It is often useful to convert equations into different forms. Let's look at how to convert a rectangular equation into polar form first.
Convert the rectangular equation $5x+y=2$5x+y=2 into a polar equation.
Think: The rectangular point $\left(x,y\right)$(x,y) corresponds with the polar point $\left(r,\theta\right)$(r,θ) where $x=r\cos\theta$x=rcosθ and $y=r\sin\theta$y=rsinθ. By substituting this information into the original rectangular equation we can convert it into polar form.
Do: Substituting the information from above we get:
$5x+y$5x+y | $=$= | $2$2 | original rectangular equation |
$5r\cos\theta+r\sin\theta$5rcosθ+rsinθ | $=$= | $2$2 | substitution |
$r\left(5\cos\theta+\sin\theta\right)$r(5cosθ+sinθ) | $=$= | $2$2 | factor out $r$r as the GCF |
$\frac{r\left(5\cos\theta+\sin\theta\right)}{5\cos\theta+\sin\theta}$r(5cosθ+sinθ)5cosθ+sinθ | $=$= | $\frac{2}{5\cos\theta+\sin\theta}$25cosθ+sinθ | solve for $r$r by dividing |
$r$r | $=$= | $\frac{2}{5\cos\theta+\sin\theta}$25cosθ+sinθ | simplify the left hand side |
Reflect: Thus, the polar form of $5x+y=2$5x+y=2 is $r=\frac{2}{5\cos\theta+\sin\theta}$r=25cosθ+sinθ.
Now let's try converting an equation in polar form into rectangular form.
Convert the polar equation $r=8\cos\theta-4\sin\theta$r=8cosθ−4sinθ into a rectangular equation. State the equation with all terms on one side.
Think: Recall that the rectangular point $\left(x,y\right)$(x,y) corresponds with the polar point $\left(r,\theta\right)$(r,θ) where $x=r\cos\theta$x=rcosθ and $y=r\sin\theta$y=rsinθ and $x^2+y^2=r^2$x2+y2=r2.
Do: We need to make the original equation $r=8\cos\theta-4\sin\theta$r=8cosθ−4sinθ look more like $x^2+y^2=r^2$x2+y2=r2. We can start by creating an $r^2$r2 term on the left hand side.
$r$r | $=$= | $8\cos\theta-4\sin\theta$8cosθ−4sinθ | original polar equation |
$r^2$r2 | $=$= | $r\left(8\cos\theta-4\sin\theta\right)$r(8cosθ−4sinθ) | multiply both sides by $r$r |
$r^2$r2 | $=$= | $8r\cos\theta-4r\sin\theta$8rcosθ−4rsinθ | distribute $r$r |
$r^2$r2 | $=$= | $8x-4y$8x−4y | substitute $x=r\cos\theta$x=rcosθ and $y=r\sin\theta$y=rsinθ |
$x^2+y^2$x2+y2 | $=$= | $8x-4y$8x−4y | substitute $x^2+y^2=r^2$x2+y2=r2 |
$x^2-8x+y^2+4y$x2−8x+y2+4y | $=$= | $0$0 | move all terms to one side |
Reflect: Thus the rectangular form of $r=8\cos\theta-4\sin\theta$r=8cosθ−4sinθ is $x^2-8x+y^2+4y=0$x2−8x+y2+4y=0.
Convert the rectangular equation $5x-4y=3$5x−4y=3 into a polar equation.
Convert the polar equation $r=\frac{8}{1+\sin\theta}$r=81+sinθ into a rectangular equation.
Convert the rectangular equation $x^2+y^2=64$x2+y2=64 into a polar equation.