Question

Find all points of intersection of the given curves.

\({{\rm{r}}^{\rm{2}}}{\rm{ = sin2\theta ,}}\;\;\;{{\rm{r}}^{\rm{2}}}{\rm{ = cos2\theta }}\).

Short answer

Therefore, the results of the intersection points of the curve is\(\left( {\sqrt {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} {\rm{,}}\frac{{\rm{\pi }}}{{\rm{8}}}} \right){\rm{,}}\left( {\sqrt {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} {\rm{,}}\frac{{{\rm{9\pi }}}}{{\rm{8}}}} \right)\)and the pole.

Step-by-step solution

Intersection points.

Given

\(\begin{aligned}{l}{{\rm{r}}^{\rm{2}}}{\rm{ = sin2\theta }}\\{{\rm{r}}^{\rm{2}}}{\rm{ = cos2\theta }}\end{aligned}\)

The figures are finished\({\rm{0}} \le {\rm{\theta < 2\pi }}\)(excluding the intervals where cos or sin are negative)

By setting the equations equal to each other, locate certain crossings. Only look for angles in which cos and sin are positive. Also, because\({\rm{2\theta }}\)appears in the equations, check for angles from\({\rm{2\theta = 0to4\pi }}\)that direction.

\(\begin{aligned}{l}{\rm{sin2\theta = cos2\theta }}\\{\rm{2\theta = }}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{,}}\frac{{{\rm{9\pi }}}}{{\rm{4}}}\\{\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{8}}}{\rm{,}}\frac{{{\rm{9\pi }}}}{{\rm{8}}}\end{aligned}\)

Resulting.

The graph, that they also meet at the pole, which both graphs pass through at different times\({\rm{\theta }}\). Look for areas where\({\rm{r = 0}}\) angles\({\rm{4\pi }}\) can be found.

\(\begin{aligned}{l}{\rm{0 = cos2\theta }}\\{\rm{2\theta = }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{,}}\frac{{{\rm{3\pi }}}}{{\rm{2}}}{\rm{,}}\frac{{{\rm{5\pi }}}}{{\rm{2}}}{\rm{,}}\frac{{{\rm{6\pi }}}}{{\rm{2}}}\\{\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{4}}}{\rm{,}}\frac{{{\rm{3\pi }}}}{{\rm{4}}}{\rm{,}}\frac{{{\rm{5\pi }}}}{{\rm{4}}}{\rm{,}}\frac{{{\rm{6\pi }}}}{{\rm{4}}}\\{\rm{0 = sin2\theta }}\end{aligned}\)

(\({\rm{2\theta = 0,\pi ,3\pi }}\)),\({\rm{4\pi }}\)is the beginning of a new complete cycle.

\({\rm{\theta = 0,}}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{,}}\frac{{{\rm{3\pi }}}}{{\rm{2}}}\)

The sites of the intersection are the same\({\rm{\theta }}\). Note that the equations are written as\({{\rm{r}}^{\rm{2}}}\).

\(\left( {\sqrt {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} {\rm{,}}\frac{{\rm{\pi }}}{{\rm{8}}}} \right){\rm{,}}\left( {\sqrt {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} {\rm{,}}\frac{{{\rm{9\pi }}}}{{\rm{8}}}} \right)\)

Negative values\({\rm{r}}\)are also possible, but the positive values already cover those points. Junctions of poles at various locations\({\rm{\theta }}\).

\(\left( {{\rm{0,}}\frac{{\rm{\pi }}}{{\rm{4}}}} \right){\rm{,}}\left( {{\rm{0,}}\frac{{{\rm{3\pi }}}}{{\rm{4}}}} \right){\rm{,}}\left( {{\rm{0,}}\frac{{{\rm{5\pi }}}}{{\rm{4}}}} \right){\rm{,}}\left( {{\rm{0,}}\frac{{{\rm{6\pi }}}}{{\rm{4}}}} \right){\rm{,(0,0),}}\left( {{\rm{0,}}\frac{{\rm{\pi }}}{{\rm{2}}}} \right){\rm{,}}\left( {{\rm{0,}}\frac{{{\rm{3\pi }}}}{{\rm{2}}}} \right)\)

Thus, the resulting is \(\left( {\sqrt {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} {\rm{,}}\frac{{\rm{\pi }}}{{\rm{8}}}} \right){\rm{,}}\left( {\sqrt {\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}} {\rm{,}}\frac{{{\rm{9\pi }}}}{{\rm{8}}}} \right)\)and the pole.