# Find the vectors T, N, and B, at the given point.

• $R(t) = < t^{2}, \frac{2}{3} t^{3} , t > \text {and point} < 4, \frac{-16}{3}, -2 >$

This question aims to determine the tangent vector, normal vector, and the binormal vector of any given vector. The tangent vector $T$ is a vector that is tangent to the given surface or vector at any particular point. The normal vector $N$ is a vector that is normal or perpendicular to a surface at any given point. And finally, the binormal vector $B$ is the vector obtained by calculating the cross-product of the unit tangent vector and the unit normal vector.

The 3 kinds of said vectors can easily be calculated for any given vector by simply calculating its derivative and applying some standard formulas. These standard formulas are stated in the solution of the question.

## Expert Solution

In the question, the vector whose $T$ and $N$ need to be determined is mentioned below:

$R(t) = < t^{2}, \frac{2}{3} t^{3}, t >$

The point specified in the question is point $< 4, \frac{-16}{3}, -2 >$

By comparing the vector $R(t)$ with the point, it becomes evident that this point exists at $t = -2$. This value of t can be counterchecked by inserting it in the vector $R(t)$. Upon inserting the value of t in the given vector $R(t)$:

$< (-2)^{2}, \frac{2}{3} (-2)^{3}, -2 >$

$< 4, \frac{-16}{3}, -2 >$

Hence, it is proved that the point exists at $t$ = $-2$.

The formula for determining the tangent vector $T$ is:

$T = \frac{R'(t)}{|R'(t)|}$

So the next thing to do is to calculate the derivative of the vector $R(t)$.

Calculating the derivative of the vector $R(t)$:

$R’(t) = \frac{d}{dt} < t^{2}, \frac{2}{3}t^{3}, t>$

$R’(t) = < 2t, 2t^{2}, 1 >$

Now, for the distance of the derivative:

$|R’(t)| = \sqrt{(2t)^{2} + (2t^{2})^{2}+ 1^{2}}$

$|R’(t)| = \sqrt{4t^{2} + 4t^{4} + 1}$

$|R’(t)| = \sqrt{(2t^{2} + 1)^{2}}$

$|R’(t)| = 2t^{2} + 1$

The formula for determining the tangent vector $T$ is:

$T = \frac{R’(t)}{|R’(t)|}$

Inserting values into this formula gives us the tangent vector $T$:

$T = \frac{1}{2t^{2} + 1} . < 2t, 2t^{2}, 1 >$

$T = < \frac{2t}{2t^{2} + 1}, \frac{2t^{2}}{2t^{2} + 1}, \frac{1}{2t^{2} + 1} >$

Tangent vector $T$ at $t = -2$:

$T = < \frac{-4}{9}, \frac{8}{9}, \frac{1}{9} >$

Now, let’s determine the normal vector $N$. The formula for determining the vector $N$ is:

$N = \frac{T’(t)}{|T’(t)|}$

The next thing to do is to calculate the derivative of the tangent vector $T$:

$T’(t) = \frac{d}{dt} < \frac{2t}{2t^{2} + 1}, \frac{2t^{2}}{2t^{2} + 1}, \frac{1}{2t^{2} + 1} >$

$T’(t) = < \frac{(2t^{2} + 1) \times(2) – (2t) \times (4t)}{(2t^{2} + 1)^{2}}, \frac{(2t^{2} + 1) \times(4t) – (2t^{2}) \times(4t)}{(2t^{2} + 1)^{2}}, \frac{(2t^{2} + 1) \times(0) – (1) \times(4t)}{ (2t^{2} + 1)^{2}} >$

$T’(t) = \frac{1}{(2t^{2} + 1)^{2}} < 4t^{2} + 2 -8t^{2}, 8t^{3} + 4t – 8t^{3}, -4t >$

$T’(t) = \frac{1}{(2t^{2} + 1)^{2}} < 2 – 4t^{2}, 4t, -4t >$

$T’(t) = < \frac{2 – 4t^{2}}{(2t^{2} + 1)^{2}}, \frac{4t}{(2t^{2} + 1)^{2}}, \frac{-4t}{(2t^{2} + 1)^{2}} >$

Now, for the distance of the tangent vector $T$ derivative:

$|T’(t)| = \frac{1}{(2t^{2} + 1)^{2}} \sqrt{(2 – 4t^{2})^{2} + (4t)^{2} + (-4t)^{2}}$

$|T’(t)| = \frac{1}{(2t^{2} + 1)^{2}} \sqrt{4 – 16t^{2} + 16t^{4} + 16t^{2} + 16t^{2}}$

$|T’(t)| = \frac{1}{(2t^{2} + 1)^{2}} \sqrt{4 +16t^{2} + 16t^{4}}$

$|T’(t)| = \frac{1}{(2t^{2} + 1)^{2}} \sqrt{(2 + 4t^{2})^{2}}$

$|T’(t)| = \frac{2 + 4t^{2}}{(2t^{2} + 1)^{2}}$

$|T’(t)| = \frac {2( 2t^{2} + 1)}{(2t^{2} + 1)^{2}}$

$|T’(t)| = \frac {2}{2t^{2} + 1}$

The formula for determining the normal vector $N$ is:

$N = \frac{T’(t)}{|T’(t)|}$

Inserting the values:

$N = \frac{< 2 – 4t^{2}, 4t, -4t >}{(2t^{2} + 1)^{2}} \times \frac{(2t^{2} + 1)}{2}$

$N = \frac{< 2 – 4t^{2}, 4t, -4t >}{2t^{2} + 1} \times \frac{1}{2}$

$N = \frac{2 < 1 – 2t^{2}, 2t, -2t >}{2t^{2} + 1} \times \frac{1}{2}$

$N = < \frac{1 – 2t^{2}}{2t^{2} + 1}, \frac{2t}{2t^{2} + 1}, \frac{-2t}{2t^{2} + 1} >$

Normal vector $N$ at $t = -2$:

$N = < \frac{-7}{9}, \frac{-4}{9}, \frac{4}{9} >$

## Example

Find the vector $B$ for the above question.

The binormal vector $B$ refers to the cross-product of vectors $T$ and $N$.

$B(-2) = T(-2) x N(-2)$

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

$B = (\frac{32}{81} + \frac{4}{81})i – (\frac{-16}{81} + \frac{7}{81})j + (\frac{16}{81} + \frac{56}{81})k$

$B = < \frac{36}{81}, \frac{9}{81}, \frac{72}{81} >$

$B = < \frac{4}{9}, \frac{1}{9}, \frac{8}{9} >$