2D Vectors and Polar Coordinates: Study Notes

Coordinate Systems: Rectangular (x, y) and Polar (r, θ)

In the xy-plane, x-axis goes to the right and y-axis goes up. A point P with coordinates (x, y) means: drop a perpendicular to the x-axis giving distance x, and drop a perpendicular to the y-axis giving distance y. Thus two common representations of the same location:
- Rectangular: P = (x, y)
- Polar: P = (r, θ), where r is the distance from the origin to P and θ is the angle the line from the origin to P makes with the +x axis.
Example: If the point has coordinates (3, 4) in rectangular form, then the distance from the origin is r = √(3² + 4²) = 5, and θ is found from tan(θ) = opposite/adjacent = 4/3.
Polar convention: the line from the origin to P has length r and forms an angle θ with the +x axis.
The two representations are interchangeable: you can convert back and forth.
Trig refresher (relevant for conversions):
- tan(θ) = y/x (for the angle between the line to P and the +x axis)
- θ = arctan(y/x) (inverse tangent) gives the angle in radians or degrees depending on calculator mode
- In radians: 1 rad ≈ 57.2958°, and 2π radians = 360°; π radians = 180°
- Conversion between units:
- 2π rad = 360°
- 1 rad ≈ 57.3°; 1° ≈ π/180 rad
Example: Given (x, y) = (3, 4):
- r = √(3² + 4²) = 5
- θ = arctan(4/3) ≈ 0.9273 rad ≈ 53.13°
- So P = (3, 4) in rectangular form or P = (5, 53.13°) in polar form
Polar coordinates to rectangular: x = r cos θ, y = r sin θ
- Example: r = 5, θ = 53.13° → x = 5 cos(53.13°) ≈ 3, y = 5 sin(53.13°) ≈ 4
Rectangular to polar: r = √(x² + y²), θ = arctan(y/x)
- Note: quadrant considerations may adjust θ by adding π or adjusting by 2π as needed
Summary: you can express a point in either form and convert between them using the above formulas

Two-Dimensional Vectors: Displacement, Path, and Time

A vector represents both magnitude and direction; in 2D we work in a plane (xy-plane).
Displacement vs. path:
- Displacement is the net change in position from start to end, regardless of the path taken.
- The actual path is the curve or route followed to get from start to end.
Displacement vector d (arrow), with magnitude |d| and direction θ.
Direction is specified with respect to the x-axis (east): e.g., “north of east” or “east of north.”
Time notation:
- t1, t2, t3 denote successive times
- Δt = t2 − t1 is the time interval between two instants
For a moving object (e.g., a bug) at time t1 it is at some position; at time t3 it is at another position; the displacement from t1 to t3 is a single vector, regardless of intermediate path
A vector can be translated parallel to itself without changing its meaning (same magnitude and direction)
If you go from time t1 to t2 with displacement vector d, and then from t2 to t3 with displacement a, the total displacement from t1 to t3 is d + a
Vectors are directional quantities; their addition follows a geometric rule (parallelogram or triangle rule) and is easiest to handle with components

Scalars vs. Vectors

Scalar quantities have magnitude only (no direction): examples include mass, time interval, length, distance (magnitude only). Units are required (kg, s, m, etc.).
Vector quantities have magnitude and direction: e.g., displacement, velocity, force.
Velocity is a vector (has speed and direction); speed is a scalar (magnitude only, no direction).
Key point: when working with vectors, you must keep track of both magnitude and direction

Vector Addition, Subtraction, and Scaling

Visual rule: to add vectors a and b, place them tip-to-tail or form a parallelogram; the resultant c goes from the initial point to the final point.
Algebraic rule (component-wise):
- If a = ax î + ay ĵ and b = bx î + by ĵ, then
- a + b = (ax + bx) î + (ay + by) ĵ
- a − b = (ax − bx) î + (ay − by) ĵ
Scaling: for a scalar c, c a = (c ax) î + (c ay) ĵ
In 2D, you typically do not mix î and ĵ components during algebra; you keep them separate
Subtraction is addition with the negative: a − b = a + (−b)
If ax, ay, bx, by are known, all vector operations reduce to simple algebra on components
Note on intuition: you can imagine moving the vectors parallel to themselves; the sum corresponds to traversing the first vector, then the second; the final endpoint gives the resultant vector

Unit Vectors and the Component Form of a Vector

Unit vectors along the axes in 2D:
- î: unit vector in +x direction (to the right), length 1
- ĵ: unit vector in +y direction (up), length 1
Any vector a can be written as a linear combination of these unit vectors:
- a = ax î + ay ĵ
The quantities ax and ay are the components of a along the x and y axes, respectively
If a has magnitude |a| and makes angle θ with the +x axis, then
- a_x = |a| cos θ
- a_y = |a| sin θ
Conversely, given the components ax and ay, the vector is a = ax î + ay ĵ

From Polar/Angle Form to Rectangular Components (and Vice Versa: Worked Examples)

Given a vector with magnitude |a| and angle θ (with respect to the +x axis):
- a_x = |a| cos θ
- a_y = |a| sin θ
- So a = ax î + ay ĵ
Given the components (ax, ay):
- |a| = √(ax² + ay²)
- θ = arctan(ay / ax) (with quadrant checks as needed)
Example setup from a spoken problem:
- Vector a = 3 miles at 40° north of east →
- ax = 3 cos 40°, ay = 3 sin 40°
- Vector b = 4 miles at 30° east of north (i.e., from north toward east):
- relative to +x axis, angle is 60° (since 90° − 30° = 60°)
- bx = 4 cos 60°, by = 4 sin 60°
- Net displacement d = a + b has components
- dx = ax + b_x
- dy = ay + b_y
Practical calculation example (numerical):
- a_x = 3 cos 40° ≈ 3 × 0.7660 ≈ 2.298
- a_y = 3 sin 40° ≈ 3 × 0.6428 ≈ 1.929
- b_x = 4 cos 60° = 4 × 0.5 = 2.000
- b_y = 4 sin 60° ≈ 4 × 0.8660 ≈ 3.464
- Net displacement: dx ≈ 2.298 + 2.000 = 4.298, dy ≈ 1.929 + 3.464 = 5.393
- Magnitude: |d| ≈ √(4.298² + 5.393²) ≈ √(18.48 + 29.07) ≈ √47.55 ≈ 6.90
- Direction: θ_d ≈ arctan(5.393 / 4.298) ≈ arctan(1.254) ≈ 51.6° above the +x axis

Magnitude, Direction, and Trigonometric Relationships

For a vector a with magnitude |a| and angle θ: components are
- ax = |a| cos θ, ay = |a| sin θ
For a vector given by components (ax, ay):
- Magnitude: $|a| = \, \sqrt{ax^2 + ay^2}$
- Angle relative to +x axis: $\theta = \tan^{-1}\left(\frac{ay}{ax}\right)$
Inverse relationships rely on the usual trigonometric identities (cos, sin, tan) and understanding of the coordinate quadrants

The Scalar (Dot) Product and the Vector (Cross) Product

Two distinct ways to “multiply” vectors exist:
- Dot product (scalar product): yields a scalar
- Cross product (vector product): yields a vector (defined in 3D; discussed later for 3D)
In 2D, we primarily use the dot product; the cross product is defined in 3D and yields a vector perpendicular to the plane
Dot product definition (component form):
- For vectors a and b with components ax, ay and bx, by:
- $\vec{a} \cdot \vec{b} = ax bx + ay by$
Geometric interpretation: $\vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \, \cos \phi$ where φ is the angle between a and b
Dot product is a scalar (no î or ĵ in the result)
Cross product (brief note for 3D): defined as a × b producing a vector perpendicular to the plane containing a and b; in 3D, |a × b| = |a| |b| sin φ and direction given by right-hand rule; will be introduced in 3D context
In 2D problems you can often rely on the dot product to relate magnitudes and angles, and on component form for algebraic manipulation

3D Extension (Preview)

3D vectors introduce a third axis k̂ (along +z)
A 3D vector can be written as: $\vec{a} = ax \hat{i} + ay \hat{j} + a_z \hat{k}$
Unit vectors in 3D: $\hat{i}, \hat{j}, \hat{k}$ corresponding to +x, +y, +z directions
The same component approach extends to 3D for addition, subtraction, and scalar multiplication; cross product becomes meaningful and is used to compute a vector perpendicular to both inputs

Quick Recap: How to Solve 2D Vector Problems

Pick a representation: rectangular (x, y) or polar (r, θ)
If given magnitude and angle: compute components using
- $ax = |a| \cos \theta, \quad ay = |a| \sin \theta$
If given components: compute magnitude and angle using
- $|a| = \sqrt{ax^2 + ay^2}, \quad \theta = \tan^{-1}\left(\frac{ay}{ax}\right)$
To add vectors: add components separately; i.e., for a and b, a + b = (ax + bx) \hat{i} + (ay + by) \hat{j}
To scale a vector: c a = (c ax) \hat{i} + (c ay) \hat{j}
Dot product is a convenient algebraic tool: $\vec{a} \cdot \vec{b} = ax bx + ay by$
Always keep track of units and the distinction between scalars and vectors; velocity is a vector, speed is a scalar
Remember the difference between displacement (net change in position) and the actual path taken
In calculator work, radian vs degree mode matters for angle outputs; be mindful of conversions and quadrant placement when using arctan

Worked Conceptual Example (Summary)

Given two displacement vectors in 2D, express each in components, add them component-wise, and then interpret the resultant magnitude and direction
Example steps from earlier discussion: combine a = 3 miles at 40° north of east and b = 4 miles at 60° from +x axis (30° east of north) to obtain the net displacement d, then compute |
- d_x = 3 cos 40° + 4 cos 60°
- d_y = 3 sin 40° + 4 sin 60°
- |d| = √(dx² + dy²)
- θd = arctan(dy / d_x)
This illustrates how to translate directional language into components and then back to magnitude/direction