Arc Length And Curvature

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1. Find the length of the curve

$\mathbf{r}\left(t\right)=⟨2\mathrm{sin}\left(t\right),5t,2\mathrm{cos}\left(t\right)⟩$

for $-10\le t\le 10$.

$20\sqrt{29}$
2. Find the length of the curve

$\mathbf{r}\left(t\right)=12t\mathbf{i}+8{t}^{3/2}\mathbf{j}+3{t}^{2}\mathbf{k}$

for $0\le t\le 1$.

$\mathbf{r}\prime \left(t\right)=⟨12,8\frac{3}{2}\sqrt{t},6t⟩=⟨12,12\sqrt{t},6t⟩$.

So the arc length

$L={\int }_{0}^{1}\left|\mathbf{r}\prime \left(t\right)\right|\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{1}\sqrt{{12}^{2}+{12}^{2}t+36{t}^{2}}\phantom{\rule{0.2em}{0ex}}dt=6{\int }_{0}^{1}\left(2+t\right)\phantom{\rule{0.2em}{0ex}}dt=15$.

3. Find the length of the curve

$\mathbf{r}\left(t\right)=⟨\frac{{t}^{2}}{2},\frac{{\left(2t+1\right)}^{3/2}}{3}⟩$

for $0\le t\le 2$.

4. Parametrize the curve

$\mathbf{r}\left(t\right)=2t\mathbf{i}+\left(1-3t\right)\mathbf{j}+\left(5+4t\right)\mathbf{k}$

with respect to arc length measured from the point where $t=0$ in the direction of increasing $t$.

$\mathbf{r}\left(t\left(s\right)\right)=\frac{2}{\sqrt{29}}s\mathbf{i}+\left(1-\frac{3}{\sqrt{29}}s\right)\mathbf{j}+\left(5+\frac{4}{\sqrt{29}}s\right)\mathbf{k}$
5. Parametrize the curve

$\mathbf{r}\left(t\right)={e}^{2t}\mathrm{cos}\left(2t\right)\mathbf{i}+2\mathbf{j}+{e}^{2t}\mathrm{sin}\left(2t\right)\mathbf{k}$

with respect to arc length measured from the point where $t=0$ in the direction of increasing $t$.

6. A particle starts out at the point $\left(0,0,3\right)$ and moves $5$ units along the the curve $x=3\mathrm{sin}\left(t\right),\phantom{\rule{0.5em}{0ex}}y=4t,\phantom{\rule{0.5em}{0ex}}z=3\mathrm{cos}\left(t\right)$ in the positive direction. Where is it now?
$\left(3\mathrm{sin}\left(1\right),4,3\mathrm{cos}\left(1\right)\right)$
7. Compute the arc length of the curve defined by the vector-valued function

$\mathbf{r}\left(t\right)={t}^{2}\mathbf{i}-2t\mathbf{j}+\mathrm{ln}\left(t\right)\mathbf{k}$

from the point $\left(1,-2,0\right)$ to the point $\left({e}^{2},-2e,1\right)$.

${e}^{2}$