Question

Sketch the curve with the given polar equation by first sketching the graph of as a function of\({\rm{\theta }}\) in Cartesian coordinates.

\({\rm{r = - 2sin\theta }}\)

Short answer

A circle having a radius of \({\rm{1}}\)and a center of\(\left( {{\rm{0, - 1}}} \right){\rm{.}}\)

\({{\rm{x}}^{\rm{2}}}{\rm{ + (y + 1}}{{\rm{)}}^{\rm{2}}}{\rm{ = }}{{\rm{1}}^{\rm{2}}}\)

Step-by-step solution

Step 1: Both sides should be multiplied by \({\rm{r}}{\rm{.}}\)

\({\rm{r = - 2sin\theta }}\)

\({{\rm{r}}^{\rm{2}}}{\rm{ = - 2rsin\theta }}\)

Keep in mind

\(\begin{array}{c}{{\rm{r}}^{\rm{2}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}\\\;\;\;{\rm{rsin\theta = y}}\\{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = - 2y}}\\{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + 2y = 0}}\end{array}\)

Add \({\rm{1}}\) to each side.

\(\begin{array}{c}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + 2y + 1 = 1}}\\{{\rm{x}}^{\rm{2}}}{\rm{ + (y + 1}}{{\rm{)}}^{\rm{2}}}{\rm{ = }}{{\rm{1}}^{\rm{2}}}\end{array}\)

This is a circle with a radius of \({\rm{1}}\) and a center of \(\left( {{\rm{0, - 1}}} \right){\rm{.}}\)

Mark the position \(\left( {{\rm{0, - 1}}} \right)\)on the graph, then creates a circle with this point as the centre.

Step 3: A circle with a radius of \({\rm{1}}\)and a center of \(\left( {{\rm{0, - 1}}} \right){\rm{.}}\)

\({{\rm{x}}^{\rm{2}}}{\rm{ + (y + 1}}{{\rm{)}}^{\rm{2}}}{\rm{ = }}{{\rm{1}}^{\rm{2}}}\)

This is a circle with a radius of \({\rm{1}}\) and a center of \(\left( {{\rm{0, - 1}}} \right){\rm{.}}\)