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Compute the derivatives of the vector-valued functions.
$\text{r}(t)={t}^{3}\phantom{\rule{0.1em}{0ex}}\text{i}+3{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{{t}^{3}}{6}\phantom{\rule{0.1em}{0ex}}\text{k}$
$\u27e83{t}^{2},6t,\frac{1}{2}{t}^{2}\u27e9$
$\text{r}(t)=\text{sin}(t)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}(t)\phantom{\rule{0.1em}{0ex}}\text{j}+{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{k}$
$\text{r}(t)={e}^{\text{\u2212}t}\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}(3t)\text{}\phantom{\rule{0.1em}{0ex}}\text{j}+10\sqrt{t}\phantom{\rule{0.1em}{0ex}}\text{k}.$ A sketch of the graph is shown here. Notice the varying periodic nature of the graph.
$\u27e8\text{\u2212}{e}^{\text{\u2212}t},3\phantom{\rule{0.1em}{0ex}}\text{cos}(3t),\frac{5}{\sqrt{t}}\u27e9$
$\text{r}(t)={e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+2{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{j}+\phantom{\rule{0.1em}{0ex}}\text{k}$
$\text{r}(t)=\text{i}+\phantom{\rule{0.1em}{0ex}}\text{j}+\phantom{\rule{0.1em}{0ex}}\text{k}$
$\u27e80,0,0\u27e9$
$\text{r}(t)=t{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+t\phantom{\rule{0.1em}{0ex}}\text{ln}(t)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{sin}(3t)\phantom{\rule{0.1em}{0ex}}\text{k}$
$\text{r}(t)=\frac{1}{t+1}\phantom{\rule{0.1em}{0ex}}\text{i}+\text{arctan}(t)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{ln}\phantom{\rule{0.1em}{0ex}}{t}^{3}\phantom{\rule{0.1em}{0ex}}\text{k}$
$\u27e8\frac{\mathrm{-1}}{{\left(t+1\right)}^{2}},\frac{1}{1+{t}^{2}},\frac{3}{t}\u27e9$
$\text{r}(t)=\text{tan}(2t)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sec}(2t)\phantom{\rule{0.1em}{0ex}}\text{j}+{\text{sin}}^{2}(t)\phantom{\rule{0.1em}{0ex}}\text{k}$
$\text{r}(t)=3\phantom{\rule{0.1em}{0ex}}\text{i}+4\phantom{\rule{0.1em}{0ex}}\text{sin}(3t)\phantom{\rule{0.1em}{0ex}}\text{j}+t\phantom{\rule{0.1em}{0ex}}\text{cos}(t)\phantom{\rule{0.1em}{0ex}}\text{k}$
$\u27e80,12\phantom{\rule{0.1em}{0ex}}\text{cos}(3t),\text{cos}\phantom{\rule{0.1em}{0ex}}t-t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\u27e9$
$\text{r}(t)={t}^{2}\phantom{\rule{0.1em}{0ex}}\text{i}+t{e}^{\mathrm{-2}t}\phantom{\rule{0.1em}{0ex}}\text{j}-5{e}^{\mathrm{-4}t}\phantom{\rule{0.1em}{0ex}}\text{k}$
For the following problems, find a tangent vector at the indicated value of t .
$\text{r}(t)=t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}(2t)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{cos}(3t)\phantom{\rule{0.1em}{0ex}}\text{k};t=\frac{\pi}{3}$
$\frac{1}{\sqrt{2}}\u27e81,\mathrm{-1},0\u27e9$
$\text{r}(t)=3{t}^{3}\phantom{\rule{0.1em}{0ex}}\text{i}+2{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{1}{t}\phantom{\rule{0.1em}{0ex}}\text{k};t=1$
$\text{r}(t)=3{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+2{e}^{\mathrm{-3}t}\phantom{\rule{0.1em}{0ex}}\text{j}+4{e}^{2t}\phantom{\rule{0.1em}{0ex}}\text{k};$ $t=\text{ln}(2)$
$\frac{1}{\sqrt{1060.5625}}\u27e86,-\frac{3}{4},32\u27e9$
$\text{r}(t)=\text{cos}(2t)\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k};t=\frac{\pi}{2}$
Find the unit tangent vector for the following parameterized curves.
$\text{r}(t)=6\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}(3t)\phantom{\rule{0.1em}{0ex}}\text{j}+3\phantom{\rule{0.1em}{0ex}}\text{sin}(4t)\phantom{\rule{0.1em}{0ex}}\text{k},$ $0\le t<2\pi $
$\frac{1}{\sqrt{9\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{2}(3t)+144\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}(4t)}}\u27e80,\mathrm{-3}\phantom{\rule{0.1em}{0ex}}\text{sin}(3t),12\phantom{\rule{0.1em}{0ex}}\text{cos}(4t)\u27e9$
$\text{r}(t)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{k},$ $0\le t<2\pi .$ Two views of this curve are presented here:
$\text{r}(t)=3\phantom{\rule{0.1em}{0ex}}\text{cos}(4t)\phantom{\rule{0.1em}{0ex}}\text{i}+3\phantom{\rule{0.1em}{0ex}}\text{sin}(4t)\phantom{\rule{0.1em}{0ex}}\text{j}+5t\phantom{\rule{0.1em}{0ex}}\text{k},1\le t\le 2$
$\text{T}(t)=\frac{\mathrm{-12}}{13}\text{sin}(4t)\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{12}{13}\text{cos}(4t)\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{5}{13}\phantom{\rule{0.1em}{0ex}}\text{k}$
$\text{r}(t)=t\phantom{\rule{0.1em}{0ex}}\text{i}+3t\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}$
Let $\text{r}(t)=t\phantom{\rule{0.1em}{0ex}}\text{i}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}-{t}^{4}\phantom{\rule{0.1em}{0ex}}\text{k}$ and $s(t)=\text{sin}(t)\phantom{\rule{0.1em}{0ex}}\text{i}+{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{j}+\text{cos}(t)\phantom{\rule{0.1em}{0ex}}\text{k}.$ Here is the graph of the function:
Find the following.
$\frac{d}{dt}\left[r\left({t}^{2}\right)\right]$
$\u27e82t,4{t}^{3},\mathrm{-8}{t}^{7}\u27e9$
$\frac{d}{dt}\left[{t}^{2}\xb7s(t)\right]$
$\frac{d}{dt}\left[r(t)\xb7s(t)\right]$
$\text{sin}(t)+2t{e}^{t}-4{t}^{3}\text{cos}(t)+t\phantom{\rule{0.1em}{0ex}}\text{cos}(t)+{t}^{2}{e}^{t}+{t}^{4}\text{sin}(t)$
Compute the first, second, and third derivatives of $\text{r}(t)=3t\phantom{\rule{0.1em}{0ex}}\text{i}+6\phantom{\rule{0.1em}{0ex}}\text{ln}(t)\phantom{\rule{0.1em}{0ex}}\text{j}+5{e}^{\mathrm{-3}t}\phantom{\rule{0.1em}{0ex}}\text{k}.$
Find $\text{r}\prime (t)\xb7\phantom{\rule{0.1em}{0ex}}\text{r}\text{\u2033}(t)\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}\text{r}(t)=\mathrm{-3}{t}^{5}\phantom{\rule{0.1em}{0ex}}\text{i}+5t\phantom{\rule{0.1em}{0ex}}\text{j}+2{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}.$
$900{t}^{7}+16t$
The acceleration function, initial velocity, and initial position of a particle are
$\text{a}(t)=\mathrm{-5}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j},\phantom{\rule{0.1em}{0ex}}\text{v}(0)=9\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{r}(0)=5\phantom{\rule{0.1em}{0ex}}\text{i}.$
Find
$\text{v}(t)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{r}(t).$
The position vector of a particle is $\text{r}(t)=5\phantom{\rule{0.1em}{0ex}}\text{sec}(2t)\phantom{\rule{0.1em}{0ex}}\text{i}-4\phantom{\rule{0.1em}{0ex}}\text{tan}(t)\phantom{\rule{0.1em}{0ex}}\text{j}+7{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}.$
Find the velocity and the speed of a particle with the position function $\text{r}(t)=\left(\frac{2t-1}{2t+1}\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{ln}(1-4{t}^{2})\phantom{\rule{0.1em}{0ex}}\text{j}.$ The speed of a particle is the magnitude of the velocity and is represented by $\text{\Vert}{r}^{\text{'}}(t)\text{\Vert}.$
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