# Find the vectors T, N, and B at the given point. r(t)=< t^2,2/3 t^3,t > and point < 4,-16/3,-2 >.

This question aims to find the Tangent, Normal, and Binormal vectors by using the given point and a function.

Consider a vector function, $\vec{r}(t)$. If $\vec{r}'(t)\neq 0$ and $\vec{r}'(t)$ exist then $\vec{r}'(t)$ is called a tangent vector. The line that passes through the point $P$ and is parallel to the tangent vector, $\vec{r}'(t)$, is the line tangent to $\vec{r}(t)$ at $P$. It is worth noting that we do need $\vec{r}'(t)\neq 0$ to have a tangent vector. If $\vec{r}'(t)=0$, then it will be a vector with no magnitude and hence it will be impossible to know the direction of the tangent.

Furthermore, if $\vec{r}'(t)\neq0$, the unit tangent vector to the curve is given by:

$\vec{T}(t)=\dfrac{\vec{r}'(t)}{||\vec{r}'(t)||}$

The unit normal is orthogonal/ perpendicular to the unit tangent vector and, by extension, to the curve.

Mathematically:

$\vec{N}(t)=\dfrac{\vec{T}'(t)}{||\vec{T}'(t)||}$

The binormal vector is defined as the cross-product of the unit tangent and unit normal vectors and is hence orthogonal to both the tangent and normal vectors.

Mathematically:

$\vec{B}(t)=\vec{T}(t)\times \vec{N}(t)$

Given $\vec{r}(t)=\left\langle t^2,\dfrac{2}{3}t^3,t\right\rangle$ and point $\left\langle 4,-\dfrac{16}{3},-2\right\rangle$.

Since $\left\langle 4,-\dfrac{16}{3},-2\right\rangle$ occurs at $t=-2$, so to find the tangent we compute:

$\vec{r}'(t)=\langle 2t, 2t^2,1\rangle$

$||\vec{r}'(t)||=\sqrt{(2t)^2+(2t^2)^2+(1)^2}$

$=\sqrt{4t^2+4t^4+1}$

$=\sqrt{(2t^2+1)^2}$

$=2t^2+1$

The Tangent vector is given as:

$\vec{T}(t)=\dfrac{\vec{r}'(t)}{||\vec{r}'(t)||}$

$=\dfrac{1}{2t^2+1}\langle 2t, 2t^2,1\rangle$

At $t=-2$:

$\vec{T}(-2)=\dfrac{1}{2(-2)^2+1}\langle 2(-2), 2(-2)^2,1\rangle$

$\vec{T}(-2)=\left\langle -\dfrac{4}{3}, \dfrac{8}{9},\dfrac{1}{9}\right\rangle$

Now, for the Normal vector:

$\vec{T}'(t)=\left\langle \dfrac{(2t^2+1)2-2t(4t)}{(2t^2+1)^2},\dfrac{(2t^2+1)4t-(2t^2)(4t)}{(2t^2+1)^2},-\dfrac{4t}{(2t^2+1)^2}\right\rangle$

$=\left\langle \dfrac{4t^2+2-8t^2}{(2t^2+1)^2},\dfrac{8t^3+4t-8t^3}{(2t^2+1)^2},-\dfrac{4t}{(2t^2+1)^2}\right\rangle$

$=\left\langle \dfrac{2-4t^2}{(2t^2+1)^2},\dfrac{4t}{(2t^2+1)^2},-\dfrac{4t}{(2t^2+1)^2}\right\rangle$

$||\vec{T}'(t)||=\sqrt{\dfrac{(2-4t^2)^2+(4t)^2+(-4t)^2}{(2t^2+1)^4}}$

$=\dfrac{\sqrt{4-16t^2+16t^4+16t^2+16t^2}}{(2t^2+1)^2}$

$=\dfrac{\sqrt{16t^4+16t^2+4}}{(2t^2+1)^2}$

$=\dfrac{2(2t^2+1)}{(2t^2+1)^2}$

$=\dfrac{2}{(2t^2+1)}$

The Normal vector is:

$\vec{N}(t)=\dfrac{\vec{T}'(t)}{||\vec{T}'(t)||}$

$=\dfrac{(2t^2+1)}{2}\cdot\dfrac{1}{(2t^2+1)^2}\langle 2-4t^2, 4t , -4t\rangle$

$=\dfrac{1}{(2t^2+1)}\langle 1-2t^2, 2t , -2t\rangle$

At $t=-2$:

$\vec{N}(-2)=\dfrac{1}{(2(-2)^2+1)}\langle 1-2(-2)^2, 2(-2) , -2(-2)\rangle$

$=\left\langle -\dfrac{7}{9}, -\dfrac{4}{9},\dfrac{4}{9}\right\rangle$

And the Binormal vector at $t=-2$ is:

$\vec{B}(-2)=\vec{T}(-2)\times \vec{N}(-2)$

$=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-\dfrac{4}{9}& \dfrac{8}{9} & \dfrac{1}{9}\\ -\dfrac{7}{9}& -\dfrac{4}{9}& \dfrac{4}{9}\end{vmatrix}$

$=\left(\dfrac{32}{81}+\dfrac{4}{81}\right)\hat{i}-\left(-\dfrac{16}{81}+\dfrac{7}{81}\right)\hat{j}+\left(\dfrac{16}{81}+\dfrac{56}{81}\right)\hat{k}$

$=\left\langle \dfrac{4}{9}, \dfrac{1}{9},\dfrac{8}{9}\right\rangle$

## Example

Given $\vec{r}(t)=\langle 1, -\cos t,\sin t\rangle$, find the normal and binormal vectors.

### Solution

To find the normal and binormal vectors, we first need to work out the tangent vector.

For this:

$\vec{r}'(t)=\langle 0, \sin t ,\cos t\rangle$

$||\vec{r}'(t)||=\sqrt{(0)^2+(\sin t)^2+(\cos t)^2}$

$=\sqrt{0+\sin^2t+\cos^2t}$

$=\sqrt{1}$

$=1$

The unit tangent vector is:

$\vec{T}(t)=\langle 0, \sin t ,\cos t\rangle$

Now, for the normal vector, we need the derivative and magnitude of the tangent vector as follows:

$\vec{T}'(t)=\langle 0, \cos t ,-\sin t\rangle$

$||\vec{T}'(t)||=\sqrt{ (0)^2+(\cos t)^2 +(-\sin t)^2}$

$=\sqrt{0+\cos^2t+\sin^2t}$

$=\sqrt{1}$

$=1$

So,

$\vec{N}(t)=\langle 0, \cos t ,-\sin t\rangle$

And the binormal vector can be calculated as:

$\vec{B}(t)=\vec{T}(t)\times \vec{N}(t)$

$=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0& \sin t &\cos t\\ 0& \cos t &-\sin t\end{vmatrix}$

$=(-\sin^2t-\cos^2t)\vec{i}-(0)\vec{j}+(0)\vec{k}$

$=-\vec{i}$