To find

asked 2021-06-05

\(r=\cos^2(\frac{\theta}{2})\)

asked 2021-09-07

\(\displaystyle{x}={\sin{{\left({\frac{{\theta}}{{{2}}}}\right)}}},{y}={\cos{{\left({\frac{{\theta}}{{{2}}}}\right)}}},-\pi\leq\theta\leq\pi\) (a) Eliminate the parameter to find a Cartesian equation of the curve. and how does thecurve look

asked 2021-09-02

Find a Cartesian equation for the curve and identify it.

\(\displaystyle{r}^{{2}}{\cos{{2}}}\theta={1}\)

\(\displaystyle{r}^{{2}}{\cos{{2}}}\theta={1}\)

asked 2021-09-14

Replace the polar equations with equivalent

Cartesian equations. Then describe or identify the graph. \(\displaystyle{r}^{{{2}}}=-{4}{r}{\cos{\theta}}\)

Cartesian equations. Then describe or identify the graph. \(\displaystyle{r}^{{{2}}}=-{4}{r}{\cos{\theta}}\)

asked 2021-08-20

Find the length of the curve.

\(\displaystyle{r}{\left({t}\right)}={\left\langle{6}{t},\ {t}^{{{2}}},\ {\frac{{{1}}}{{{9}}}}{t}^{{{3}}}\right\rangle},\ {0}\leq{t}\leq{1}\)

\(\displaystyle{r}{\left({t}\right)}={\left\langle{6}{t},\ {t}^{{{2}}},\ {\frac{{{1}}}{{{9}}}}{t}^{{{3}}}\right\rangle},\ {0}\leq{t}\leq{1}\)

asked 2021-06-06

Find the length of the curve.

\(r(t)=\langle6t,\ t^{2},\ \frac{1}{9}t^{3}\rangle,\ 0\leq t\leq1\)

\(r(t)=\langle6t,\ t^{2},\ \frac{1}{9}t^{3}\rangle,\ 0\leq t\leq1\)

asked 2021-11-16

Find the length of the curve.

\(\displaystyle{r}{\left({t}\right)}={4}{t},{t}^{{2}},{t}^{{3}},\ {0}\leq{t}\leq{1}\)

\(\displaystyle{r}{\left({t}\right)}={4}{t},{t}^{{2}},{t}^{{3}},\ {0}\leq{t}\leq{1}\)