# Find the vectors T and N, and the unit binomial vector B = T \times N, for the vector-valued...

## Question:

Find the vectors {eq}T {/eq} and {eq}N {/eq}, and the unit binomial vector {eq}B = T \times N {/eq}, for the vector-valued function {eq}r(t) {/eq} and the given value of {eq}t {/eq}. {eq}r(t)= ti + t^2j + \frac{t^3}{3}k, t_0 = 1 {/eq}

## Finding Unit Vectors:

By using the derivative of the vector and its norm, we can find the needed tangent and normal unit vectors. The unit bi-normal vector is using the vector product of tangent and normal unit vectors.

Consider the vector-valued function {eq}\displaystyle r(t)= t \vec{i} + t^2 \vec{j} + \frac{t^3}{3} \vec{k} {/eq} at {eq}\displaystyle t_0 = 1 {/eq}.

Solution for the vectors {eq}\displaystyle T {/eq} and {eq}\displaystyle N {/eq}, and the unit binomial vector {eq}\displaystyle B = T \times N {/eq}:

{eq}\begin{align*} \displaystyle \text{Finding T(t)}: \\ \displaystyle \frac{dr}{dt} &=\frac{d}{dt}(t) \vec{i} + \frac{d}{dt}(t^2) \vec{j} + \frac{d}{dt}\left( \frac{t^3}{3} \right) \vec{k} \\ \displaystyle v(t) &=1 \vec{i} + 2t \vec{j} + t^2 \vec{k} \\ \displaystyle \left \| v(t) \right \| &=\sqrt{(1)^{2}+(2t)^{2}+(t^2)^{2}} \\ \displaystyle \left \| v(t) \right \| &=\sqrt{1+4t^2+t^4} \\ \displaystyle T(t) &=\frac{ v(t)}{\left \| v(t) \right \|} \\ \displaystyle T(t) &=\frac{1}{\sqrt{1+4t^2+t^4}} \vec{i} + \frac{2t}{\sqrt{1+4t^2+t^4}} \vec{j} + \frac{t^2}{\sqrt{1+4t^2+t^4}} \vec{k} \\ \displaystyle T(1) &=\frac{1}{\sqrt{1+4(1)^2+(1)^4}} \vec{i} + \frac{2(1)}{\sqrt{1+4(1)^2+(1)^4}} \vec{j} + \frac{(1)^2}{\sqrt{1+4(1)^2+(1)^4}} \vec{k} \\ \displaystyle T(1) &=\frac{1}{\sqrt{6}} \vec{i} + \frac{2}{\sqrt{6}} \vec{j} + \frac{1}{\sqrt{6}} \vec{k} \\ \\ \displaystyle \text{Finding N(t)}: \\ \displaystyle \frac{dT}{dt} &=\frac{d}{dt}\left( \frac{1}{\sqrt{1+4t^2+t^4}} \right) \vec{i} + \frac{d}{dt}\left( \frac{2t}{\sqrt{1+4t^2+t^4}} \right) \vec{j} + \frac{d}{dt}\left( \frac{t^2}{\sqrt{1+4t^2+t^4}} \right) \vec{k} \\ \displaystyle {T}'(t) &=-\frac{2t\left(t^2+2\right)}{\left(1+4t^2+t^4\right)^{\frac{3}{2}}} \vec{i} + \frac{2\left(-t^4+1\right)}{\left(1+4t^2+t^4\right)^{\frac{3}{2}}} \vec{j} + \frac{4t^3+2t}{\left(1+4t^2+t^4\right)^{\frac{3}{2}}} \vec{k} \\ \displaystyle {T}'(1) &=-\frac{2(1)\left((1)^2+2\right)}{\left(1+4(1)^2+(1)^4\right)^{\frac{3}{2}}} \vec{i} + \frac{2\left(-(1)^4+1\right)}{\left(1+4(1)^2+(1)^4\right)^{\frac{3}{2}}} \vec{j} + \frac{4(1)^3+2(1)}{\left(1+4(1)^2+(1)^4\right)^{\frac{3}{2}}} \vec{k} \\ \displaystyle {T}'(1) &=-\frac{1}{\sqrt{6}} \vec{i} + 0 \vec{j} + \frac{1}{\sqrt{6}} \vec{k} \\ \displaystyle \left \| {T}'(1) \right \| &=\sqrt{\left(-\frac{1}{\sqrt{6}}\right)^2+\left(\frac{1}{\sqrt{6}}\right)^2} \\ \displaystyle \left \| {T}'(1) \right \| &=\sqrt{\frac{1}{3}} \\ \displaystyle N(t) &=\frac{{T}'(t)}{ \left \| {T}'(t) \right \|} \\ \displaystyle N(1) &=\frac{-\frac{1}{\sqrt{6}}}{\sqrt{\frac{1}{3}}} \vec{i} + 0 \vec{j} + \frac{\frac{1}{\sqrt{6}}}{\sqrt{\frac{1}{3}}} \vec{k} \\ \displaystyle N(1) &=-\frac{1}{\sqrt{2}} \vec{i} + 0 \vec{j} + \frac{1}{\sqrt{2}} \vec{k} \\ \\ \displaystyle \text{Finding B(t)}: \\ \displaystyle B(t) &= T(t) \times N(t) \\ \displaystyle &=\begin{vmatrix} i & j & k\\ \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}}\\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{vmatrix} \\ \displaystyle &=\left( \left( \frac{1}{\sqrt{2}} \right)\left( \frac{2}{\sqrt{6}} \right)-\left( 0 \right)\left( \frac{1}{\sqrt{6}} \right) \right) \vec{i}-\left( \left( \frac{1}{\sqrt{2}} \right)\left( \frac{1}{\sqrt{6}} \right)-\left( -\frac{1}{\sqrt{2}} \right)\left( \frac{1}{\sqrt{6}} \right) \right)\vec{j}+\left( \left( 0 \right)\left( \frac{1}{\sqrt{6}} \right)-\left( -\frac{1}{\sqrt{2}} \right)\left( \frac{2}{\sqrt{6}} \right) \right)\vec{k} \\ \displaystyle B(1) &=\frac{1}{\sqrt{3}} \vec{i} -\frac{1}{\sqrt{3}} \vec{j}+ \frac{1}{\sqrt{3}} \vec{k} \end{align*} {/eq}

The unit tangent vector is {eq}\ \displaystyle \mathbf{\color{blue}{ T(1)=\frac{1}{\sqrt{6}} \vec{i} + \frac{2}{\sqrt{6}} \vec{j} + \frac{1}{\sqrt{6}} \vec{k} }} {/eq}.

The unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ N(1) =-\frac{1}{\sqrt{2}} \vec{i} + \frac{1}{\sqrt{2}} \vec{k} }} {/eq}.

The bi-normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ B(1)=\frac{1}{\sqrt{3}} \vec{i} -\frac{1}{\sqrt{3}} \vec{j}+ \frac{1}{\sqrt{3}} \vec{k} }} {/eq}.