# Question 3e769

Nov 11, 2016

$\rho = \frac{- \sin \theta + \sqrt[2]{{\sin}^{2} \theta + 80 {\cos}^{2} \theta}}{2 {\cos}^{2} \theta}$
for $0 < \theta < 2 \pi$

#### Explanation:

Starting from the equations of passage from cartesian (rectangular) to polar coordinates
$x = \rho \cos \theta$, $y = \rho \sin \theta$ and $\rho = \sqrt[2]{{x}^{2} + {y}^{2}}$

we can replace them inside the given equation and get

${\rho}^{2} {\cos}^{2} \theta + \rho \sin \theta - 20 = 0$ this can be seen as a second degree equation in the unknown $\rho$ that can be solved like this

$\rho = \frac{- \sin \theta \pm \sqrt[2]{{\sin}^{2} \theta + 80 {\cos}^{2} \theta}}{2 {\cos}^{2} \theta}$
that describes our function for $0 < \theta < 2 \pi$.

As long as the sign in front of the root is concerned, we have to consider that, according to its defintion $\rho$ must be positive for any $0 < \theta < 2 \pi$.
As a matter of fact the polar function describing the given function is the one with the plus sign linking the two terms at the numerator

$\rho = \frac{- \sin \theta + \sqrt[2]{{\sin}^{2} \theta + 80 {\cos}^{2} \theta}}{2 {\cos}^{2} \theta}$
for $0 < \theta < 2 \pi$

Nov 11, 2016

Assuming that intended equation is ${x}^{2} + {\left(y - 4\right)}^{2} = {4}^{2}$, $r = 8 \sin \theta , \theta \in \left[0 , \pi\right]$. .

#### Explanation:

${x}^{2} + {\left(y - 4\right)}^{2} = {4}^{2}$ represents the circle , with center C(0, 4) and

Let O be the pole and $P \left(r , \theta\right)$ be any point on the circle.

OC=CP=4 and, easily, anglePOC=angleCOP-pi/2-theta.angle

OCP=2theta.

It is immediate from the isosceles $\triangle$,

$r = O P = \left(C P + C O\right) \cos \left(\frac{\pi}{2} - \theta\right) = 8 \sin \theta .$

However, the following method befits any circle, with given center

$O {P}^{2} = O {C}^{2} + C {P}^{2} - 2 \left(O C\right) \left(C P\right) \cos \angle O C P$

So,

r^2=4^2+4^2-2(4)(4)(2 cos^2theta-1))=64(1-cos^2theta)=64sin^2theta#

This is simply, $r = 8 \sin \theta , \theta \in \left[0 , \pi\right]$, for the whole circle.

not used $\left(x , y\right) = r \left(\cos \theta , \sin \theta\right)$ at all. If you use this, it is

relatively a short method, for this problem. I agree. Yet, what I did

was for vector orientation

Nov 11, 2016

Assuming the equation was ${x}^{2} + {\left(y - 4\right)}^{4} = 16$

$r = 8 \sin \theta$

#### Explanation:

$x = r \cos \theta$
$y = r \sin \theta$

${r}^{2} {\cos}^{2} \theta + {\left(r \sin \theta - 4\right)}^{2} = 16$
${r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta - 8 r \sin \theta + \cancel{16} = \cancel{16}$
${r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = 8 r \sin \theta$
${r}^{\cancel{2}} = 8 \cancel{r} \sin \theta$
$r = 8 \sin \theta$