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i hate this class

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Mars Ultor
2025-10-04 16:31:27 -05:00
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@@ -544,5 +544,59 @@ This corresponds with answer choice B.
This corresponds with answer choice D.
}
\qs{}{
\begin{align*}
\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 1, \frac{\cancel{\sec{t}}\tan{t}}{\cancel{\sec{t}}},0 \rangle \\
\Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &=\sqrt{1+\tan^2{t}}=\sec{t} \\
\vec{T}(t) &= \langle \cos{t},\sin{t} ,0 \rangle \\
\frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &= \langle -\sin{t},\cos{t},0\rangle \\
\Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= 1 \\
\vec{N}(t) &= \langle -\sin{t},\cos{t},0\rangle \\
\vec{B}(t) &= \vec{T}(t) \times \vec{N}(t) = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\cos{t} & \sin{t} & 0 \\
-\sin{t} & \cos{t} & 0
\end{vmatrix} \\
&= \hat{k}\left[\cos^2{t}+\sin^2{t}\right ] \\
&= \hat{k}
\end{align*}
This corresponds with answer choice C.
}
\dfn{Torsion}{
\begin{align*}
\tau = \frac{(\vec{r}^\prime \times \vec{r}^{\prime\prime})\cdot \vec{r}^{\prime\prime\prime}}{
\left \| \vec{r}^\prime \times \vec{r}^{\prime\prime} \right \|^2
}
\end{align*}
}
\qs{}{
\begin{align*}
\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 4\cos{\tfrac{2}{5}t},3,-4\sin{\tfrac{2}{5}t} \rangle \\
\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} &= \langle -\tfrac{8}{5}\sin{\tfrac{2}{5}t},0,-\tfrac{8}{5}\cos{\tfrac{2}{5}t} \rangle \\
\frac{\mathrm{d}^3\vec{r}}{\mathrm{d}t^3} &= \langle -\tfrac{16}{25}\cos{\tfrac{2}{5}t},0,\tfrac{16}{25}\sin{\tfrac{2}{5}t} \rangle \\
\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} &= \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
4\cos{\tfrac{2}{5}t}&3&-4\sin{\tfrac{2}{5}t}\\
-\tfrac{8}{5}\sin{\tfrac{2}{5}t}& 0 & -\tfrac{8}{5}\cos{\tfrac{2}{5}t}
\end{vmatrix} \\
&= \hat{i}\left [-\tfrac{24}{5}\cos{\tfrac{2}{5}t} \right]-\hat{j}\left [-\tfrac{32}{5}\cos^2{\tfrac{2}{5}t}-\tfrac{32}{5}\sin^2{\tfrac{2}{5}t}\right]+\hat{k}\left[\tfrac{24}{5}\sin{\tfrac{2}{5}t}] \\
\left \| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \right \| &= \sqrt{(\tfrac{24}{5})^2 +(\tfrac{32}{5})^2}=8 \\
\tau = \frac{(\vec{r}^\prime \times \vec{r}^{\prime\prime})\cdot \vec{r}^{\prime\prime\prime}}{
\left \| \vec{r}^\prime \times \vec{r}^{\prime\prime} \right \|^2
} &= \frac{\langle -\tfrac{24}{5}\cos{\tfrac{2}{5}t}, \frac{32}{5}, \tfrac{24}{5}\sin{\tfrac{2}{5}t} \rangle \cdot \langle -\tfrac{-16}{25}\cos{\tfrac{2}{5}t},0,\tfrac{16}{25}\sin{\tfrac{2}{5}t} \rangle }{8^2}\\
&= \left(-\frac{24}{5}\cos{\tfrac{2}{5}t}\right)\left(-\frac{16}{25}\cos{\tfrac{2}{5}t}\right)
+ \left(\frac{32}{5}\right)(0)
+ \left(\frac{24}{5}\sin{\tfrac{2}{5}t}\right)\left(\frac{16}{25}\sin{\tfrac{2}{5}t}\right) \\
&= \frac{384}{125}\cos^2{\tfrac{2}{5}t} + \frac{384}{125}\sin^2{\tfrac{2}{5}t} \\
&= \frac{384}{125} \\
\tau &= \frac{(\vec{r}^\prime \times \vec{r}^{\prime\prime})\cdot \vec{r}^{\prime\prime\prime}}{\| \vec{r}^\prime \times \vec{r}^{\prime\prime} \|^2}
= \frac{384/125}{8^2}
= \frac{384}{10000}
= \frac{6}{125}.
\end{align*}
This corresponds with answer choice A.
}
\end{document}