Unit Tangent Vector
How To Find 'Em w/ Detailed Examples!

Did you know that there are three special vectors that play a vital role in understanding the motion of an object along a space curve?

These three critical vectors are the:

1. Unit Tangent
2. Unit Normal
3. Binormal vectors

or the TNB vectors.

Let’s look at each of these vectors more closely.

Unit Tangent Vector

If we let \(\mathrm{C}\) be a smooth curve with position vector \(\vec{r}(t)\), then the Unit Tangent Vector, denoted \(\vec{T}(t)\), is defined to be \(\vec{T}(t)=\frac{\vec{r}^{\prime}(t)}{\left\|\vec{r}^{\prime}(t)\right\|}\) and represents the unit vector in the direction of the velocity vector.

Unit Normal Vector

Now the Unit Normal Vector, sometimes called the principal normal vector, denoted \(\vec{N}(t)\), is defined as \(\vec{N}(t)=\frac{\vec{T}^{\prime}(t)}{\left\|\vec{T}^{\prime}(t)\right\|}\), and is orthogonal (i.e., perpendicular) to the unit tangent vector and the curve and points in the direction where the curve is bending.

Binormal Vector

And finally, if we \(\mathrm{C}\) be a smooth curve with position vector \(\vec{r}(t)\), then the Binormal Vector, denoted \(\vec{B}(t)\), is defined to be \(\vec{B}(t)=\vec{T}(t) \times \vec{N}(t)\), representing a unit vector that is perpendicular to both \(\vec{T}(t)\) and \(\vec{N}(t)\), and registers the tilt of the plane determined by \(\vec{T}(t)\) and \(\vec{N}(t)\).

Curvature

And here’s another super cool thing to know about these unit vectors, they are all related to curvature!

Let’s see how.

Since \(\vec{T}(t)\) is a constant length then \(\frac{d}{d s} \vec{T}\) is orthogonal to \(\vec{T}(t)\) and points in the direction in which \(\vec{T}(t)\) is turning.

Therefore, the length of \(\frac{d}{d s} \vec{T}(t)\) is the curvature \(\kappa\) and can be written as:

\begin{equation}
\frac{d}{d s} \vec{T}=\kappa \vec{N} \text { where } \vec{N}(t) \text { is the unit normal vector }
\end{equation}

And since we know that the binormal vector is perpendicular to the unit tangent vector and the unit normal vector, then we can concluded that \(\frac{d}{d s} \vec{B} \perp \vec{T}\) and \(\frac{d}{d s} \vec{B} \perp \vec{N}\) and \(\frac{d}{d s} \vec{B}=-\tau \vec{N}\) where the number \(\tau(s)\) is the measure of the degree of twisting of a curve called the torsion of the curve.

Frenet Serret Formulas

And knowing the fact that \(\vec{B}(t)=\vec{T}(t) \times \vec{N}(t)\) this leads us to three rather vital formulas in differential geometry called the Frenet-Serret formulas.

\begin{equation}
\begin{aligned}
&\frac{d}{d s} \vec{T}=\kappa \vec{N} \\
&\frac{d}{d s} \vec{N}=-\kappa \vec{T}+\tau \vec{B} \\
&\frac{d}{d s} \vec{B}=-\tau \vec{N}
\end{aligned}
\end{equation}

Example – Unit Tangent Vector Of A Helix

Alright, so now that we know what the TNB vectors are, let’s look at an example of how to find them.

Suppose we are given the circular helix \(\vec{r}(t)=\langle t, \cos t, \sin t\rangle\).

First, we need to find the unit tangent for our vector-valued function by calculating \(\vec{r}^{\prime}(t)\) and \(\left\|\vec{r}^{\prime}(t)\right\|\).

\begin{equation}
\begin{aligned}
&\vec{r}^{\prime}(t)=\langle 1,-\sin t, \cos t\rangle \\
&\left\|\vec{r}^{\prime}(t)\right\|=\sqrt{1^{2}+(-\sin t)^{2}+(\cos t)^{2}}=\sqrt{1+\underbrace{\sin ^{2} t+\cos ^{2} t}_{1}}=\sqrt{1+1}=\sqrt{2} \\
&\vec{T}(t)=\frac{\vec{r}^{\prime}(t)}{\left\|\vec{r}^{\prime}(t)\right\|}=\frac{\langle 1,-\sin t, \cos t\rangle}{\sqrt{2}} \\
&\vec{T}(t)=\left\langle\frac{1}{\sqrt{2}}, \frac{-\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}\right\rangle
\end{aligned}
\end{equation}

Next, we will find the principal normal vector by computing \(\vec{T}^{\prime}(t)\) and its magnitude \(\left\|\vec{T}^{\prime}(t)\right\|\).

\begin{equation}
\begin{aligned}
\vec{T}^{\prime}(t) &=\left\langle 0, \frac{-1}{\sqrt{2}} \cos t, \frac{-1}{\sqrt{2}} \sin t\right\rangle \\
\left\|\vec{T}^{\prime}(t)\right\| &=\sqrt{0^{2}+\left(\frac{-1}{\sqrt{2}} \cos t\right)^{2}+\left(\frac{-1}{\sqrt{2}} \sin t\right)^{2}} \\
&=\sqrt{\frac{\cos ^{2} t}{2}+\frac{\sin ^{2} t}{2}}=\sqrt{\frac{1}{2} \underbrace{\left(\cos ^{2} t+\sin ^{2} t\right)}_{1}} \\
&=\sqrt{\frac{1}{2}(1)}=\frac{1}{\sqrt{2}} \\
\vec{N}(t) &=\frac{\vec{T}^{\prime}(t)}{|| \vec{T}^{\prime}(t) \mid}=\frac{\left\langle 0, \frac{-1}{\sqrt{2}} \cos t, \frac{-1}{\sqrt{2}} \sin t\right\rangle}{\frac{1}{\sqrt{2}}} \\
\vec{N}(t) &=\langle 0,-\cos t,-\sin t\rangle
\end{aligned}
\end{equation}

Lastly, we will find the binormal vector by calculating the cross-product of the unit tangent and unit normal vectors.

\begin{equation}
\begin{aligned}
&\vec{B}(t)=\vec{T}(t) \times \vec{N}(t) \\
&\vec{B}(t)=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
\frac{1}{\sqrt{2}} & \frac{-\sin t}{\sqrt{2}} & \frac{\cos t}{\sqrt{2}} \\
0 & -\cos t & -\sin t
\end{array}\right|=\vec{i}\left(\frac{\sin ^{2} t}{\sqrt{2}}+\frac{\cos ^{2} t}{\sqrt{2}}\right)-\vec{j}\left(\frac{-\sin t}{\sqrt{2}}\right)+\vec{k}\left(\frac{-\cos t}{\sqrt{2}}\right) \\
&\vec{B}(t)=\left\langle\frac{1}{\sqrt{2}}, \frac{\sin t}{\sqrt{2}}, \frac{-\cos t}{\sqrt{2}}\right\rangle
\end{aligned}
\end{equation}

Not bad, right?

So, together we will look at how these three important vectors are represented geometrically along a space curve. And we will learn how to calculate the unit tangent, unit normal, and binormal vectors for various position vectors and how to represent these vectors as a given point.

It’s going to be fun, so let’s get to it!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

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