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.

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