r(t) = \left \langle 2cos(4t), 15t, 2sin(4t) \right \rangle . a) find the curvature function...

Question:

{eq}r(t) = \left \langle 2cos(4t), 15t, 2sin(4t) \right \rangle {/eq}.

a) find the curvature function {eq}k(t){/eq}.

b) for {eq}t=\frac{\pi }{4}{/eq} write the acceleration vectors tangential coefeciant {eq}a_T{/eq} and normal coefficient {eq}a_N{/eq}

Curvature and Components of Acceleration Vectors:

Given vector function is used to find the curvature, component of acceleration vectors. Curvature and normal component of acceleration vectors are similar. The only difference is the cube of the norm of derivative is used in the curvature formula.

Answer and Explanation:

Let us consider the given function {eq}\displaystyle r(t) = \left \langle 2 \cos (4t), 15t, 2 \sin (4t) \right \rangle {/eq}.

a) Finding the curvature function {eq}\displaystyle k(t){/eq}:

{eq}\begin{align*} \displaystyle {r}'(t) &= \left \langle -8\sin \left(4t\right), 15, 8\cos \left(4t\right) \right \rangle \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{(-8\sin \left(4t\right))^{2}+(15)^{2}+(8\cos \left(4t\right))^{2}} \\ \displaystyle \left \| {r}'(t) \right \| &=17 \\ \displaystyle {r}''(t) &= \left \langle -32\cos \left(4t\right), 0, -32\sin \left(4t\right) \right \rangle \\ \displaystyle {r}'(t) \times {r}''(t) &=\begin{vmatrix} i & j & k\\ -8\sin \left(4t\right) & 15 & 8\cos \left(4t\right)\\ -32\cos \left(4t\right) & 0 & -32\sin \left(4t\right) \end{vmatrix} \\ \displaystyle &=((-32\sin \left(4t\right))(15)-(0)(8\cos \left(4t\right))) \vec{i}-((-32\sin \left(4t\right))(-8\sin \left(4t\right))-(-32\cos \left(4t\right))(8\cos \left(4t\right)))\vec{j}+((0)(-8\sin \left(4t\right))-(-32\cos \left(4t\right))(15)) \vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=-480\sin \left(4t\right) \vec{i} -256 \vec{j}+ 480\cos \left(4t\right) \vec{k} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{(-480\sin \left(4t\right))^{2}+(-256)^{2}+(480\cos \left(4t\right))^{2}} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=544 \\ \displaystyle \kappa &= \frac{ \left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{ \left \| {\vec{r}}'(t) \right \|^{3}} \\ \displaystyle \kappa &= \frac{544}{(17)^{3}} \\ \displaystyle \kappa &= \frac{32}{289} \end{align*} {/eq}

The curvature function is {eq}\ \displaystyle \mathbf{\color{blue}{ \kappa = \frac{32}{289} }} {/eq}.


b) Finding the tangential component of acceleration vector {eq}\displaystyle a_T{/eq} and normal component of acceleration vector {eq}\displaystyle a_N{/eq} for {eq}\displaystyle t=\frac{\pi }{4}{/eq} :

{eq}\begin{align*} \displaystyle \text{Tangential Component}: \\ \displaystyle {r}'(t) &= \left \langle -8\sin \left(4t\right), 15, 8\cos \left(4t\right) \right \rangle \\ \displaystyle {r}'\left( \frac{\pi }{4} \right) &= \left \langle -8\sin \left(4\left( \frac{\pi }{4} \right)\right), 15, 8\cos \left(4\left( \frac{\pi }{4} \right)\right) \right \rangle \\ \displaystyle {r}'\left( \frac{\pi }{4} \right) &= \left \langle 0, 15, -8 \right \rangle \\ \displaystyle \left \| {r}'\left( \frac{\pi }{4} \right) \right \| &=17 \\ \displaystyle {r}''(t) &= \left \langle -32\cos \left(4t\right), 0, -32\sin \left(4t\right) \right \rangle \\ \displaystyle {r}''\left( \frac{\pi }{4} \right) &= \left \langle -32\cos \left(4\left( \frac{\pi }{4} \right)\right), 0, -32\sin \left(4\left( \frac{\pi }{4} \right)\right) \right \rangle \\ \displaystyle {r}''\left( \frac{\pi }{4} \right) &= \left \langle 32, 0, 0 \right \rangle \\ \displaystyle {r}'\left( \frac{\pi }{4} \right) \cdot {r}''\left( \frac{\pi }{4} \right) &= \left \langle 0, 15, -8 \right \rangle \cdot \left \langle 32, 0, 0 \right \rangle \\ \displaystyle {r}'\left( \frac{\pi }{4} \right) \cdot {r}''\left( \frac{\pi }{4} \right) &=(0)(32)+(15)(0)+(-8)(0)=0 \\ \displaystyle a_{T} &=\frac{{\vec{r}}'(t)\cdot {\vec{r}}''(t)}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle a_{T}\left( \frac{\pi }{4} \right) &=\frac{0}{17} \\ \displaystyle a_{T}\left( \frac{\pi }{4} \right) &=0 \\ \\ \displaystyle \text{Normal Component}: \\ \displaystyle a_{N} &=\frac{\left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle a_{N}\left( \frac{\pi }{4} \right) &=\frac{544}{17} \\ \displaystyle a_{N}\left( \frac{\pi }{4} \right) &=32 \end{align*} {/eq}

The tangential component of acceleration vector is {eq}\ \displaystyle \mathbf{\color{blue}{ a_{T}\left( \frac{\pi }{4} \right) =0 }} {/eq} and normal component of acceleration vector is {eq}\ \displaystyle \mathbf{\color{blue}{ a_{N}\left( \frac{\pi }{4} \right) =32 }} {/eq}.


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