# Given r(t) = \left \langle 3t^2, \sin t - t\cos t, \cos t + t\sin t \right \rangle, (a) Find the...

## Question:

Given {eq}r(t) = \left \langle 3t^2, \sin t - t\cos t, \cos t + t\sin t \right \rangle, {/eq}

(a) Find the unit tangent and unit normal vectors T(t) and N(t). t > 0.

b) Find the curvature.

## Unit Tangent Vector, Unit Normal Vector And Curvature:

For finding the given unit vectors, mainly we have to use the derivative of the vector and their magnitude. Curvature is evaluated by using the norm of unit tangent vector and given vector derivatives.

We take the vector function {eq}\displaystyle r(t) = \left \langle 3t^2, \sin t - t\cos t, \cos t + t\sin t \right \rangle {/eq} to find some given conditions.

(a) Finding the unit tangent and unit normal vectors T(t) and N(t):

{eq}\begin{align*} \displaystyle \text{Unit tangent vector}: \\ \displaystyle \frac{dr}{dt} &=\frac{d}{dt} \left \langle 3t^2, \sin t - t\cos t, \cos t + t\sin t \right \rangle \\ \displaystyle {r}'(t) &=\left \langle 6t, t\sin \left(t\right), t\cos \left(t\right) \right \rangle \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{(6t)^{2}+(t\sin \left(t\right))^{2}+(t\cos \left(t\right))^{2}} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{37}t \\ \displaystyle \vec{T}(t) &=\frac{{\vec{r}}'(t)}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle \vec{T}(t) &=\left \langle \frac{6t}{\sqrt{37}t}, \frac{t\sin \left(t\right)}{\sqrt{37}t}, \frac{t\cos \left(t\right)}{\sqrt{37}t} \right \rangle \\ \displaystyle \vec{T}(t) &=\left \langle \frac{6}{\sqrt{37}}, \frac{\sin \left(t\right)}{\sqrt{37}}, \frac{\cos \left(t\right)}{\sqrt{37}} \right \rangle \\ \\ \displaystyle \text{Unit normal vector}: \\ \displaystyle \frac{d\vec{T}}{dt} &=\frac{d}{dt}\left \langle \frac{6}{\sqrt{37}}, \frac{\sin \left(t\right)}{\sqrt{37}}, \frac{\cos \left(t\right)}{\sqrt{37}} \right \rangle \\ \displaystyle \vec{T}'(t) &=\left \langle 0, \frac{1}{\sqrt{37}}\cos \left(t\right), -\frac{1}{\sqrt{37}}\sin \left(t\right) \right \rangle \\ \displaystyle \left \| \vec{T}'(t) \right \| &=\sqrt{\left( \frac{1}{\sqrt{37}}\cos \left(t\right) \right)^{2}+\left( -\frac{1}{\sqrt{37}}\sin \left(t\right) \right)^{2}} \\ \displaystyle \left \| \vec{T}'(t) \right \| &=\sqrt{\frac{1}{37}} \\ \displaystyle \vec{N}(t) &=\frac{{\vec{T}}'(t)}{ \left \| {\vec{T}}'(t) \right \|} \\ \displaystyle \vec{N}(t) &=\left \langle 0, \frac{\frac{1}{\sqrt{37}}\cos \left(t\right)}{\sqrt{\frac{1}{37}}}, \frac{-\frac{1}{\sqrt{37}}\sin \left(t\right)}{\sqrt{\frac{1}{37}}} \right \rangle \\ \displaystyle \vec{N}(t) &=\left \langle 0, \cos \left(t\right), -\sin \left(t\right) \right \rangle \end{align*} {/eq}

Unit tangent vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{T}(t)=\left \langle \frac{6}{\sqrt{37}}, \frac{\sin \left(t\right)}{\sqrt{37}}, \frac{\cos \left(t\right)}{\sqrt{37}} \right \rangle }} {/eq} and unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{N}(t) =\left \langle 0, \cos \left(t\right), -\sin \left(t\right) \right \rangle }} {/eq}.

b) Finding the curvature:

{eq}\begin{align*} \displaystyle \kappa &=\frac{ \left \| {\vec{T}}'(t) \right \|}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle \kappa &=\frac{\sqrt{\frac{1}{37}}}{\sqrt{37}t} \\ \displaystyle \kappa &=\frac{1}{37t} \end{align*} {/eq}

Curvature is {eq}\ \displaystyle \mathbf{\color{blue}{ \kappa=\frac{1}{37t} }} {/eq}.