miniliv4

2022-09-30

Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph.

$$r=\mathrm{cot}\theta \mathrm{csc}\theta $$

$$r=\mathrm{cot}\theta \mathrm{csc}\theta $$

seppegettde

Beginner2022-10-01Added 7 answers

Since $$\mathrm{cot}\theta =\mathrm{cos}\theta /\mathrm{sin}\theta $$ and $$\mathrm{csc}\theta =1/\mathrm{sin}\theta $$ ,multiply $${\mathrm{sin}}^{2}\theta $$ to both sides of the equation and obtain $$r{\mathrm{sin}}^{2}\theta =\mathrm{cos}\theta $$ . Multiply r to both sides of the equation again, we obtain $${r}^{2}{\mathrm{sin}}^{2}\theta =r\mathrm{cos}\theta $$ . Since $$x=r\mathrm{cos}\theta $$ and $$y=r\mathrm{sin}\theta $$ , so the Cartesian equation $${y}^{2}=x$$, which is a side-way parabola with vertex at the origin opening to the right.

Polar coordinates and Cartesian coordinates conversion:

$$r=\sqrt{{x}^{2}+{y}^{2}}$$

$$\theta ={\mathrm{tan}}^{-1}(\frac{y}{x})$$

$$x=r\mathrm{cos}\theta $$

$$y=r\mathrm{sin}\theta $$

Result:

$$x={y}^{2}$$

Polar coordinates and Cartesian coordinates conversion:

$$r=\sqrt{{x}^{2}+{y}^{2}}$$

$$\theta ={\mathrm{tan}}^{-1}(\frac{y}{x})$$

$$x=r\mathrm{cos}\theta $$

$$y=r\mathrm{sin}\theta $$

Result:

$$x={y}^{2}$$

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