Let \vec{r}(t)= \left \langle 4\cos (2t)e^t, 4e^t\sin (2t), 0 \right \rangle, find \left \|...


Let {eq}\vec{r}(t)= \left \langle 4\cos (2t)e^t, 4e^t\sin (2t), 0 \right \rangle {/eq}, find {eq}\left \| \vec{a}_T \right \|\ and\ \left \| \vec{a}_N,\ \right \|, {/eq} the components of the acceleration vector for the curve {eq}\vec{r}(t) {/eq} at {eq}t= \frac{2}{3}\pi. {/eq}

Finding the Components of Acceleration Vectors:

The components of acceleration vectors are classified into tangential and normal.

The scalar product and the vector product are used to find these components respectively.

Answer and Explanation:

Consider the given vector function {eq}\displaystyle \vec{r}(t)= \left \langle 4\cos (2t)e^t, 4e^t\sin (2t), 0 \right \rangle {/eq} at {eq}\displaystyle t= \frac{2}{3}\pi {/eq}.

Finding the tangent and acceleration vectors at the parameter value:

{eq}\begin{align*} \displaystyle \frac{d\vec{r}}{dt} &=\frac{d}{dt} \left \langle 4\cos (2t)e^t, 4e^t\sin (2t), 0 \right \rangle \\ \displaystyle {\vec{r}}'(t) &=\left \langle 4\left(-2e^t\sin \left(2t\right)+e^t\cos \left(2t\right)\right), 4\left(e^t\sin \left(2t\right)+2e^t\cos \left(2t\right)\right), 0 \right \rangle \\ \displaystyle {\vec{r}}'\left( \frac{2}{3}\pi \right) &=\left \langle 4\left(-2e^{\frac{2}{3}\pi}\sin \left(2\left( \frac{2}{3}\pi \right)\right)+e^{\frac{2}{3}\pi}\cos \left(2\left( \frac{2}{3}\pi \right)\right)\right), 4\left(e^{\frac{2}{3}\pi}\sin \left(2\left( \frac{2}{3}\pi \right)\right)+2e^{\frac{2}{3}\pi}\cos \left(2\left( \frac{2}{3}\pi \right)\right)\right), 0 \right \rangle \\ \displaystyle {\vec{r}}'\left( \frac{2}{3}\pi \right) &=\left \langle 4\sqrt{3}e^{\frac{2\pi }{3}}-2e^{\frac{2\pi }{3}}, -2\sqrt{3}e^{\frac{2\pi }{3}}-4e^{\frac{2\pi }{3}}, 0 \right \rangle \\ \displaystyle \frac{d{\vec{r}}'}{dt} &=\frac{d}{dt}\left \langle 4\left(-2e^t\sin \left(2t\right)+e^t\cos \left(2t\right)\right), 4\left(e^t\sin \left(2t\right)+2e^t\cos \left(2t\right)\right), 0 \right \rangle \\ \displaystyle {\vec{r}}''(t) &=\left \langle 4\left(-3e^t\cos \left(2t\right)-4e^t\sin \left(2t\right)\right), 4\left(-3e^t\sin \left(2t\right)+4e^t\cos \left(2t\right)\right), 0 \right \rangle \\ \displaystyle {\vec{r}}''\left( \frac{2}{3}\pi \right) &=\left \langle 4\left(-3e^{\frac{2}{3}\pi}\cos \left(2\left( \frac{2}{3}\pi \right)\right)-4e^{\frac{2}{3}\pi}\sin \left(2\left( \frac{2}{3}\pi \right)\right)\right), 4\left(-3e^{\frac{2}{3}\pi}\sin \left(2\left( \frac{2}{3}\pi \right)\right)+4e^{\frac{2}{3}\pi}\cos \left(2\left( \frac{2}{3}\pi \right)\right)\right), 0 \right \rangle \\ \displaystyle {\vec{r}}''\left( \frac{2}{3}\pi \right) &=\left \langle 6e^{\frac{2\pi }{3}}+8\sqrt{3}e^{\frac{2\pi }{3}}, 4\left(\frac{3^{\frac{3}{2}}e^{\frac{2\pi }{3}}}{2}-2e^{\frac{2\pi }{3}}\right), 0 \right \rangle \end{align*} {/eq}

Finding the components of the acceleration vectors {eq}\displaystyle \left \| \vec{a}_T \right \|\ and\ \left \| \vec{a}_N,\ \right \| {/eq}:

{eq}\begin{align*} \displaystyle \text{Finding} \ \ \left \| \vec{a}_T(t) \right \|\ : \\ \displaystyle {\vec{r}}'\left( \frac{2}{3}\pi \right) \cdot {\vec{r}}''\left( \frac{2}{3}\pi \right) &=\left \langle 4\sqrt{3}e^{\frac{2\pi }{3}}-2e^{\frac{2\pi }{3}}, -2\sqrt{3}e^{\frac{2\pi }{3}}-4e^{\frac{2\pi }{3}}, 0 \right \rangle \cdot \left \langle 6e^{\frac{2\pi }{3}}+8\sqrt{3}e^{\frac{2\pi }{3}}, 4\left(\frac{3^{\frac{3}{2}}e^{\frac{2\pi }{3}}}{2}-2e^{\frac{2\pi }{3}}\right), 0 \right \rangle \\ \displaystyle &=\left( 4\sqrt{3}e^{\frac{2\pi }{3}}-2e^{\frac{2\pi }{3}} \right)\left( 6e^{\frac{2\pi }{3}}+8\sqrt{3}e^{\frac{2\pi }{3}} \right)+\left( -2\sqrt{3}e^{\frac{2\pi }{3}}-4e^{\frac{2\pi }{3}} \right)\left( 4\left(\frac{3^{\frac{3}{2}}e^{\frac{2\pi }{3}}}{2}-2e^{\frac{2\pi }{3}}\right) \right)+(0)(0) \\ \displaystyle {\vec{r}}'\left( \frac{2}{3}\pi \right) \cdot {\vec{r}}''\left( \frac{2}{3}\pi \right) &=80e^{\frac{4\pi }{3}} \\ \displaystyle \left \| {\vec{r}}'\left( \frac{2}{3}\pi \right) \right \| &=\sqrt{\left(4\sqrt{3}e^{\frac{2\pi }{3}}-2e^{\frac{2\pi }{3}}\right)^2+\left(-2\sqrt{3}e^{\frac{2\pi }{3}}-4e^{\frac{2\pi }{3}}\right)^2} \\ \displaystyle \left \| {\vec{r}}'\left( \frac{2}{3}\pi \right) \right \| &=4\sqrt{5}e^{\frac{2\pi }{3}} \\ \displaystyle a_{T}(t) &=\frac{{\vec{r}}'(t)\cdot {\vec{r}}''(t)}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle a_{T}\left( \frac{2}{3}\pi \right) &=\frac{80e^{\frac{4\pi }{3}}}{4\sqrt{5}e^{\frac{2\pi }{3}}} \\ \displaystyle \left \| a_{T}\left( \frac{2}{3}\pi \right) \right \| &=4\sqrt{5}e^{\frac{2\pi }{3}} \\ \\ \displaystyle \text{Finding} \ \ \left \| \vec{a}_N (t)\ \right \| : \\ \displaystyle {\vec{r}}'\left( \frac{2}{3}\pi \right) \times {\vec{r}}''\left( \frac{2}{3}\pi \right) &=\begin{vmatrix} i & j & k\\ 4\sqrt{3}e^{\frac{2\pi }{3}}-2e^{\frac{2\pi }{3}} & -2\sqrt{3}e^{\frac{2\pi }{3}}-4e^{\frac{2\pi }{3}} & 0\\ 6e^{\frac{2\pi }{3}}+8\sqrt{3}e^{\frac{2\pi }{3}} & 4\left(\frac{3^{\frac{3}{2}}e^{\frac{2\pi }{3}}}{2}-2e^{\frac{2\pi }{3}}\right) & 0 \end{vmatrix} \\ \displaystyle &=(0-0) \vec{i}-(0-0)\vec{j}+\left( \left( 4\left(\frac{3^{\frac{3}{2}}e^{\frac{2\pi }{3}}}{2}-2e^{\frac{2\pi }{3}}\right) \right)\left( 4\sqrt{3}e^{\frac{2\pi }{3}}-2e^{\frac{2\pi }{3}} \right)-\left( 6e^{\frac{2\pi }{3}}+8\sqrt{3}e^{\frac{2\pi }{3}} \right)\left( -2\sqrt{3}e^{\frac{2\pi }{3}}-4e^{\frac{2\pi }{3}} \right) \right)\vec{k} \\ \displaystyle {\vec{r}}'\left( \frac{2}{3}\pi \right) \times {\vec{r}}''\left( \frac{2}{3}\pi \right) &=0 \vec{i}+0\vec{j}+160e^{\frac{4\pi }{3}}\vec{k} \\ \displaystyle \left \| {\vec{r}}'\left( \frac{2}{3}\pi \right) \times {\vec{r}}''\left( \frac{2}{3}\pi \right) \right \| &=160e^{\frac{4\pi }{3}} \\ \displaystyle a_{N}(t) &=\frac{\left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle a_{N} \left( \frac{2}{3}\pi \right) &=\frac{160e^{\frac{4\pi }{3}}}{4\sqrt{5}e^{\frac{2\pi }{3}}} \\ \displaystyle \left \| a_{N} \left( \frac{2}{3}\pi \right) \right \| &=8\sqrt{5}e^{\frac{2\pi }{3}} \end{align*} {/eq}

The component of the acceleration vector {eq}\displaystyle \left \| \vec{a}_T \right \|\ {/eq} is {eq}\ \displaystyle \mathbf{\color{blue}{ 4\sqrt{5}e^{\frac{2\pi }{3}} }} {/eq} and the component of the acceleration vector {eq}\displaystyle \left \| \vec{a}_N \ \right \| {/eq} is {eq}\ \displaystyle \mathbf{\color{blue}{ 8\sqrt{5}e^{\frac{2\pi }{3}} }} {/eq}.

Learn more about this topic:

Vector Components: The Magnitude of a Vector

from UExcel Physics: Study Guide & Test Prep

Chapter 2 / Lesson 8

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