7.3 Polar coordinates  (Page 3/16)

Because the function is a multiple of a sine function, it is periodic with period $2\pi ,$ so use values for $\theta$ between 0 and $2\pi .$ The result of steps 1–3 appear in the following table. [link] shows the graph based on this table.

This is the graph of a circle. The equation $r=4\text{sin}\theta$ can be converted into rectangular coordinates by first multiplying both sides by $r.$ This gives the equation $r^2=4r\text{sin}\theta .$ Next use the facts that $r^2=x^2+y^2$ and $y=r\text{sin}\theta .$ This gives $x^2+y^2=4y.$ To put this equation into standard form, subtract $4y$ from both sides of the equation and complete the square:

This is the equation of a circle with radius 2 and center $\left(0,2\right)$ in the rectangular coordinate system.

1. Take the tangent of both sides. This gives $\text{tan}\theta =\text{tan}(\pi \text{/}3)=\sqrt{3}.$ Since $\text{tan}\theta =y\text{/}x$ we can replace the left-hand side of this equation by $y\text{/}x.$ This gives $y\text{/}x=\sqrt{3},$ which can be rewritten as $y=x\sqrt{3}.$ This is the equation of a straight line passing through the origin with slope $\sqrt{3}.$ In general, any polar equation of the form $\theta =K$ represents a straight line through the pole with slope equal to $\text{tan}K.$
2. First, square both sides of the equation. This gives $r^2=9.$ Next replace $r^2$ with $x^2+y^2.$ This gives the equation $x^2+y^2=9,$ which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form $r=k$ where k is a positive constant represents a circle of radius k centered at the origin. ( Note : when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, $\left(\mathrm{-3},\frac{\pi }{3}\right)$ is the same point as $\left(3,\frac{4\pi }{3}\right).)$
3. Multiply both sides of the equation by $r.$ This leads to $r^2=6r\text{cos}\theta -8r\text{sin}\theta .$ Next use the formulas
   
   $r^2=x^2+y^2,x=r\text{cos}\theta ,y=r\text{sin}\theta .$
   
   This gives
   
   $\begin{array}{ccc}\hfill r^2& =\hfill & 6\left(r\text{cos}\theta \right)-8\left(r\text{sin}\theta \right)\hfill \\ \hfill x^2+y^2& =\hfill & 6x-8y.\hfill \end{array}$
   
   To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.
   
   $\begin{array}{ccc}\hfill x^2+y^2& =\hfill & 6x-8y\hfill \\ \hfill x^2-6x+y^2+8y& =\hfill & 0\hfill \\ \hfill \left(x^2-6x\right)+\left(y^2+8y\right)& =\hfill & 0\hfill \\ \hfill \left(x^2-6x+9\right)+\left(y^2+8y+16\right)& =\hfill & 9+16\hfill \\ \hfill {\left(x-3\right)}^2+{\left(y+4\right)}^2& =\hfill & 25.\hfill \end{array}$
   
   This is the equation of a circle with center at $\left(3,\mathrm{-4}\right)$ and radius 5. Notice that the circle passes through the origin since the center is 5 units away.