# For the position function: r(t) = t cos (t) i + t sin(t) j + t^2 k , t_0 = 0. Find the four...

## Question:

For the position function: {eq}r(t) = t cos (t) i + t sin(t) j + t^2 k , t_0 = 0. {/eq} Find the four acceleration components {eq}T(t_0), a T(t_0), a N(t_0), \ and \ N(t_0) {/eq}

## Finding the Acceleration Components:

The acceleration components of the tangential and normal vectors, then the unit tangent vector and the unit normal vectors are evaluated by using the velocity and acceleration vectors. The parameter value {eq}\displaystyle t {/eq} should be considered as {eq}\displaystyle t_0 {/eq}.

We consider the given position function {eq}\displaystyle r(t) = t \cos (t) \vec{i} + t \sin(t) \vec{j} + t^2 \vec{k} , {/eq} at {eq}\displaystyle t_0 = 0 {/eq}.

Finding {eq}\displaystyle T(t_0) {/eq}:

{eq}\begin{align*} \displaystyle T(t) &=\frac{ v(t)}{\left \| v(t) \right \|} \\ \displaystyle v(t) &= (\cos \left(t\right)-t\sin \left(t\right)) \vec{i} + (\sin \left(t\right)+t\cos \left(t\right)) \vec{j} + (2t) \vec{k} \\ \displaystyle \left \| v(t) \right \| &=\sqrt{1+5t^2} \\ \displaystyle T(t) &= \frac{\cos \left(t\right)-t\sin \left(t\right)}{\sqrt{1+5t^2}} \vec{i} + \frac{\sin \left(t\right)+t\cos \left(t\right)}{\sqrt{1+5t^2}} \vec{j} + \frac{2t}{\sqrt{1+5t^2}} \vec{k} \\ \displaystyle T(0) &= \frac{\cos \left(0\right)-(0)\sin \left(0\right)}{\sqrt{1+5(0)^2}} \vec{i} + \frac{\sin \left(0\right)+(0)\cos \left(0\right)}{\sqrt{1+5(0)^2}} \vec{j} + \frac{2(0)}{\sqrt{1+5(0)^2}} \vec{k} \\ \displaystyle T(0) &= 1 \vec{i} + 0 \vec{j} + 0 \vec{k} \end{align*} {/eq}

Answer for the unit tangent vector is {eq}\ \displaystyle \mathbf{\color{blue}{ T(0) = \vec{i} }} {/eq}.

Finding {eq}\displaystyle N(t_0) {/eq}:

{eq}\begin{align*} \displaystyle N(t) &=\frac{{T}'(t)}{ \left \| {T}'(t) \right \|} \\ \displaystyle {T}'(t) &= \left( \frac{-5t^3\cos \left(t\right)-5t^2\sin \left(t\right)-6t\cos \left(t\right)-2\sin \left(t\right)}{\left(1+5t^2\right)^{\frac{3}{2}}} \right) \vec{i} + \left( \frac{-5t^3\sin \left(t\right)+5t^2\cos \left(t\right)-6t\sin \left(t\right)+2\cos \left(t\right)}{\left(1+5t^2\right)^{\frac{3}{2}}} \right) \vec{j} + \left( \frac{2}{\left(1+5t^2\right)^{\frac{3}{2}}} \right) \vec{k} \\ \displaystyle {T}'(0) &= \left( \frac{-5(0)^3\cos \left(0\right)-5(0)^2\sin \left(0\right)-6(0)\cos \left(0\right)-2\sin \left(0\right)}{\left(1+5(0)^2\right)^{\frac{3}{2}}} \right) \vec{i} + \left( \frac{-5(0)^3\sin \left(0\right)+5(0)^2\cos \left(0\right)-6(0)\sin \left(0\right)+2\cos \left(0\right)}{\left(1+5(0)^2\right)^{\frac{3}{2}}} \right) \vec{j} + \left( \frac{2}{\left(1+5(0)^2\right)^{\frac{3}{2}}} \right) \vec{k} \\ \displaystyle {T}'(0) &= 0 \vec{i} + 2 \vec{j} + 2 \vec{k} \\ \displaystyle \left \| {T}'(0) \right \| &=\sqrt{\left(2\right)^2+\left(2\right)^2} \\ \displaystyle \left \| {T}'(0) \right \| &=2\sqrt{2} \\ \displaystyle N(0) &= 0 \vec{i} + \frac{2}{2\sqrt{2}} \vec{j} + \frac{2}{2\sqrt{2}} \vec{k} \\ \displaystyle N(0) &= 0 \vec{i} + \frac{1}{\sqrt{2}} \vec{j} + \frac{1}{\sqrt{2}} \vec{k} \end{align*} {/eq}

Answer for the unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ N(0) = \frac{1}{\sqrt{2}} \vec{j} + \frac{1}{\sqrt{2}} \vec{k} }} {/eq}.

Finding {eq}\displaystyle a_{T}(t_0){/eq}:

{eq}\begin{align*} \displaystyle a_{T} &=\frac{ v(t) \cdot a(t)}{\left \| v(t) \right \|} \\ \displaystyle v(t) &= (\cos \left(t\right)-t\sin \left(t\right)) \vec{i} + (\sin \left(t\right)+t\cos \left(t\right)) \vec{j} + (2t) \vec{k} \\ \displaystyle v(0) &= (\cos \left(0\right)-(0)\sin \left(0\right)) \vec{i} + (\sin \left(0\right)+(0)\cos \left(0\right)) \vec{j} + (2(0)) \vec{k} \\ \displaystyle v(0) &= 1 \vec{i} + 0 \vec{j} + 0 \vec{k} \\ \displaystyle a(t) &= (-2\sin \left(t\right)-t\cos \left(t\right)) \vec{i} + (2\cos \left(t\right)-t\sin \left(t\right)) \vec{j} + 2 \vec{k} \\ \displaystyle a(0) &= (-2\sin \left(0\right)-(0)\cos \left(0\right)) \vec{i} + (2\cos \left(0\right)-(0)\sin \left(0\right)) \vec{j} + 2 \vec{k} \\ \displaystyle a(0) &= 0 \vec{i} + 2 \vec{j} + 2 \vec{k} \\ \displaystyle \left \| v(0) \right \| &=1 \\ \displaystyle v(0) \cdot a(0) &=\left \langle 1, 0, 0 \right \rangle \cdot \left \langle 0, 2, 2 \right \rangle \\ \displaystyle &=(1)(0)+(0)(2)+(0)(2) \\ \displaystyle v(0) \cdot a(0) &=0 \\ \displaystyle a_{T}(0) &=\frac{0}{1} \\ \displaystyle a_{T}(0) &=0 \end{align*} {/eq}

Answer for the tangential component of acceleration vector is {eq}\ \displaystyle \mathbf{\color{blue}{ a_{T}(0) =0 }} {/eq}.

Finding {eq}\displaystyle a_{N}(t_0){/eq}:

{eq}\begin{align*} \displaystyle a_{N} &=\frac{\left \| v(t) \times a(t) \right \|}{\left \| v(t) \right \|} \\ \displaystyle v(0) \times a(0) &=\begin{vmatrix} i & j & k\\ 1 & 0 & 0\\ 0 & 2 & 2 \end{vmatrix} \\ \displaystyle &=(0-0)\vec{i}-(2-0)\vec{j}+(2-0)\vec{k} \\ \displaystyle v(0) \times a(0) &=0\vec{i}-2\vec{j}+2\vec{k} \\ \displaystyle \left \| v(0) \times a(0) \right \| &=\sqrt{(-2)^{2}+(2)^{2}} \\ \displaystyle \left \| v(0) \times a(0) \right \| &=2\sqrt{2} \\ \displaystyle a_{N}(0) &=\frac{2\sqrt{2}}{1} \\ \displaystyle a_{N}(0) &=2\sqrt{2} \end{align*} {/eq}

Answer for the normal component of acceleration vector is {eq}\ \displaystyle \mathbf{\color{blue}{ a_{N}(0) =2\sqrt{2} }} {/eq}.