# Calculus/Vector Functions
## Introduction

If we have a function ${\vec {f}}:\mathbb {R} \to \mathbb {R} ^{n}$ , we say that ${\vec {f}}$'s image (the set ${\Big \{\vec {f}}(t){\Big |}t\in \mathbb {R} {\Big \}}$ - or some subset of $\mathbb {R}$) is a curve in $\mathbb {R} ^{n}$ and ${\vec {f}}$ is its parametrization.
Parameterizations are not necessarily unique - for example, ${\vec {f}}(t)=(\cos(t),\sin(t))$ such that $t\in [0,2\pi )$ is one parametrization of the unit circle, and ${\vec {g}}(t)=(\cos(at),\sin(at))$ such that $t\in \left[0,{\frac {2\pi }{a}}\right)$ is a whole family of parameterizations of that circle.

### Collision and intersection points

Say we have two different curves. It may be important to consider
* points the two curves share - where they intersect
* intersections which occur for the same value of $t$ - where they collide.

#### Intersection points

Firstly, we have two parameterizations ${\vec {f}}(t)$ and ${\vec {g}}(t)$ , and we want to find out when they intersect, this means that we want to know when the function values of each parametrization are the same. This means that we need to solve
    ${\vec {f}}(t)={\vec {g}}(s)$
because we're seeking the function values independent of the times they intersect.
For example, if we have ${\vec {f}}(t)=(t,3t)$ and ${\vec {g}}(t)=(t,t^{2})$ , and we want to find intersection points:
    ${\vec {f}}(t)={\vec {g}}(s)$
    $(t,3t)=(s,s^{2})$
    $t=s\ ,\ 3t=s^{2}$
with solutions $(t,s)=(0,0)\ ,\ (3,9)$
So, the two curves intersect at the points $(0,0)\ ,\ (3,9)$ .

#### Collision points

However, if we want to know when the points "collide", with ${\vec {f}}(t)$ and ${\vec {g}}(t)$ , we need to know when both the function values and the times are the same, so we need to solve instead
    ${\vec {f}}(t)={\vec {g}}(t)$
For example, using the same functions as before, ${\vec {f}}(t)=(t,3t)$ and ${\vec {g}}(t)=(t,t^{2})$ , and we want to find collision points:
    ${\vec {f}}(t)={\vec {g}}(t)$
    $(t,3t)=(t,t^{2})$
    $t=t\ ,\ 3t=t^{2}$
which gives solutions $t=0,3$ So the collision points are $(0,0)\ ,\ (3,9)$ .
We may want to do this to actually model physical problems, such as in ballistics.

### Intersection vector functions

We can also use vector functions to represent the curve of intersection of two surfaces. For example, we want to know the curve of intersection of the cylinder $x^{2}+y^{2}=1$ and the plane $y+z=2$.
   Vector functions rely on parameterizations, so we can rewrite the equation of the cylinder into: $\cos ^{2}t+\sin ^{2}t=1$, where $x=\cos t,y=\sin t$.
   From the equation of the plane, we know that $z=2-y=2-\sin t$. Thus the corresponding vector equation is:
   $\mathbf {r} (t)=\langle \cos t,\sin t,2-\sin t\rangle$

### Limits and Continuity

The limit of a vector function $\mathbf {r}$ is defined by taking the limits of its component functions.
There is a vector function $\mathbf {r} (t)=\langle f(t),g(t),h(t)\rangle$. If $\lim _{t\rightarrow a}f(t),\lim _{t\rightarrow a}g(t),\lim _{t\rightarrow a}h(t)$ exist, then
    $\lim _{t\rightarrow a}\mathbf {r} (t)=\langle \lim _{t\rightarrow a}f(t),\lim _{t\rightarrow a}g(t),\lim _{t\rightarrow a}h(t)\rangle$
And the requirement for continuity is also simple:
   A vector function $\mathbf {r}$ is continuous at $a$ if $\lim _{t\rightarrow a}\mathbf {r} (t)=\mathbf {r} (a)$.

## Derivatives and Integrals
### Differentiation

Recall that the first derivative of a scalar function $f(x)$ is defined as:
   $f'(x)=\lim _{\Delta x\rightarrow 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$
The first derivative of a vector function $\mathbf {r} (t)$ is defined in much the same way:
   $\mathbf {r} '(t)=\lim _{\Delta t\rightarrow 0}{\frac {\mathbf {r} (t+\Delta t)-\mathbf {r} (t)}{\Delta t}}$
We can use this definition to prove that the derivative of a vector function can be presented as the derivative of its component functions.
   ${\begin{aligned}\mathbf {r} '(t)&=\lim _{\Delta t\rightarrow 0}{\frac {1}{\Delta t}}[\mathbf {r} (t+\Delta t)-\mathbf {r} (t)]\quad {\text{definition}}\\&=\lim _{\Delta t\rightarrow 0}{\frac {1}{\Delta t}}[\langle f(t+\Delta t),g(t+\Delta t),h(t+\Delta t)\rangle -\langle f(t),g(t),h(t)\rangle ]\quad {\text{component form}}\\&=\lim _{\Delta t\rightarrow 0}\langle {\frac {f(t+\Delta t)-f(t)}{\Delta t}},{\frac {g(t+\Delta t)-g(t)}{\Delta t}},{\frac {h(t+\Delta t)-h(t)}{\Delta t}}\rangle \quad {\text{vector addition and multiplication}}\\&=\langle \lim _{\Delta t\rightarrow 0}{\frac {f(t+\Delta t)-f(t)}{\Delta t}},\lim _{\Delta t\rightarrow 0}{\frac {g(t+\Delta t)-g(t)}{\Delta t}},\lim _{\Delta t\rightarrow 0}{\frac {h(t+\Delta t)-h(t)}{\Delta t}}\rangle \quad {\text{definition}}\\&=\langle f'(t),g'(t),h'(t)\rangle \quad {\text{definition}}\\\end{aligned}}$
Thus, using the same method, we can derive the second derivative and so on.
There is a vector function $\mathbf {r} (t)=\langle f(t),g(t),h(t)\rangle$. The first derivative of this vector function is:
    $\mathbf {r} '(t)=\lim _{\Delta t\rightarrow 0}{\frac {\mathbf {r} (t+\Delta t)-\mathbf {r} (t)}{\Delta t}}=\langle f'(t),g'(t),h'(t)\rangle$
So the $n$th-order derivative should look like this:
   $\mathbf {r} ^{(n)}(t)=\langle f^{(n)}(t),g^{(n)}(t),h^{(n)}(t)\rangle$

### Differentiation rules

Just like real-valued functions, there are some differentiation rules in the world of vector functions. The factor that makes vector differentiation rules slight more complicated is the product rule because there are two kinds of multiplication in vectors: dot product and cross product.
Suppose $\mathbf {u} ,\mathbf {v}$ are differentiable vector functions and $f$ is a real-valued function. Then
* ${\frac {d}{dt}}[\mathbf {u} (t)+\mathbf {v} (t)]=\mathbf {u} '(t)+\mathbf {v} '(t)\quad$ (addition)
* ${\frac {d}{dt}}[f(t)\mathbf {u} (t)]=f'(t)\mathbf {u} (t)+f(t)\mathbf {u} '(t)\quad$ (scalar multiplication)
* ${\frac {d}{dt}}[\mathbf {u} (t)\cdot \mathbf {v} (t)]=\mathbf {u} '(t)\cdot \mathbf {v} (t)+\mathbf {u} (t)\cdot \mathbf {v} '(t)\quad$ (dot product)
* ${\frac {d}{dt}}[\mathbf {u} (t)\times \mathbf {v} (t)]=\mathbf {u} '(t)\times \mathbf {v} (t)+\mathbf {u} (t)\times \mathbf {v} '(t)\quad$ (cross product)
* ${\frac {d}{dt}}[\mathbf {u} (f(t))]=f'(t)\mathbf {u} '(f(t))\quad$ (chain rule)
Naturally, we will prove that those rules are correct. Let us assume that $\mathbf {u} (t)=\langle f_{1}(t),f_{2}(t),f_{3}(t)\rangle$ and $\mathbf {v} (t)=\langle g_{1}(t),g_{2}(t),g_{3}(t)\rangle$.
Rule 1: the addition rule
   ${\begin{aligned}{\frac {d}{dt}}[\mathbf {u} (t)+\mathbf {v} (t)]&={\frac {d}{dt}}[\langle f_{1}(t),f_{2}(t),f_{3}(t)\rangle +\langle g_{1}(t),g_{2}(t),g_{3}(t)\rangle ]\quad {\text{component form}}\\&={\frac {d}{dt}}\langle f_{1}(t)+g_{1}(t),f_{2}(t)+g_{2}(t),f_{3}(t)+g_{3}(t)\rangle \quad {\text{vector addition}}\\&=\langle {\frac {d}{dt}}(f_{1}(t)+g_{1}(t)),{\frac {d}{dt}}(f_{2}(t)+g_{2}(t)),{\frac {d}{dt}}(f_{3}(t)+g_{3}(t))\rangle \quad {\text{distribution}}\\&=\langle f_{1}'(t)+g_{1}'(t),f_{2}'(t)+g_{2}'(t),f_{3}'(t)+g_{3}'(t)\rangle \quad {\text{addition rule for real-valued functions}}\\&=\langle f_{1}'(t),f_{2}'(t),f_{3}'(t)\rangle +\langle g_{1}'(t),g_{2}'(t),g_{3}'(t)\rangle \quad {\text{vector addition}}\\&=\mathbf {u} '(t)+\mathbf {v} '(t)\quad {\text{vector form}}\\\end{aligned}}$
Rule 2: the scalar multiplication rule
   ${\begin{aligned}{\frac {d}{dt}}[f(t)\mathbf {u} (t)]&={\frac {d}{dt}}[f(t){\begin{pmatrix}f_{1}(t)\\f_{2}(t)\\f_{3}(t)\end{pmatrix}}]\quad {\text{component form}}\\&={\frac {d}{dt}}{\begin{pmatrix}f(t)f_{1}(t)\\f(t)f_{2}(t)\\f(t)f_{3}(t)\end{pmatrix}}\quad {\text{scalar multiplication}}\\&={\begin{pmatrix}{\frac {d}{dt}}(f(t)f_{1}(t))\\{\frac {d}{dt}}(f(t)f_{2}(t))\\{\frac {d}{dt}}(f(t)f_{3}(t))\\\end{pmatrix}}\quad {\text{distribution}}\\&={\begin{pmatrix}f'(t)f_{1}(t)+f(t)f_{1}'(t)\\f'(t)f_{2}(t)+f(t)f_{2}'(t)\\f'(t)f_{3}(t)+f(t)f_{3}'(t)\\\end{pmatrix}}\quad {\text{multiplication rule for real-valued functions}}\\&={\begin{pmatrix}f'(t)f_{1}(t)\\f'(t)f_{2}(t)\\f'(t)f_{3}(t)\end{pmatrix}}+{\begin{pmatrix}f(t)f_{1}'(t)\\f(t)f_{2}'(t)\\f(t)f_{3}'(t)\end{pmatrix}}\quad {\text{vector addition}}\\&=f'(t){\begin{pmatrix}f_{1}(t)\\f_{2}(t)\\f_{3}(t)\end{pmatrix}}+f(t){\begin{pmatrix}f_{1}'(t)\\f_{2}'(t)\\f_{3}'(t)\end{pmatrix}}\quad {\text{scalar multiplication}}\\&=f'(t)\mathbf {u} (t)+f(t)\mathbf {u} '(t)\quad {\text{vector form}}\\\end{aligned}}$
Rule 3: the dot product rule
   ${\begin{aligned}{\frac {d}{dt}}[\mathbf {u} (t)\cdot \mathbf {v} (t)]&={\frac {d}{dt}}[\langle f_{1}(t),f_{2}(t),f_{3}(t)\rangle \cdot \langle g_{1}(t),g_{2}(t),g_{3}(t)\rangle ]\quad {\text{component form}}\\&={\frac {d}{dt}}(f_{1}(t)g_{1}(t)+f_{2}(t)g_{2}(t)+f_{3}(t)g_{3}(t))\quad {\text{dot product}}\\&={\frac {d}{dt}}\sum _{i=1}^{3}f_{i}(t)g_{i}(t)\quad {\text{simplification}}\\&=\sum _{i=1}^{3}{\frac {d}{dt}}f_{i}(t)g_{i}(t)\quad {\text{distribution}}\\&=\sum _{i=1}^{3}[f_{i}'(t)g_{i}(t)+f_{i}(t)g_{i}'(t)]\quad {\text{multiplication rule for real-valued functions}}\\&=\sum _{i=1}^{3}f_{i}(t)g_{i}(t)+\sum _{i=1}^{3}f_{i}(t)g_{i}'(t)\quad {\text{vector addition}}\\&=\mathbf {u} '(t)\cdot \mathbf {v} (t)+\mathbf {u} (t)\cdot \mathbf {v} '(t)\quad {\text{vector form}}\\\end{aligned}}$
Rule 4: the cross product rule
   ${\begin{aligned}{\frac {d}{dt}}[\mathbf {u} (t)\times \mathbf {v} (t)]&={\frac {d}{dt}}[\langle f_{1}(t),f_{2}(t),f_{3}(t)\rangle \times \langle g_{1}(t),g_{2}(t),g_{3}(t)\rangle ]\quad {\text{component form}}\\&={\frac {d}{dt}}{\begin{vmatrix}\mathbf {\hat {i}} &\mathbf {\hat {j}} &\mathbf {\hat {k}} \\f_{1}(t)&f_{2}(t)&f_{3}(t)\\g_{1}(t)&g_{2}(t)&g_{3}(t)\\\end{vmatrix}}\quad {\text{cross product}}\\&={\frac {d}{dt}}{\begin{pmatrix}f_{2}(t)g_{3}(t)-f_{3}(t)g_{2}(t)\\f_{3}(t)g_{1}(t)-f_{1}(t)g_{3}(t)\\f_{1}(t)g_{2}(t)-f_{2}(t)g_{1}(t)\\\end{pmatrix}}\quad \\&={\begin{pmatrix}{\frac {d}{dt}}(f_{2}(t)g_{3}(t)-f_{3}(t)g_{2}(t))\\{\frac {d}{dt}}(f_{3}(t)g_{1}(t)-f_{1}(t)g_{3}(t))\\{\frac {d}{dt}}(f_{1}(t)g_{2}(t)-f_{2}(t)g_{1}(t))\\\end{pmatrix}}\quad {\text{distribution}}\\&={\begin{pmatrix}f_{2}'(t)g_{3}(t)+f_{2}(t)g_{3}'(t)-f_{3}'(t)g_{2}(t)-f_{3}(t)g_{2}'(t)\\f_{3}'(t)g_{1}(t)+f_{3}(t)g_{1}'(t)-f_{1}'(t)g_{3}(t)-f_{1}(t)g_{3}'(t)\\f_{2}'(t)g_{3}(t)+f_{2}(t)g_{3}'(t)-f_{3}'(t)g_{2}(t)-f_{3}(t)g_{2}'(t)\\\end{pmatrix}}\quad {\text{multiplication rule for real-valued functions}}\\&={\begin{pmatrix}f_{2}'(t)g_{3}(t)-f_{3}'(t)g_{2}(t)+f_{2}(t)g_{3}'(t)-f_{3}(t)g_{2}'(t)\\f_{3}'(t)g_{1}(t)-f_{1}'(t)g_{3}(t)+f_{3}(t)g_{1}'(t)-f_{1}(t)g_{3}'(t)\\f_{2}'(t)g_{3}(t)-f_{3}'(t)g_{2}(t)+f_{2}(t)g_{3}'(t)-f_{3}(t)g_{2}'(t)\\\end{pmatrix}}\quad {\text{rearrangement}}\\&={\begin{pmatrix}f_{2}'(t)g_{3}(t)-f_{3}'(t)g_{2}(t)\\f_{3}'(t)g_{1}(t)-f_{1}'(t)g_{3}(t)\\f_{2}'(t)g_{3}(t)-f_{3}'(t)g_{2}(t)\\\end{pmatrix}}+{\begin{pmatrix}f_{2}(t)g_{3}'(t)-f_{3}(t)g_{2}'(t)\\f_{3}(t)g_{1}'(t)-f_{1}(t)g_{3}'(t)\\f_{2}(t)g_{3}'(t)-f_{3}(t)g_{2}'(t)\\\end{pmatrix}}\quad {\text{vector addtion}}\\&={\begin{vmatrix}\mathbf {\hat {i}} &\mathbf {\hat {j}} &\mathbf {\hat {k}} \\f_{1}'(t)&f_{2}'(t)&f_{3}'(t)\\g_{1}(t)&g_{2}(t)&g_{3}(t)\\\end{vmatrix}}+{\begin{vmatrix}\mathbf {\hat {i}} &\mathbf {\hat {j}} &\mathbf {\hat {k}} \\f_{1}(t)&f_{2}(t)&f_{3}(t)\\g_{1}'(t)&g_{2}'(t)&g_{3}'(t)\\\end{vmatrix}}\quad {\text{cross product}}\\&={\begin{pmatrix}f_{1}'(t)\\f_{2}'(t)\\f_{3}'(t)\\\end{pmatrix}}\times {\begin{pmatrix}g_{1}(t)\\g_{2}(t)\\g_{3}(t)\\\end{pmatrix}}+{\begin{pmatrix}f_{1}(t)\\f_{2}(t)\\f_{3}(t)\\\end{pmatrix}}\times {\begin{pmatrix}g_{1}'(t)\\g_{2}'(t)\\g_{3}'(t)\\\end{pmatrix}}\quad \\&=\mathbf {u} '(t)\times \mathbf {v} (t)+\mathbf {u} (t)\times \mathbf {v} '(t)\quad {\text{vector form}}\\\end{aligned}}$
Rule 5: the chain rule
   ${\begin{aligned}{\frac {d}{dt}}[\mathbf {u} (f(t))]&={\frac {d}{dt}}{\begin{pmatrix}f_{1}(f(t))\\f_{2}(f(t))\\f_{3}(f(t))\\\end{pmatrix}}\quad {\text{component form}}\\&={\begin{pmatrix}{\frac {d}{dt}}f_{1}(f(t))\\{\frac {d}{dt}}f_{2}(f(t))\\{\frac {d}{dt}}f_{3}(f(t))\\\end{pmatrix}}\quad {\text{distribution}}\\&={\begin{pmatrix}f'(t)f_{1}'(f(t))\\f'(t)f_{2}'(f(t))\\f'(t)f_{3}'(f(t))\\\end{pmatrix}}\quad {\text{chain rule for real-valued functions}}\\&=f'(t){\begin{pmatrix}f_{1}'(f(t))\\f_{2}'(f(t))\\f_{3}'(f(t))\\\end{pmatrix}}\quad {\text{scalar multiplication}}\\&=f'(t)\mathbf {u} '(f(t))\quad {\text{vector form}}\\\end{aligned}}$
$\blacksquare$

#### Angle between curves

We can then formulate the concept of the angle between two curves by considering the angle between the two tangent vectors. If two curves, parametrized by ${\vec {f_{1}}}$ and ${\vec {f_{2}}}$ intersect at some point, which means that
    ${\vec {f_{1}}}(s)={\vec {f_{2}}}(t)=c$
the angle between these two curves at $c$ is the angle between the tangent vectors ${\vec {f_{1}'}}(s)$ and ${\vec {f_{2}'}}(t)$, using the dot product, is given by
    $\arccos \left({\frac {{\vec {f_{1}'}}(s)\cdot {\vec {f_{2}'}}(t)}{{\Big \|}{\vec {f_{1}'}}(s){\Big \|}{\Big \|}{\vec {f_{2}'}}(t){\Big \|}}}\right)$

### Integration

Similar to real-valued functions, the definite integral of a vector function $\mathbf {r} (t)$ is defined as:
   ${\begin{aligned}\int _{a}^{b}\mathbf {r} (t)dt&=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}\mathbf {r} (t_{i}^{*})\Delta t\\&=\lim _{n\rightarrow \infty }{\biggl{\biggl\sum _{i=1}^{n}f(t_{i}^{*})\Delta t{\biggr )}\mathbf {i} +{\biggl\sum _{i=1}^{n}g(t_{i}^{*})\Delta t{\biggr )}\mathbf {j} +{\biggl\sum _{i=1}^{n}h(t_{i}^{*})\Delta t{\biggr )}\mathbf {k} {\biggr ]}\\&={\biggl\int _{a}^{b}f(t)dt{\biggr )}\mathbf {i} +{\biggl\int _{a}^{b}g(t)dt{\biggr )}\mathbf {j} +{\biggl\int _{a}^{b}h(t)dt{\biggr )}\mathbf {k} \end{aligned}}$
We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:
    $\int _{a}^{b}\mathbf {r} (t)dt=\mathbf {R} (t){\biggr ]}_{a}^{b}=\mathbf {R} (b)-\mathbf {R} (a)$
For indefinite integrals, the definition is:
   $\int \mathbf {r} (t)dt={\biggl\int f(t)dt{\biggr )}\mathbf {i} +{\biggl\int g(t)dt{\biggr )}\mathbf {j} +{\biggl\int h(t)dt{\biggr )}\mathbf {k} +\mathbf {C}$

### Arc length

Recall that in Chapter 5.3, we deduced the length of a curve with parametric equations ${\begin{cases}x=f(t)\\y=g(t)\end{cases}}$, $a\leq t\leq b$ should be:
   $L=\int _{a}^{b}{\sqrt {{\biggl{\frac {dx}{dt}}{\biggr )}^{2}+{\biggl{\frac {dy}{dt}}{\biggr )}^{2}}}dt$
Since vector functions are fundamentally parametric equations with directions, we can utilize the formula above into the length of a space curve.
If the curve has the vector equation $\mathbf {r} (t)=\langle f(t),g(t),h(t)\rangle ,a\leq t\leq b$, or, equivalently, the parametric equations $x=f(t),y=g(t),z=h(t)$, where $f',g',h'$ are continuous, then the length of the curve from $t=a$ to $t=b$ is:
    $L=\int _{a}^{b}{\sqrt {[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}}dt=\int _{a}^{b}{\sqrt {{\biggl{\frac {dx}{dt}}{\biggr )}^{2}+{\biggl{\frac {dy}{dt}}{\biggr )}^{2}+{\biggl{\frac {dx}{dz}}{\biggr )}^{2}}}dt$
For those who prefer simplicity, the formula can be rewritten into:
   $L=\int _{a}^{b}|\mathbf {r} '(t)|dt\quad$ or $\quad {\frac {dL}{dt}}=|\mathbf {r} '(t)|$

### Reparametrization

Suppose that there is a vector function which describes the displacement of a particle with respect to time and has the equation:
   $\mathbf {r} (t)=\cos t\ \mathbf {i} +\sin t\ \mathbf {j} +t\ \mathbf {k}$
However, for some reason, we do not want to know the displacement of this particle with respect to time. Instead, we want to know its displacement with respect to its traveled distance ($s$) from $(1,0,0)$ in the direction of increasing $t$. In order to do so, we need to find a way to describe time as a function of distance. In other words, we need to find $t=t(s)$. We can use the formula for the arc length to establish the relationship between time and distance because arc length in this case describes the distance traveled by the particle.
Before we start calculating, we need to introduce the arc length function.
Suppose that a piecewise-smooth curve with a vector function $\mathbf {r} (t)=\langle f(t),g(t),h(t)\rangle ,a\leq t\leq b$, and the curve is traversed exactly once as $t$ increases from $a$ to $b$. The arc length function $s(t)$ is:
    $s(t)=\int _{a}^{t}{\sqrt {{\biggl{\frac {dx}{du}}{\biggr )}^{2}+{\biggl{\frac {dy}{du}}{\biggr )}^{2}+{\biggl{\frac {dz}{du}}{\biggr )}^{2}}}du$
According to the definition, the arc length function for our curve $\mathbf {r} (t)$ should be
   $s(t)=\int _{0}^{t}|\mathbf {r} '(u)|du=\int _{0}^{t}{\sqrt {(-\sin u)^{2}+(\cos u)^{2}+1^{2}}}du={\sqrt {2}}u{\bigg ]}_{0}^{t}={\sqrt {2}}t$
   Note that $a=0$ because the initial point $(1,0,0)$ corresponds to the parameter value $t=0$. Since it is in the direction of increasing $t$, the integration direction should be from $0$ to $t$.
   $t(s)={\frac {s}{\sqrt {2}}}$
Then we substitute the original function and get the answer:
   $\mathbf {r} (s)=\cos {\frac {s}{\sqrt {2}}}\ \mathbf {i} +\sin {\frac {s}{\sqrt {2}}}\ \mathbf {j} +{\frac {s}{\sqrt {2}}}\ \mathbf {k}$
Reparametrization has important applications in real life because sometimes we want to know a value with respect to different variables. In this case, instead of describing the path of a particle with respect to time, we described its path with respect to its distance, which will be very useful in certain situations.

## Curvature
### Terminology

Before we start discussing curvature, there are some important vectors and concepts we need to be at least aware of.

#### The unit tangent vector

In the differentiation section of this chapter, we discussed the derivatives of a vector function. We know that $\mathbf {v} (t)=\mathbf {r} '(t)$ at $t=t_{0}$ is tangent to the curve $\mathbf {r} (t)$ at $t=t_{0}$. $\mathbf {r} '(t)$ is called the tangent vector. The unit tangent vector, however, eliminates the aspect of magnitude because it is defined as:
   $\mathbf {T} (t)={\frac {\mathbf {r} '(t)}{|\mathbf {r} '(t)|}}$
As we can see, the magnitude of the unit tangent vector is always $1$. We can imagine that $\mathbf {r} (t)$ as the displacement of a particle with respect to time. So, the unit tangent vector can be perceived as the direction of the velocity of the particle with respect to time. It can also be perceived as the direction of the tangential acceleration of the particle with respect to time. We will discuss motion in space in the next section, but this is a useful method to intuitively understand some vectors.

#### The unit normal vector

The unit normal vector is defined as
   $\mathbf {N} (t)={\frac {\mathbf {T} '(t)}{|\mathbf {T} '(t)|}}$
The unit normal is orthogonal to the unit tangent because since $|\mathbf {T} (t)|=1$, we can get that:
   ${\frac {d}{dt}}|\mathbf {T} (t)|^{2}=0={\frac {d}{dt}}[\mathbf {T} (t)\cdot \mathbf {T} (t)]=2\mathbf {T} '(t)\cdot \mathbf {T} (t)$
   $\Leftrightarrow \mathbf {T} '(t)\cdot \mathbf {T} (t)=0$
This means that $\mathbf {T} '(t)$ is orthogonal to $\mathbf {T} (t)$. Therefore, $\mathbf {N} (t)$ is orthogonal to $\mathbf {T} (t)$. We can imagine that the unit normal vector is the direction of the normal acceleration of the particle with respect to time.

#### The binormal vector

The binormal vector is defined as
   $\mathbf {B} (t)=\mathbf {T} (t)\times \mathbf {N} (t)$
The binormal vector is perpendicular to both the unit tangent and the unit normal because of the properties of the cross product. The magnitude of the binormal is always 1 because
   $|\mathbf {B} (t)|=|\mathbf {T} (t)||\mathbf {N} (t)|\sin \theta =1$

#### The normal plane, osculating plane, and the osculating circle

* The normal plane is the plane determined by the normal and binormal vectors $\mathbf {N} {\text{ and }}\mathbf {B}$. The normal plane consists of all lines that are orthogonal to the tangent vector $\mathbf {T}$.
* The osculating plane is the plane determined by the unit tangent and unit normal $\mathbf {T} {\text{ and }}\mathbf {N}$. It is the plane that comes closest to containing the part of the curve near a point where $t=t_{0}$.
* The osculating circle is the circle that lies in the osculating plane towards the direction of $\mathbf {N}$ with a radius $r={\frac {1}{\kappa }}$ (the inverse of the curvature, which we will immediately discuss after this). It best describes how the curve behaves near the point where $t=t_{0}$ because it shares the same tangent, normal, and curvature at that point.
These concepts are very important in the branch of differential geometry and in its applications to the motion of spacecraft.

### Curvature

The curvature of a curve at a given point is a measure of how quickly the curve changes direction at that point. We define it to be the magnitude of the rate of change of the unit tangent with respect to arc length. We use arc length so that the curvature will be independent of parametrization.
   Suppose that a space curve has vector function $\mathbf {r}$, unit tangent vector $\mathbf {T}$, and arc length $s$. The curvature of this curve is: $\kappa ={\biggl |}{\frac {d\mathbf {T} (s)}{ds}}{\biggr |}$.
There are two other ways to express the curvature. We can express curvature in terms of $t$ instead of $s$ by utilizing the chain rule (recall that ${\frac {dL}{dt}}=|\mathbf {r} '(t)|$):
   $\kappa ={\biggl |}{\frac {d\mathbf {T} }{ds}}{\biggr |}={\Biggl |}{\frac {{\frac {d\mathbf {T} }{ds}}{\frac {ds}{dt}}}{\frac {ds}{dt}}}{\Biggr |}={\Biggl |}{\frac {\frac {d\mathbf {T} (t)}{dt}}{\frac {ds}{dt}}}{\Biggr |}={\frac {|\mathbf {T} '(t)|}{|\mathbf {r} '(t)|}}$
The third way is more complicated to deduce, but it is often more convenient to apply because it only requires $\mathbf {r} (t)$ and its derivatives.
   $\kappa (t)={\frac {|\mathbf {r} '(t)\times \mathbf {r} ''(t)|}{|\mathbf {r} '(t)|^{3}}}$
And now for the proof for this theorem:
   According to the definition of the unit tangent vector, we know that $\mathbf {r} '=|\mathbf {r} '|\mathbf {T}$. So the second derivative of $\mathbf {r}$ should be:
      ${\begin{aligned}\mathbf {r} ''&=(|\mathbf {r} '|\ \mathbf {T} )'\\&=|\mathbf {r} ''|\ \mathbf {T} +|\mathbf {r} '|\ \mathbf {T} '\quad {\text{the product rule}}\\\end{aligned}}$
   Now we calculate $\mathbf {r} '\times \mathbf {r} ''$.
      ${\begin{aligned}\mathbf {r} '\times \mathbf {r} ''&={\bigl|\mathbf {r} '|\mathbf {T} {\bigr )}\times {\bigl|\mathbf {r} ''|\ \mathbf {T} +|\mathbf {r} '|\ \mathbf {T} '{\bigr )}\quad {\text{substitution}}\\&={\bigl|\mathbf {r} '|\mathbf {T} \times |\mathbf {r} ''|\mathbf {T} {\bigr )}+{\bigl|\mathbf {r} '|\mathbf {T} \times |\mathbf {r} '|\mathbf {T} '{\bigr )}\quad {\text{distribution}}\\&=|\mathbf {r} '||\mathbf {r} ''|(\mathbf {T} \times \mathbf {T} )+|\mathbf {r} '|^{2}(\mathbf {T} \times \mathbf {T} ')\quad {\text{rearrangement}}\\&=|\mathbf {r} '|^{2}(\mathbf {T} \times \mathbf {T} ')\quad {\text{realizing that }}\mathbf {T} \times \mathbf {T} =0\\\end{aligned}}$
   And then we calculate $|\mathbf {r} '\times \mathbf {r} ''|$.
      ${\begin{aligned}|\mathbf {r} '\times \mathbf {r} ''|&=|\mathbf {r} '|^{2}|\mathbf {T} \times \mathbf {T} '|\quad {\text{substitution}}\\&=|\mathbf {r} '|^{2}|\mathbf {T} ||\mathbf {T} '|\sin \theta \quad {\text{the magnitude for the cross product}}\\&=|\mathbf {r} '|^{2}|\mathbf {T} '|\quad {\text{when you realize }}|\mathbf {T} |=1{\text{ and }}\mathbf {T} \perp \mathbf {T} '\\\end{aligned}}$
   We rearrange the equation into:
      $|\mathbf {T} '|={\frac {|\mathbf {r} '\times \mathbf {r} ''|}{|\mathbf {r} '|^{2}}}$
   Since $\kappa ={\frac {|\mathbf {T} '|}{|\mathbf {r} '|}}$, we can substitute $|\mathbf {T} '|$ with ${\frac {|\mathbf {r} '\times \mathbf {r} ''|}{|\mathbf {r} '|^{2}}}$ and get: $\kappa ={\frac {|\mathbf {r} '\times \mathbf {r} ''|}{|\mathbf {r} '|^{3}}}$
Here is a little summary on ways to calculate the curvature.
| Definition | With parametrization with respect to $\mathbf {t}$ | In terms of $\mathbf {r} (t)$ and its derivatives |
| $\kappa ={\biggl |}{\frac {d\mathbf {T} }{ds}}{\biggr |}$ | $\kappa ={\frac {|\mathbf {T} '|}{|\mathbf {r} '|}}$ | $\kappa ={\frac {|\mathbf {r} '\times \mathbf {r} ''|}{|\mathbf {r} '|^{3}}}$ |

## Motion in space
### Velocity and acceleration

Remember in 2-dimensional calculus, we mentioned that a particle with displacement function $f(x)$ has velocity $f'(x)$ and acceleration $f''(x)$. In vector functions, the definition is basically the same. Suppose a particle moves through space so that its position vector at time $t$ is $\mathbf {r} (t)$, its velocity function and acceleration function are:
   $\mathbf {v} (t)=\lim _{h\rightarrow 0}{\frac {\mathbf {r} (t+h)-\mathbf {r} (t)}{h}}=\mathbf {r} '(t)\quad$ and $\quad \mathbf {a} (t)=\lim _{h\rightarrow 0}{\frac {\mathbf {v} (t+h)-\mathbf {v} (t)}{h}}=\mathbf {v} '(t)=\mathbf {r} ''(t)$
Simply put: $\mathbf {r} ''(t)=\mathbf {v} '(t)=\mathbf {a} (t)$.
The speed of the particle ignores the direction. It is the magnitude of the velocity vector: $|\mathbf {v} (t)|$. The distance traveled by the particle from $t=a{\text{ to }}b$ is $\int _{a}^{b}|\mathbf {v} (t)|dt$, which is also the formula for the arc length.
With the help of the Fundamental Theorem of Calculus, we can deduce the velocity function and position function when we know the particle's acceleration.
   $\mathbf {v} (t)=\mathbf {v} (t_{0})+\int _{t_{0}}^{t}\mathbf {a} (u)du\quad$ and $\quad \mathbf {r} (t)=\mathbf {r} (t_{0})+\int _{t_{0}}^{t}\mathbf {v} (u)du$

### Tangential and normal acceleration

We can split the acceleration vector into two components: the tangential acceleration $a_{T}$ and the normal acceleration $a_{N}$. The tangential acceleration faces the same direction as the unit tangent vector ($\mathbf {T}$), and the normal acceleration faces the same direction as the unit normal vector ($\mathbf {N}$). Since both $\mathbf {T}$ and $\mathbf {N}$ are unit vectors, the acceleration vector can be written as the sum of two vectors:
   $\mathbf {a} =a_{T}\mathbf {T} +a_{N}\mathbf {N}$
Our goal is to figure out how to describe the two components.
   Recall that $\mathbf {T} ={\frac {\mathbf {v} }{|\mathbf {v} |}}$, thus
      $\mathbf {v} =|\mathbf {v} |\mathbf {T}$
   Now we differentiate both sides of the equation,
      ${\begin{aligned}\mathbf {v} '&={\bigl|\mathbf {v} |\mathbf {T} {\bigr )}'\\&=|\mathbf {v} |'\mathbf {T} +|\mathbf {v} |\mathbf {T} '\\\end{aligned}}$
   So we get $\mathbf {a} =|\mathbf {v} |'\mathbf {T} +|\mathbf {v} |\mathbf {T} '$.
      $\kappa ={\frac {{\big |}\mathbf {T} '{\big |}}{{\big |}\mathbf {r} '{\big |}}}={\frac {{\big |}\mathbf {T} '{\big |}}{{\big |}\mathbf {v} {\big |}}}$
      ${\big |}\mathbf {T} '{\big |}=\kappa |\mathbf {v} |$
      $\mathbf {N} ={\frac {\mathbf {T} '}{{\big |}\mathbf {T} '{\big |}}}$
      $\mathbf {T} '={\big |}\mathbf {T} '{\big |}\mathbf {N} =\kappa |\mathbf {v} |\mathbf {N}$
   We substitute $\mathbf {T} '$ for $\kappa |\mathbf {v} |\mathbf {N}$ to yield: $\mathbf {a} =|\mathbf {v} |'\mathbf {T} +\kappa |\mathbf {v} |^{2}\mathbf {N}$
That leaves us with:
   $a_{T}=|\mathbf {v} |'\quad$ and $\quad a_{N}=\kappa |\mathbf {v} |^{2}$
$\blacksquare$
Of course, it will be more convenient if those components can be written in terms if $\mathbf {r} (t)$ and its derivatives. Suppose that $\theta$ is the angle between $\mathbf {T}$ and $\mathbf {a}$, then we can write $a_{T},a_{N}$ like this:
   ${\begin{aligned}a_{T}&=|\mathbf {a} (t)|\cos \theta \\&={\frac {|\mathbf {v} (t)||\mathbf {a} (t)|\cos \theta }{|\mathbf {v} (t)|}}\quad {\text{algebraic manipulation}}\\&={\frac {\mathbf {v} (t)\cdot \mathbf {a} (t)}{|\mathbf {v} (t)|}}\quad {\text{dot product}}\\&={\frac {\mathbf {r} '(t)\cdot \mathbf {r} ''(t)}{|\mathbf {r} '(t)|}}\quad {\text{in terms of }}\mathbf {r} (t){\text{ and its derivatives}}\\\end{aligned}}$${\begin{aligned}a_{N}&=|\mathbf {a} (t)|\sin \theta \\&={\frac {|\mathbf {v} (t)||\mathbf {a} (t)|\sin \theta }{|\mathbf {v} (t)|}}\quad {\text{algebraic manipulation}}\\&={\frac {|\mathbf {v} (t)\times \mathbf {a} (t)|}{|\mathbf {v} (t)|}}\quad {\text{cross product}}\\&={\frac {|\mathbf {r} '(t)\times \mathbf {r} ''(t)|}{|\mathbf {r} '(t)|}}\quad {\text{in terms of }}\mathbf {r} (t){\text{ and its derivatives}}\\\end{aligned}}$
$\blacksquare$
To sum it up:
   $a_{T}=|\mathbf {v} |'={\frac {\mathbf {r} '(t)\cdot \mathbf {r} ''(t)}{|\mathbf {r} '(t)|}}\quad$ and $\quad a_{N}=\kappa |\mathbf {v} |^{2}={\frac {|\mathbf {r} '(t)\times \mathbf {r} ''(t)|}{|\mathbf {r} '(t)|}}$

## Expansion

We only discussed vector functions with three variables $\mathbf {r} (t)=\langle f(t),g(t),h(t)\rangle$. How about expanding our understanding of vector functions into $n$ variables? Suppose we have a curve with vector function:
   $\mathbf {r} (t)={\begin{pmatrix}f_{1}(t)\\f_{2}(t)\\\vdots \\f_{n}(t)\\\end{pmatrix}}$

### Limits

The limit of a vector function is defined as:
   $\lim _{t\rightarrow c}\mathbf {r} (t)={\begin{pmatrix}\lim _{t\rightarrow c}f_{1}(t)\\\lim _{t\rightarrow c}f_{2}(t)\\\vdots \\\lim _{t\rightarrow c}f_{n}(t)\\\end{pmatrix}}$

### Differentiation and integration

The derivative of a vector function is defined as:
   $\mathbf {r} '(t)=\lim _{\Delta t\rightarrow 0}{\frac {\mathbf {r} (t+\Delta t)-\mathbf {r} (t)}{\Delta t}}$, thus $\quad \mathbf {r} '(t)={\begin{pmatrix}f_{1}'(t)\\f_{2}'(t)\\\vdots \\f_{n}'(t)\\\end{pmatrix}}$
All differentiation rules apply.
We can expand integration into:
   $\int \mathbf {r} (t)dt={\begin{pmatrix}\int f_{1}(t)dt\\\int f_{2}(t)dt\\\vdots \\\int f_{n}(t)dt\\\end{pmatrix}}+\mathbf {C}$
Then, the arc length will become:
   $L=\int _{a}^{b}|\mathbf {r} '(t)|dt\quad$or $\quad L=\int _{a}^{b}{\sqrt {[f_{1}'(t)]^{2}+[f_{2}'(t)]^{2}+\cdots +[f_{n}'(t)]^{2}}}dt$