivative of position.

9. Kinematics of a Particle

Kinematics is the study of motion without considering the forces that cause the motion.

If s is displacement and t is time, then:

v = ^ds⁄_dt

a = ^dv⁄_dt = ^d2s⁄_dt²

When acceleration is expressed in terms of displacement, another useful formula is:

a = v^dv⁄_ds

Velocity in Two Dimensions

If a particle has position (x(t), y(t)), then its velocity components are:

v_x = ^dx⁄_dt, v_y = ^dy⁄_dt

The speed is:

v = √[(^dx⁄_dt)² + (^dy⁄_dt)²]

The direction is given by:

tan θ = ^dy/dt⁄_dx/dt

Example

A particle has displacement s = e^4tcos 5t. Find its velocity.

Solution:

v = ^ds⁄_dt

= e^4t(4cos 5t - 5sin 5t)

Exam Points

• Velocity is the rate of change of displacement.
• Acceleration is the rate of change of velocity.
• For two-dimensional motion, resolve velocity into components.

10. Force and Momentum

A force is an action that tends to change the state of rest or motion of a body, or deform it.

Newton's second law gives:

F = ma

where F is force, m is mass, and a is acceleration.

Momentum

Momentum is the product of mass and velocity:

p = mv

Momentum is a vector quantity.

Conservation of Linear Momentum

If no external force acts on a system, the total momentum before collision equals the total momentum after collision:

total initial momentum = total final momentum

Example

A 3 kg mass has acceleration a = 2i + 5j m/s². Find the force.

Solution:

F = ma = 3(2i + 5j) = 6i + 15j N

|F| = √(6² + 15²) = √261 = 16.2 N approximately.

Key Takeaways

• Force, velocity, acceleration, and momentum are vectors.
• Use F = ma for resultant force problems.
• Use conservation of momentum when collision systems have no external resultant force.

11. Motion Under Gravity

When a particle moves vertically under gravity, its acceleration is g downward.

If upward motion is taken as positive, the standard formulas are:

v = u - gt

h = ut - ¹⁄₂gt²

v² = u² - 2gh

H = ^u2⁄_2g

T = ^2u⁄_g

where u is initial velocity, v is final velocity, h is height, H is maximum height, and T is total time of flight for vertical projection returning to the same level.

Example

A particle is projected vertically upward with velocity 57 m/s. Find its velocity after 3 s using g = 9.8 m/s².

Solution:

v = u - gt = 57 - 9.8(3) = 27.6 m/s

Exam Points

• Choose a positive direction before using equations of motion.
• At maximum height, the vertical velocity is zero.
• For bodies falling downward, signs must be handled consistently.

12. Projectile Motion

Projectile motion is the motion of an object projected into the air and moving under the influence of gravity alone, neglecting air resistance.

If a projectile is launched with initial speed u at angle θ above the horizontal:

u_x = ucos θ

u_y = usin θ

Horizontal motion has constant velocity, while vertical motion has acceleration -g.

Standard Results

Time of flight: T = ^{2usin θ}⁄_g

Maximum height: H = ^u2sin2θ⁄_2g

Horizontal range: R = ^{u2sin 2θ}⁄_g

Example

A projectile is launched at 20 m/s at an angle of 25°. Take g = 10 m/s². Find the time of flight.

Solution:

T = ^{2usin θ}⁄_g = ^{2(20)sin 25°}⁄₁₀ = 1.69 s approximately.

Key Takeaways

• Separate projectile motion into horizontal and vertical components.
• Horizontal velocity remains constant when air resistance is ignored.
• Vertical velocity changes because of gravity.
• Use standard formulas only when the projectile lands at the same vertical level from which it was projected.

13. Restricted Vertical Motion and Resistance

In restricted vertical motion, a particle moves vertically while a resisting force opposes its motion. A common model assumes that resistance is proportional to velocity.

If the retardation due to resistance is kv, where k is constant and v is velocity, then the formulas for upward projection include:

t₁ = ¹⁄_k ln(^{ku + g}⁄_g)

H = ^u⁄_k - ^g⁄_k² ln(^{ku + g}⁄_g)

where t₁ is the time to reach maximum height and H is the maximum height.

Key Takeaways

• Resistance changes the motion from the simple constant-acceleration model.
• The natural logarithm appears because the velocity equation involves exponential decay.
• At maximum height, vertical velocity is zero.

14. Elastic Strings and Simple Pendulum

Elastic Strings

An elastic string is a string that can stretch under tension and return to its original length when the force is removed, provided its elastic limit is not exceeded.

Hooke's law for an elastic string is:

T = λ^x⁄_l

where T is tension, λ is the modulus of elasticity, x is extension, and l is the natural length.

Example

An elastic string of natural length 1 m is stretched to 1.4 m by a force of 4 N. Find λ.

Solution:

x = 1.4 - 1 = 0.4 m

T = λ^x⁄_l

4 = λ(^0.4⁄₁)

λ = 10 N

Simple Pendulum

A simple pendulum consists of a small heavy particle suspended from a fixed point by a light inextensible string. For small oscillations, its period is:

T = 2π√(^l⁄_g)

where l is the length of the pendulum and g is acceleration due to gravity.

Key Takeaways

• For elastic strings, extension is stretched length minus natural length.
• The modulus of elasticity has the dimension of force.
• The period of a simple pendulum depends on length and gravity, not on mass.

15. Impulse and Impact

Impulse

Impulse is the product of force and the time for which the force acts. It equals the change in momentum.

For constant force:

I = Ft = m(v - u)

For variable force:

I = ∫ F(t)dt

Example

A mass of 3 kg moving at 5 m/s is acted on by a force of 10 N for 0.3 s in the direction of motion. Find the final velocity.

Solution:

I = Ft = 10(0.3) = 3 Ns

I = m(v - u)

3 = 3(v - 5)

v = 6 m/s

Impact

Impact is a collision between bodies over a very short time. For direct impact between smooth bodies, two principles are commonly used:

• Conservation of momentum along the line of impact.
• Newton's law of restitution.

The coefficient of restitution e is:

e = ^{relative speed of separation}⁄_{relative speed of approach}

For two bodies moving along a line:

e = ^{v₂ - v₁}⁄_{u1 - u2}

where u₁, u₂ are velocities before impact and v₁, v₂ are velocities after impact, chosen along the same line.

Exam Points

• Impulse equals change in momentum.
• Use integration when force varies with time.
• In collision problems, write momentum equations before and after impact.
• The coefficient of restitution compares separation speed with approach speed.

16. Exam-Focused Formula Sheet

Vectors

PQ = ⟨x₂ - x₁, y₂ - y₁, z₂ - z₁⟩

|v| = √(v₁² + v₂² + v₃²)

â = ^a⁄_|a|

u · v = |u||v|cos θ

proj_vu = (^{u · v}⁄_|v|²)v

|u × v| = |u||v|sin θ

W = F · d

M = r × F

v = ω × r

Dynamics

v = ^ds⁄_dt

a = ^dv⁄_dt = ^d2s⁄_dt²

F = ma

p = mv

I = Ft = m(v - u)

v = u - gt

h = ut - ¹⁄₂gt²

v² = u² - 2gh

R = ^{u2sin 2θ}⁄_g

H = ^u2sin2θ⁄_2g

T = ^{2usin θ}⁄_g

T_pendulum = 2π√(^l⁄_g)

T_elastic = λ^x⁄_l

17. Practice Questions for Revision

A. Vectors

1. Find the component form and magnitude of the vector from P(1, 3, -2) to Q(4, -1, 5).
2. Find the unit vector in the direction of 3i - 4j + 12k.
3. Find the direction cosines of v = 4i + 2j - k.
4. Given u = ⟨3, -2⟩ and v = ⟨-2, 5⟩, compute 2u - 3v.
5. Determine whether ⟨2, -4⟩ and ⟨-1, 2⟩ are linearly dependent.
6. Find the angle between A = 3i + 2j - 6k and B = 4i - 3j + k.
7. Find A × B for A = 2i + j - 3k and B = i - j + k.
8. Find the work done by F = 4i - 3j + 2k when an object moves from ⟨3, 2, -1⟩ to ⟨2, -1, 4⟩.

B. Dynamics

1. A particle has displacement s = 5t³ + 9t² + 7t. Find its velocity and acceleration.
2. A mass of 4 kg has acceleration 3i - 2j m/s². Find the force and its magnitude.
3. A particle of mass 2 kg moves with velocity 5i - 3j m/s. Find its momentum and magnitude.
4. A particle is projected vertically upward with velocity 40 m/s. Find its maximum height, taking g = 10 m/s².
5. A projectile is launched at 30 m/s at 45°. Find its time of flight and range.
6. An elastic string of natural length 2 m is extended by 0.5 m under tension 12 N. Find its modulus of elasticity.
7. A pendulum has length 1.5 m. Find its period using g = 9.8 m/s².
8. A force of 20 N acts on a body for 0.4 s. Find the impulse.

18. Final Revision Checklist

• Can you convert between coordinate form and unit-vector form?
• Can you compute magnitude, unit vector, direction cosines, and projections?
• Can you use dot product for angles and work done?
• Can you use cross product for moments, perpendicular vectors, and rotational velocity?
• Can you differentiate and integrate vector functions component by component?
• Can you derive velocity and acceleration from displacement?
• Can you apply F = ma and p = mv correctly?
• Can you solve vertical motion and projectile motion problems using consistent signs?
• Can you apply Hooke's law for elastic strings?
• Can you use impulse and momentum principles in collision problems?