Method for Calculating Unit Normal and Unit Binormal Vectors

Method for Calculating Unit Normal and Unit Binormal Vectors

Recall from the Unit Normal and Unit Binormal Vectors to a Space Curve page that the unit normal vector denoted $\hat{N}(t)$ is a vector that is perpendicular to $\hat{T}(t)$ and points in the direction to which the curve $C$ bends towards.

Furthermore, the unit binormal vector $\hat{B}(t)$ is a vector that is perpendicular to both $\hat{T}(t)$ and $\hat{N}(t)$. The following formulas provide a method for calculating the unit normal and unit binormal vectors:

• Unit Normal Vector: $\hat{N}(t) = \frac{\hat{T'}(t)}{\| \hat{T'}(t) \|}$.
• Unit Binormal Vector: $\hat{B}(t) = \hat{T}(t) \times \hat{N}(t)$.

Often times it is difficult to calculate $\hat{N}(t)$ since $\hat{T}(t)$ often has an annoying square root in the denominator to deal with, and so differentiating $\hat{T}(t)$ to get $\hat{T'}(t)$ is cumbersome. Consequentially, if we do the work to get $\hat{N}(t)$ and the result is rather messy, then calculating $\hat{B}(t) = \hat{T}(t) \times \hat{N}(t)$ also proves to be rather annoying.

Fortunate, we can greatly reduce the amount of work we need to do with the following theorem that gives us a method for calculating $\hat{B}(t)$ more easily.

 Theorem 1: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$ for $t \in [a, b]$. Then the unit binormal vector can be calculated with the formula $\hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|}$.
• We will note that $\vec{r'}(t) \times \vec{r''}(t) = \| \vec{r'}(t) \| ^3 \kappa (t) \hat{B}(t)$. Notice that $\| \vec{r'}(t) \| ^3 > 0$ and $\kappa (t) > 0$, and so $\hat{B}(t)$ points in a scalar multiple of $\vec{r'}(t) \times \vec{r''}(t)$ that points in the same direction as $\vec{r'}(t) \times \vec{r''}(t)$. Therefore we have that:
(1)
\begin{align} \hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|} \quad \blacksquare \end{align}
 Remark: We have a relatively easy method for calculating both $\hat{T}(t)$ and $\hat{B}(t)$. From the properties of the cross product of vectors, we can then calculate $\hat{N}(t)$ with the formula $\hat{N}(t) = \hat{B}(t) \times \hat{T}(t)$.

We will now look at some examples of calculating some unit normal and unit binormal vectors.

Example 1

Find the unit normal and unit binormal vectors for the vector-valued function $\vec{r}(t) = \left (t, \frac{t^2}{2}, \frac{t^3}{3} \right )$.

We will find $\hat{B}(t)$ first. We have that $\vec{r'}(t) = (1, t, t^2)$ and $\vec{r''}(t) = (0, 1, 2t)$. Computing $\vec{r'}(t) \times \vec{r''}(t)$ and we have that:

(2)
\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 1 & t & t^2 \\ 0 & 1 & 2t \end{vmatrix} = (2t^2 - t^2) \vec{i} - (2t) \vec{j} + (1) \vec{k} = (t^2, -2t, 1) \end{align}

Now we calculate $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{t^4 + 4t^2 + 1}$, and so $\hat{B}(t) = \frac{1}{\sqrt{t^4 + 4t^2 + 1}}(t^2, -2t, 1) = \left ( \frac{t^2}{\sqrt{t^4 + 4t^2 + 1}}, \frac{-2t}{\sqrt{t^4 + 4t^2 + 1}}, \frac{1}{\sqrt{t^4 + 4t^2 + 1}} \right )$.

Now to calculate $\hat{N}(t)$, we will evaluate $\hat{B}(t) \times \hat{T}(t)$. First we need to calculate $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. We have that $\| \vec{r'}(t) \| = \sqrt{t^4 + t^2 + 1}$. Therefore $\hat{T}(t) = \frac{1}{\sqrt{t^4 + t^2 + 1}} (1, t, t^2) = \left (\frac{1}{\sqrt{t^4 + t^2 + 1}}, \frac{t}{\sqrt{t^4 + t^2 + 1}}, \frac{t^2}{\sqrt{t^4 + t^2 + 1}}\right )$. Now:

(3)
\begin{align} \hat{N}(t) = \hat{B}(t) \times \hat{T}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{t^2}{\sqrt{t^4 + 4t^2 + 1}} & \frac{-2t}{\sqrt{t^4 + 4t^2 + 1}} & \frac{1}{\sqrt{t^4 + 4t^2 + 1}} \\ \frac{1}{\sqrt{t^4 + t^2 + 1}} & \frac{t}{\sqrt{t^4 + t^2 + 1}} & \frac{t^2}{\sqrt{t^4 + t^2 + 1}} \end{vmatrix} \\ = \left [\left ( \frac{-2t}{\sqrt{t^4 + 4t^2 + 1}} \cdot \frac{t^2}{\sqrt{t^4 + t^2 + 1}}\right ) - \left ( \frac{1}{\sqrt{t^4 + 4t^2 + 1}} \cdot \frac{t}{\sqrt{t^4 + t^2 + 1}} \right ) \right ] \vec{i} \\ - \left [ \left ( \frac{t^2}{\sqrt{t^4 + 4t^2 + 1}} \cdot \frac{t^2}{\sqrt{t^4 + t^2 + 1}} \right ) - \left ( \frac{1}{\sqrt{t^4 + 4t^2 + 1}} \cdot \frac{1}{\sqrt{t^4 + t^2 + 1}}\right )\right ] \vec{j} \\ + \left [ \left ( \frac{t^2}{\sqrt{t^4 + 4t^2 + 1}} \cdot \frac{t}{\sqrt{t^4 + t^2 + 1}}\right ) - \left ( \frac{-2t}{\sqrt{t^4 + 4t^2 + 1}} \cdot \frac{1}{\sqrt{t^4 + t^2 + 1}} \right ) \right ] \vec{k} \\ = \frac{1}{\sqrt{t^4 + 4t^2 + 1} \cdot \sqrt{t^4 + t^2 + 1}} \left (-2t^3 - t, 1 - t^4, t^3 + 2t \right ) \end{align}