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Doing your best means never stop trying.
Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.
Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.
Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.
Vector Subtraction: The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.
Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.
The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$
Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$
The projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.
The cross product, denoted by $\vec{A} \times \vec{B}$, is a binary operation on two vectors in three-dimensional space. It yields a new vector that is perpendicular (orthogonal, at a right angle) to both original vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
The cross product $\vec{A}x\vec{B}$ can be computed using the determinant of a $3 \times 3$ matrix formed by the standard basis vectors ($\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$) and the components of the two vectors: $\vec{A} \times \vec{B} = det(\begin{smallmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|\vec{k}$.
Example: If $\vec{A} = \langle 1, 2, 3 \rangle$ and $\vec{B} = \langle 4, 5, 6 \rangle$: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = (2\cdot 6 - 3\cdot 5)\hat{\mathbf{i}} - (1\cdot 6 - 3\cdot 4)\hat{\mathbf{j}} + (1\cdot 5 - 2\cdot 4)\hat{\mathbf{k}} = -3\hat{\mathbf{i}} + 6\hat{\mathbf{j}} - 3\hat{\mathbf{k}} = \langle -3, 6, -3 \rangle$
Definition. A position vector indicates the position or location of a point relative to an arbitrary reference point. or a point $P$ in space moving over time $t$, this is written as a vector-valued function: $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$
For example, consider a cycloid traced by a point on a wheel rolling along the x-axis at a constant unit speed. If the wheel’s radius is 1, the position vector is: $\vec{r}(t) = \langle t - \sin(t), 1 - \cos(t) \rangle$
Velocity is a vector quantity that measures the rate of change of position. It always points in the direction of motion (tangent to the trajectory). It has both magnitude and direction, and is given by the derivative of the position vector: $\vec{v}(t) = \frac{d\vec{r}}{dt} = \left\langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right\rangle$.
For our cycloid, the velocity is $\vec{v}(t) = \langle 1 - \cos(t), \sin(t) \rangle$.
The speed is a scalar value representing the magnitude (length) of the velocity vector: $\vert{}\vec{v}\vert{} = \sqrt{(1 - \cos(t))^2 + \sin^2(t)} = \sqrt{1 - 2\cos(t) + \cos^2(t) + \sin^2(t)} = \sqrt{2 - 2\cos(t)}$
Acceleration is the derivative of the velocity vector with respect to time: $\vec{a}(t) = \frac{d\vec{v}}{dt}$. For our cycloid example, $\vec{a}(t) = \langle \sin(t), \cos(t) \rangle$. At $t = 0$, the velocity is $\vec{v}(0) = \vec{0}$, and the acceleration is $\vec{a}(0) = \langle 0, 1 \rangle$. This means the point momentarily rests at the origin before accelerating strictly straight up.
The arc length is the distance traveled by an object from one point to another point along a curve. Since the speed of a moving object is the length of its velocity vector, the distance the object travels from t = a to t = b is the integral of |r’(t)| over the interval [a, b] (the integral of speed), e.g., 🌀 length of an arch in our cycloid is $\int_{a}^{b} \vert{}\frac{d\vec{r}}{dt}\vert{}dt = \int_{a}^{b} \vert{}\vec{v}(t)\vert{}dt =$[For our cycloid example] $\int_{0}^{2π} \sqrt{2 -2cos(t)}dt$
🧭 Why arc length = integral of speed? Imagine a particle moving along a curve described by a vector function: $\vec{r}(t) = \langle t - \sin(t), 1 - \cos(t) \rangle$. Speed tells you “how fast distance is accumulating.” So over a tiny time interval dt, the particle travels approximately: $\text{distance} \approx \vert{}\vec{v}(t)\vert{}dt$. Add up all these tiny distances from t = a to t = b, $L = \int_{a}^{b} \vert{}\vec{v}(t)\vert{}dt$
Definition. The unit tangent vector is a vector that points in the direction of the tangent to a curve at a given point and has a magnitude or length of 1. The unit tangent vector is used to describe the direction of motion along a curve (i.e., it isolates the direction of motion from the speed) and can be used to calculate the derivative of a curve at a given point, $\vec{T} = \frac{\vec{v}}{|\vec{v}|}$. So $\vec{T}$ is the “pure direction” of the motion.
Arc length $s$ measures distance traveled. As the particle moves along the curve, the arc length $s$ increases. Its rate of change is: $\frac{ds}{dt} = speed$. This is just the definition of speed: distance per unit time.
Using the Chain Rule, we can re-express velocity in terms of arc length s: $\vec{v}=\frac{d\vec{r}}{dt}$ =[The Chain Rule] $\frac{d\vec{r}}{ds}\cdot \frac{ds}{dt} = \vec{T} \cdot \frac{ds}{dt}$. This is the key identity: Velocity = (unit tangent direction) × (speed).
To sum up, velocity, being a vector, has both a magnitude and a direction. Its direction is always directed tangent to the object’s motion or trajectory. Its length is the speed, $|\vec{v}|=\frac{ds}{dt}$. In words, the speed is the magnitude of the velocity vector and is also the derivative of the arc length with respect to time.
When we approximate the change in position $\Delta r$ as $\vec{T} \cdot \Delta s$ (Figure i and ii), the speed $\frac{\Delta s}{\Delta t}$ becomes more accurate as $\Delta t \to 0$. Dividing by $\Delta t, \frac{\Delta r}{\Delta t} ≈ \vec{T} \cdot \frac{\Delta s}{\Delta t}$ and taking the limit as $\Delta t \to 0 \implies \frac{d\vec{r}}{dt}=\vec{T}·\frac{ds}{dt}$ that is the previous formula where the approximation becomes an equality (the approximation gets better and better).
Kepler's first law states that all planets orbit the Sun in an elliptical path with the Sun located at one of the two foci of the ellipse. Because the orbit is elliptical, a planet’s distance from the Sun changes constantly.
Kepler’s second law states that planets move in planes, and the imaginary line joining a planet and the Sun (or the radius vector) sweeps out equal areas of space in equal intervals of time as the planet orbits the sun.
This means that planets do not move with constant speed along their orbits. Rather, their speed varies so that the line joining the centers of the Sun and the planet sweeps out equal areas in equal intervals of time, (Figure iii):
Isaac Newton later developed a theory, using the principles of Calculus and his laws of motion and universal gravitation, to explain why planets follow Kepler’s empirical laws.
The cross product of two vectors $\vec{u}$ and $\vec{v}$ results in a vector whose magnitude is equal to the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$. Hence, we can use the vector product to compute the area of a triangle formed by three points A, B and C in space, Area of △ABC = $\frac{1}{2}|\vec{AB} \times \vec{AC}|$ (Figure iv).
To apply this to planetary motion, consider the area swept by a planet’s radius vector $\vec{r}$ (from the Sun to the planet) in a small time interval $\Delta t$, the change in position is $\Delta \vec{r}$. This area can be approximated by the area of the parallelogram formed by $\vec{r}$ and the change in the position vector $\Delta \vec{r}$: $\text{Area} \approx \frac{1}{2}|\vec{r} \times \vec{\Delta r}|$ (Area swept by the planet in a small interval of time $\Delta t$).
Since for small intervals $\Delta \vec{r} \approx \vec{v}\Delta t$: $\text{Area} \approx \frac{1}{2}\vert{}\vec{r} \times \vec{v}\Delta t\vert{} = \left(\frac{1}{2}\vert{}\vec{r} \times \vec{v}\vert{}\right) \Delta t$.
According to Kepler’s second law, the area swept per unit time is constant. Therefore, the magnitude $\vert{}\vec{r} \times \vec{v}\vert{}$ must be constant. Furthermore, because the planet moves in a flat plane, the direction of $\vec{r} \times \vec{v}$ (which is perpendicular to the orbital plane) is also constant.
Since $\vec{r} \times \vec{v}$ is a constant vector, its derivative with respect to time is zero: $\frac{d}{dt}(\vec{r} \times \vec{v}) = 0$
Using the product rule for cross products: $\left(\frac{d\vec{r}}{dt} \times \vec{v}\right) + \left(\vec{r} \times \frac{d\vec{v}}{dt}\right) = 0$.
Substitute $\vec{v}$ for $\frac{d\vec{r}}{dt}$ and $\vec{a}$ for $\frac{d\vec{v}}{dt}$, we get: $(\vec{v} \times \vec{v}) + (\vec{r} \times \vec{a}) = 0$
The cross product of any vector with itself is zero ($\vec{v} \times \vec{v} = 0$) because the area of a degenerate parallelogram is zero. This leaves: $\vec{r} \times \vec{a} = 0$
When the cross product of two non-zero vectors is zero, the vectors must be parallel. Therefore, $\vec{a} = \frac{d\vec{v}}{dt}$ is parallel to $\vec{r}$. Gravitational acceleration is the acceleration an object experiences when gravity is the only force acting on it. This proves that the gravitational acceleration is entirely radial —it is a central force directed straight towards the Sun.
If the acceleration is non-zero when the velocity is zero, it means that there is a force acting on the particle, causing its velocity to change. This non-zero acceleration ensures that the particle doesn’t remain stationary but instead starts moving in the opposite direction.
A particle slows down when its speed (the absolute magnitude of velocity) decreases. This happens when acceleration opposes the current direction of motion. If velocity and acceleration have the same sign, the particle is speeding up. $\vec a = a_T \vec{T} + a_N \vec{N} = \frac{d^2s}{dt^2}\vec{T} + \left(\frac{ds}{dt}\right)^2\kappa \vec{N}$ where:
$\star$ $a_T = \frac{d^2s}{dt^2}\vec{T}$ = tangential acceleration (parallel to velocity). Only $a_T$ affects speed.
$\star$ $\vec{T}$. This is a vector of length 1 that points exactly in the direction you are moving at any split second. It is your “forward” direction.
$\star$ $\frac{d^2s}{dt^2}$, Scalar Tangential Acceleration, which is how fast your speed is changing.
$\star$ If $\frac{d^2s}{dt^2} \gt 0$, you are stepping on the gas. Acceleration is in the direction of $\vec{T}$ (forward), so you speed up.
$\star$ If $\frac{d^2s}{dt^2} \lt 0$, you are stepping on the brakes. Acceleration points opposite to $\vec{T}$ (backward), so you slow down.
$\star$ $a_N$ = normal acceleration (perpendicular to velocity, changes direction only). This component is entirely focused on your direction. It is the “centripetal acceleration” that pushes you sideways into your car seat during a turn.
$\star$ $\vec{N}$ = Unit Normal Vector: This is a vector of length 1 that is perpendicular to your direction of travel. It always points toward the inside of the turn.
$\star$ $\frac{ds}{dt}$ = Speed, v: Because it is squared, it is always positive. This shows that the faster you drive, the exponentially harder it is to make a turn.
$\star$ $\kappa$ = Curvature: This measures how sharp the turn is.
Now, let’s summarize the signs of velocity and acceleration in different intervals:
| Time | Velocity v(t) | Acceleration a(t) | State |
|---|---|---|---|
| $0 < t < 2$ | Positive (+) | Negative (-) | Slowing down |
| $2 < t < 4$ | Negative (-) | Negative (-) | Speeding up |
| $4 < t < 6$ | Negative (-) | Positive (+) | Slowing down |
| $t > 6$ | Positive (+) | Positive (+) | Speeding up |
From the table, we see that the particle is slowing down when $0 < t < 2$ and $4 < t < 6$, because in these intervals, the velocity and acceleration have opposite signs (Figure B).
Position Vector.
The position vector $\vec{r}(t)$ represents the exact location of a point in space at any given time t.
We can express this using standard basis vectors ($\vec{i}$ and $\vec{j}$) or angled bracket notation: $\vec{r}(t) = \langle 2t-2, t^2+t-1 \rangle$
Evaluating the position at t = -1: $\vec{r}(-1) = \langle 2(-1)-2, (-1)^2+(-1)-1 \rangle = \langle -4, -1 \rangle$
At t = -1, the object is located at the coordinates $(-4, -1)$.
Velocity Vector.
The velocity vector $\vec{v}(t)$ is the first derivative of the position vector with respect to time. It describes how fast the position is changing and in which direction.
$\vec{v}(t) = \frac{d\vec{r}}{dt} = \left\langle \frac{d}{dt}(2t-2), \frac{d}{dt}(t^2+t-1) \right\rangle = \langle 2, 2t+1 \rangle$
Evaluating the velocity at t = -1: $\vec{v}(-1) = \langle 2, 2(-1)+1 \rangle = \langle 2, -1 \rangle$
Acceleration Vector.
The acceleration vector $\vec{a}(t)$ is the derivative of the velocity vector with respect to time. It describes how fast the velocity itself is changing and in which direction.
$\vec{a}(t) = \frac{d\vec{v}}{dt} = \left\langle \frac{d}{dt}(2), \frac{d}{dt}(2t+1) \right\rangle = \langle 0, 2 \rangle$
Notice that the acceleration vector is the constant $\langle 0, 2 \rangle$ and does not depend on time t. Therefore: $\vec{a}(-1) = \langle 0, 2 \rangle$
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