# 8. Properties of Curves

## 1. Definitions

When dealing with vectors in $$\mathbb{R}^{3}$$, we use the standard basis vectors $$\hat{\imath}$$, $$\hat{\jmath}$$ and $$\hat{k}$$. These are three mutually perpendicular unit vectors which form a right handed system.

When dealing with vectors along a curve, it is nice to have three basis vectors (i.e. three mutually perpendicular unit vectors which form a right handed system) which are adapted to the curve. These are called $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$, and are collectively called a Frenet frame. To understand the definitions of the Frenet frame, first notice that the velocity, $$\vec{v}$$ determines the instantaneous direction of the curve while $$\vec{v}$$ and $$\vec{a}$$ determine the instantaneous plane of the curve.

The unit tangent vector $$\hat{T}$$ is the unit vector tangent to the curve, i.e. the direction of $$\vec{v}$$.
The unit normal vector $$\hat{N}$$ is the unit vector perpendicular to $$\vec{v}$$ in the plane of $$\vec{v}$$ and $$\vec{a}$$ on the same side of $$\vec{v}$$ as $$\vec{a}$$, i.e. $$\hat{N}$$ is the unit vector in the direction of $$\text{proj}_{\bot \vec{v}}\vec{a}$$.
The unit binormal vector $$\hat{B}$$ is the unit vector perpendicular to $$\hat{T}$$ and $$\hat{N}$$ related by the right hand rule, i.e. $\hat{B}=\hat{T}\times\hat{N}$

The figures at the right show:
a curve in blue,
the unit tangent vector, $$\hat{T}$$, in red,
the unit normal vector, $$\hat{N}$$, in cyan and
the unit binormal vector, $$\hat{B}$$ in magenta.

In the first figure, use the mouse to rotate the plot until $$\hat{N}$$ is pointing straight at you. Notice that the curve instantaneously lies in the plane of $$\hat{T}$$ and $$\hat{N}$$.

Now rotate the plot until $$\hat{B}$$ is pointing straight at you. Notice that the curve instantaneously bends in the direction of $$\hat{N}$$.

The second figure shows how $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ change as the point moves along the curve. Notice, $$\hat{T}$$ is in the direction of the curve, $$\hat{N}$$ is in the direction it is turning and $$\hat{B}$$ is perpendicular to the plane of the motion.

We first need to justify that $$\hat{B}$$ defined as $$\hat{T}\times\hat{N}$$ is in fact a unit vector. However, since $$\hat{T}$$ and $$\hat{N}$$ are perpendicular unit vectors $|\hat{T}|=1 \qquad |\hat{N}|=1 \qquad \text{and} \qquad \theta=90^\circ$ Hence $|\hat{B}| =|\hat{T}\times\hat{N}| =|\hat{T}|\,|\hat{N}|\sin\theta =1\cdot1\cdot1=1$

Recall that $$\hat{\imath}$$, $$\hat{\jmath}$$ and $$\hat{k}$$ form a right handed triplet of mutually perpendicular unit vectors in that they satisfy: $\hat{\imath}\times\hat{\jmath}=\hat{k} \qquad \hat{\jmath}\times\hat{k}=\hat{\imath} \qquad \hat{k}\times\hat{\imath}=\hat{\jmath}$ where we cyclically rotate $$\hat{\imath}$$, $$\hat{\jmath}$$ and $$\hat{k}$$ in these formulas. Similarly:

$$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ form a right handed triplet of mutually perpendicular unit vectors in that they satisfy: $\hat{T}\times\hat{N}=\hat{B} \qquad \hat{N}\times\hat{B}=\hat{T} \qquad \hat{B}\times\hat{T}=\hat{N}$ where we cyclically rotate $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ in these formulas.

To see this, use your right hand to see that $$\hat{T}\times\hat{N}$$ points along $$\hat{B}$$. Then use the mouse to rotate the figure and use your right hand to see where $$\hat{N}\times\hat{B}$$ and $$\hat{B}\times\hat{T}$$ point.

Find the unit tangent vector, $$\hat{T}$$, the unit normal vector $$\hat{N}$$, and the unit binormal vector $$\hat{B}$$ of the circle of radius $$R$$ in the $$xy$$-plane, parametrized as $$\vec{r}(\theta)=(R\cos\theta,R\sin\theta,0)$$ for general $$\theta$$ and at $$\theta=\dfrac{\pi}{4}$$.

First we find the velocity and its length: \begin{aligned} \vec{v} &=\langle -R\sin\theta,R\cos\theta,0\rangle \\ |\vec{v}| &=\sqrt{(-R\sin\theta)^2+(R\cos\theta)^2+0^2}=R \end{aligned} From these, we calculate the unit tangent vector: \begin{aligned} \hat{T} &=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{1}{R}\langle -R\sin\theta,R\cos\theta,0\rangle \\ &=\langle -\sin\theta,\cos\theta,0\rangle \end{aligned} Next, we compute the acceleration and notice that it is perpendicular to the velocity: \begin{aligned} \vec{a} &=\langle -R\cos\theta,-R\sin\theta,0\rangle \\ \vec{v}\cdot\vec{a} &=R^2\sin\theta\cos\theta-R^2\cos\theta\sin\theta=0 \end{aligned} So the unit normal is the unit vector in the direction of the acceleration. We compute the length of the acceleration and the unit normal vector: \begin{aligned} |\hat{a}| &=\sqrt{(-R\cos\theta)^2+(-R\sin\theta)^2+0^2}=R \\ \hat{N} &=\dfrac{\vec{a}}{|\vec{a}|} =\dfrac{1}{R}\langle -R\cos\theta,-R\sin\theta,0\rangle \\ &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \end{aligned} Finally, the unit binormal vector is: \begin{aligned} \hat{B} &=\hat{T}\times\hat{N} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -\sin\theta & \cos\theta & 0 \\ -\cos\theta & -\sin\theta & 0 \end{vmatrix} \\ &=\hat\imath(0)-\hat\jmath(0)+\hat k(\sin^2\theta+\cos^2\theta) =\langle 0,0,1\rangle \end{aligned}

The figure shows $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ moving around the circle. In particular, for $$\theta=\dfrac{\pi}{4}$$: \begin{aligned} \hat{T} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0\right\rangle \\ \hat{N} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}},0\right\rangle \\ \hat{B} &=\langle 0,0,1\rangle \end{aligned}

We check the three vectors are perpendicular by computing their dot products: \begin{aligned} \hat{T}\cdot\hat{N}&=\sin\theta\cos\theta-\cos\theta\sin\theta=0 \\ \hat{T}\cdot\hat{B}&=0+0+0=0 \\ \hat{B}\cdot\hat{N}&=0+0+0=0 \end{aligned}

Find the unit tangent vector, $$\hat{T}$$, the unit normal vector $$\hat{N}$$, and the unit binormal vector $$\hat{B}$$ of the helix parametrized as $$\vec{r}(\theta)=(4\cos\theta,4\sin\theta,3\theta)$$ for general $$\theta$$ and at $$\theta=\dfrac{\pi}{4}$$

Like the circle, the acceleration is again perpendicular to the velocity which makes it easy to calculate the unit normal vector.

Generally: \begin{aligned} \hat{T} &=\left\langle \dfrac{-4}{5}\sin\theta,\dfrac{4}{5}\cos\theta,\dfrac{3}{5}\right\rangle \\ \hat{N} &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \hat{B} &=\left\langle \dfrac{3}{5}\sin\theta,\dfrac{-3}{5}\cos\theta,\dfrac{4}{5}\right\rangle \end{aligned} For $$\theta=\dfrac{\pi}{4}$$: \begin{aligned} \hat{T} &=\left\langle \dfrac{-2\sqrt{2}}{5},\dfrac{2\sqrt{2}}{5},\dfrac{3}{5}\right\rangle \\ \hat{N} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}},0\right\rangle \\ \hat{B} &=\left\langle \dfrac{3}{5\sqrt{2}},\dfrac{-3}{5\sqrt{2}},\dfrac{4}{5}\right\rangle \end{aligned}

First we find the velocity and its length: \begin{aligned} \vec{v} &=\langle -4\sin\theta,4\cos\theta,3\rangle \\ |\vec{v}| &=\sqrt{16\sin^2\theta+16\cos^2\theta+9}=5 \end{aligned} From these, we calculate the unit tangent vector: $\hat{T} =\dfrac{\vec{v}}{|\vec{v}|} =\left\langle \dfrac{-4}{5}\sin\theta,\dfrac{4}{5}\cos\theta,\dfrac{3}{5}\right\rangle$ Next, we compute the acceleration and check that it is perpendicular to the velocity: \begin{aligned} \vec{a} &=\langle -4\cos\theta,-4\sin\theta,0\rangle \\ \vec{v}\cdot\vec{a} &=16\sin\theta\cos\theta-16\cos\theta\sin\theta=0 \end{aligned} So the unit normal is the unit vector in the direction of the acceleration. We compute the length of the acceleration and the unit normal vector: \begin{aligned} |\hat{a}| &=\sqrt{(-4\cos\theta)^2+(-4\sin\theta)^2+0^2}=4 \\ \hat{N} &=\dfrac{\vec{a}}{|\vec{a}|} =\langle -\cos\theta,-\sin\theta,0\rangle \\ \end{aligned} Finally, the unit binormal vector is: \begin{aligned} \hat{B} &=\hat{T}\times\hat{N} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\[3 pt] \dfrac{-4}{5}\sin\theta & \dfrac{4}{5}\cos\theta & \dfrac{3}{5} \\[3 pt] -\cos\theta & -\sin\theta & 0 \end{vmatrix} \\ &=\hat\imath\left(\dfrac{3}{5}\sin\theta\right) -\hat\jmath\left(\dfrac{3}{5}\cos\theta\right) +\hat k\left(\dfrac{4}{5}\sin^2\theta+\dfrac{4}{5}\cos^2\theta\right) \\ &=\left\langle\dfrac{3}{5}\sin\theta,\dfrac{-3}{5}\cos\theta,\dfrac{4}{5}\right\rangle \end{aligned}

The figure shows $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ moving around the helix. In particular, for $$\theta=\dfrac{\pi}{4}$$: \begin{aligned} \hat{T} &=\left\langle \dfrac{-2\sqrt{2}}{5},\dfrac{2\sqrt{2}}{5},\dfrac{3}{5}\right\rangle \\ \hat{N} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}},0\right\rangle \\ \hat{B} &=\left\langle \dfrac{3}{5\sqrt{2}},\dfrac{-3}{5\sqrt{2}},\dfrac{4}{5}\right\rangle \end{aligned} It is easy to check these are $$3$$ mutually perpendicular unit vectors.

We check the three vectors \begin{aligned} \hat{T} &=\left\langle \dfrac{-4}{5}\sin\theta,\dfrac{4}{5}\cos\theta,\dfrac{3}{5}\right\rangle \\ \hat{N} &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \hat{B} &=\left\langle\dfrac{3}{5}\sin\theta,\dfrac{-3}{5}\cos\theta,\dfrac{4}{5}\right\rangle \end{aligned} are perpendicular by computing their dot products: \begin{aligned} \hat{T}\cdot\hat{N}&=\dfrac{4}{5}\sin\theta\cos\theta-\dfrac{4}{5}\cos\theta\sin\theta=0 \\ \hat{T}\cdot\hat{B}&=-\dfrac{12}{25}\sin^2\theta-\dfrac{12}{25}\cos^2\theta+\dfrac{12}{25}=0 \\ \hat{B}\cdot\hat{N}&=-\dfrac{3}{5}\sin\theta\cos\theta+\dfrac{3}{5}\cos\theta\sin\theta=0 \end{aligned}

In the examples of a circle and a helix, the acceleration is perpendicular to the velocity. So it is easy to find the normal vector. On the next page, we will find formulas which help us compute $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ even if the acceleration is not perpendicular to the velocity.

Supported in part by NSF Grant #1123255