## Curves, Velocity, “Classical Kinematics”

1. Curves. We are interested in describing the motion of my car. Well, everyone is interested in the motion of my car. How can we describe it mathematically?

First we approximate the car as a point. The point-like car moves in time, so the value of its components are functions of time. More precisely, the position of my car is
(1)$\displaystyle \vec{r}(t) = \langle f(t),g(t),h(t)\rangle = f(t)\widehat{\textbf{\i}} + g(t)\widehat{\textbf{\j}} + h(t)\widehat{\textbf{k}}$
where the functions ${f(t)}$, ${g(t)}$, and ${h(t)}$ are sometimes called component functions. Another way to think about this is writing
(2)$\displaystyle \vec{r}\colon[0,1]\rightarrow{\mathbb R}^{3}$
where ${0\leq t\leq1}$.

Classical mechanics studies such curves under various circumstances. We will discuss some notions of kinematics, and study what it means to differentiate curves.

Example 1. Consider a point traveling in circular motion in the ${xy}$-plane. What does this look like?

Well, it’s a paramteric curve, using trigonometric functions we write
(3)$\displaystyle \vec{r}(t) = \cos(t)\widehat{\textbf{\i}}+\sin(t)\widehat{\textbf{\j}}.$
This descrivbes an anti-clockwise circular motion with radius 1, lying in the ${xy}$-plane.

Example 2. Suppose a particle travels along a parabolic curve, what does the curve look like? We can write it explicitly as
(4)$\displaystyle \vec{r}(t) = t\,\widehat{\textbf{\i}}+(t^{2}-1)\widehat{\textbf{\j}}$
This is precisely aa parabola.

2. Calculus with Vector-Valued Functions. We should recall the construction of the tangent line to a curve ${y=f(x)}$ at a point ${(x_{0},f(x_{0})=y_{0})}$ had us write
(5)$\displaystyle t(h) = y_{0} + f'(x_{0}) \cdot h.$
When we consider the situation when we work with ${\vec{r}(t)}$ instead of a function ${f(x)}$. We have ${\vec{r}_{0}=\vec{r}(t_{0})}$ be the base point for the tangent to the curve, then we have
(6)$\displaystyle \vec{T}(h) = \vec{r}_{0} + \vec{r}'(t_{0})\cdot h$
The problem: what exactly is ${\vec{r}'(t_{0})}$?

3. We can let ${\vec{r}\colon(0,1)\rightarrow{\mathbb R}^{3}}$ (or more generally the codomain can be ${{\mathbb R}^{n}}$ for any positive integer ${n\in{\mathbb N}}$). We have
(7)$\displaystyle \frac{\mathrm{d}\vec{r}(t)}{\mathrm{d} t}=\vec{v}(t) = \lim_{\Delta t\rightarrow0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}$
describe the rate of change of the position vector ${\vec{r}(t)}$ with respect to time. What does this look like? Well, writiing out
(8)$\displaystyle \vec{r}(t) = \langle f(t),g(t),h(t)\rangle = f(t)\widehat{\textbf{\i}} + g(t)\widehat{\textbf{\j}} + h(t)\widehat{\textbf{k}}$
we have
(9)$\displaystyle \frac{\mathrm{d}\vec{r}(t)}{\mathrm{d} t} = \langle f'(t), g'(t), h'(t)\rangle = f'(t)\widehat{\textbf{\i}} + g'(t)\widehat{\textbf{\j}} + h'(t)\widehat{\textbf{k}}$
where primes denote differentiation with respect to time.

Remark 1. We can keep iterating this procedure to obtain higher order derivatives of a curve.

4. Kinematics. We have ${\vec{r}(t)}$ describe the position of a particle. The velocity of the particle is a vector-valued function
(10)\displaystyle \begin{aligned} \vec{v}(t) &=\lim_{\Delta t\rightarrow 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}\\ &=\frac{\mathrm{d}\vec{r}(t)}{\mathrm{d} t} \end{aligned}
However, we also can consider the speed or the magnitude of the velocity
(11)$\displaystyle \|\vec{v}(t)\|=\frac{\mathrm{d} s}{\mathrm{d} t} = \begin{pmatrix} \mbox{rate of change of distance}\\ \mbox{with respect to time} \end{pmatrix}$
Observe the speed is a scalar quantity: it’s just some function of time. The velocity is a vector-valued function of time.

We have one last kinematical quantity to consider: the acceleration. This is just the rate of change of velocity with respect to time:
(12)$\displaystyle \vec{a}(t) = \frac{\mathrm{d}\vec{v}(t)}{\mathrm{d} t} = \frac{\mathrm{d}^{2}\vec{r}(t)}{\mathrm{d} t^{2}}$
Observe we can reconstruct the position from the velocity by considering
(13)$\displaystyle \vec{r}(t) = \vec{r}(t_{0}) + \int^{t}_{t_{0}}\frac{\mathrm{d}\vec{r}(\tau)}{\mathrm{d}\tau}\,\mathrm{d}\tau$
which when we consider ${\vec{r}(t)=\langle x(t),y(t),z(t)\rangle}$ we have the integral evaluated “component-wise”:
(14)$\displaystyle \vec{r}(t) = \vec{r}(t_{0}) + \left\langle \int^{t}_{t_{0}}\frac{\mathrm{d} x(\tau)}{\mathrm{d}\tau}\,\mathrm{d}\tau, \int^{t}_{t_{0}}\frac{\mathrm{d} y(\tau)}{\mathrm{d}\tau}\,\mathrm{d}\tau, \int^{t}_{t_{0}}\frac{\mathrm{d} z(\tau)}{\mathrm{d}\tau}\,\mathrm{d}\tau\right\rangle.$
We can similarly reconstruct velocity from acceleration.

Example 3. Consider the curve describing circular motion
(15)$\displaystyle \vec{r}(t) = \cos(t)\widehat{\textbf{\i}} +\sin(t)\widehat{\textbf{\j}}$
What is its velocity vector, acceleration vector, and speed?

Solution: We find its velocity
(16)\displaystyle \begin{aligned} \vec{v}(t) &= \frac{\mathrm{d}\vec{r}(t)}{\mathrm{d} t}\\ &=-\sin(t)(t)\widehat{\textbf{\i}} +\cos(t)\widehat{\textbf{\j}} \end{aligned}
From this we can compute its speed as
(17)\displaystyle \begin{aligned} \|\vec{v}(t)\| &= \sqrt{\vec{v}(t)\cdot\vec{v}(t)}\\ &=\sqrt{\sin^{2}(t)+\cos^{2}(t)} = 1. \end{aligned}
The acceleration is precisely the derivative of the velocity vector
(18)$\displaystyle \frac{\mathrm{d}\vec{v}(t)}{\mathrm{d} t} = -\cos(t)\widehat{\textbf{\i}}-\sin(t)\widehat{\textbf{\j}}$
That concludes our example.

Exercise 1. Find the velocity vector, speed, and acceleration of the parabolic curve ${\vec{r}(t) = t\,\widehat{\textbf{\i}}+(t^{2}-1)\widehat{\textbf{\j}}}$

Exercise 2. Let ${\vec{u}(t)}$ and ${\vec{v}(t)}$ be differentiable vector-valued functions of time. Prove or find a counter-example that
(19)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} t}\bigl[\vec{u}(t)\times\vec{v}(t)\bigr]= \frac{\mathrm{d}\vec{u}(t)}{\mathrm{d} t}\times\vec{v}(t) + \vec{u}(t)\times\frac{\mathrm{d}\vec{v}(t)}{\mathrm{d} t}.$

Exercise 3. Calculate ${\displaystyle\frac{\mathrm{d}}{\mathrm{d} t}\bigl[\vec{a}(t)\cdot\bigl(\vec{b}(t)\times\vec{c}(t)\bigr)\bigr]}$