Motion in Space

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 $\mathbf {a} =|\mathbf {v} |'\mathbf {T} +|\mathbf {v} |\mathbf {T} '$.

Recall that $\kappa ={\frac {{\big |}\mathbf {T} '{\big |}}{{\big |}\mathbf {r} '{\big |}}}={\frac {{\big |}\mathbf {T} '{\big |}}{{\big |}\mathbf {v} {\big |}}}$, thus ${\big |}\mathbf {T} '{\big |}=\kappa |\mathbf {v} |$.

Recall that $\mathbf {N} ={\frac {\mathbf {T} '}{{\big |}\mathbf {T} '{\big |}}}$, thus $\mathbf {T} '={\big |}\mathbf {T} '{\big |}\mathbf {N} =\kappa |\mathbf {v} |\mathbf {N}$.

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 $\quad a_{N}=\kappa |\mathbf {v} |^{2}={\frac {|\mathbf {r} '(t)\times \mathbf {r} ''(t)|}{|\mathbf {r} '(t)|}}$

Resources

• Vector Functions, Wikibooks: Calculus
• Motion in Space, OpenStax