# Unit Normal Vector Calculator

The calculator will find the principal unit normal vector of the vector-valued function at the given point, with steps shown.

Enter a vector-valued function:

mathbf{vec{r}(t)}= (, , )
If you don't have the third coordinate, set it to 0.
Calculate at t=
Leave empty, if you don't need the unit normal vector at a specific point.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

## Solution

Your input: find the principal unit normal vector for $\mathbf{\vec{r}(t)}=\left(\sin{\left(t \right)}, \cos{\left(t \right)}, 2 \sqrt{2} t\right)$

To find the unit normal vector, we need to find the derivative of the unit tangent vector $\mathbf{\vec{T}(t)}$ and then normalize it.

So, first of all, find the unit tangent vector: $\mathbf{\vec{T}(t)}=\left(\frac{\cos{\left(t \right)}}{3}, - \frac{\sin{\left(t \right)}}{3}, \frac{2 \sqrt{2}}{3}\right)$ (for steps, see unit tangent vector calculator).

Next, find the derivative.

$\mathbf{\vec{T}^{\prime}(t)}=\left(- \frac{\sin{\left(t \right)}}{3}, - \frac{\cos{\left(t \right)}}{3}, 0\right)$

Note. For steps in finding derivatives, see derivative calculator.

Find the norm (length) of the vector: $\lVert\mathbf{\vec{T}^{\prime}(t)}\rVert=\sqrt{\left(- \frac{\sin{\left(t \right)}}{3}\right)^2+\left(- \frac{\cos{\left(t \right)}}{3}\right)^2+\left(0\right)^2}=\frac{1}{3}$

Finally, the unit normal vector is $\mathbf{\vec{N}(t)}=\frac{\mathbf{\vec{T}^{\prime}(t)}}{\lVert\mathbf{\vec{T}^{\prime}(t)}\rVert}$

$\mathbf{\vec{N}(t)}=\frac{\left(- \frac{\sin{\left(t \right)}}{3}, - \frac{\cos{\left(t \right)}}{3}, 0\right)}{\frac{1}{3}}=\left(- \sin{\left(t \right)}, - \cos{\left(t \right)}, 0\right)$

Answer: $\mathbf{\vec{N}(t)}=\left(- \sin{\left(t \right)}, - \cos{\left(t \right)}, 0\right)$