Question #3e769

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Question #3e769

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Answer 1 · 2016-11-11T06:27:24

$ρ = \frac{- sin θ + \sqrt[2]{{sin}^{2} θ + 80 {cos}^{2} θ}}{2 {cos}^{2} θ}$ $rho=(-sintheta+root2(sin^2theta+80cos^2theta))/(2cos^2theta)$
for $0 < θ < 2 π$ $0< theta < 2pi$

Explanation:

Starting from the equations of passage from cartesian (rectangular) to polar coordinates
$x = ρ cos θ$ $x=rhocostheta$ , $y = ρ sin θ$ $y=rhosintheta$ and $ρ = \sqrt[2]{x^{2} + y^{2}}$ $rho=root2(x^2+y^2)$

we can replace them inside the given equation and get

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

$ρ = \frac{- sin θ \pm \sqrt[2]{{sin}^{2} θ + 80 {cos}^{2} θ}}{2 {cos}^{2} θ}$ $rho=(-sintheta+-root2(sin^2theta+80cos^2theta))/(2cos^2theta)$
that describes our function for $0 < θ < 2 π$ $0< theta < 2pi$ .

As long as the sign in front of the root is concerned, we have to consider that, according to its defintion $ρ$ $rho$ must be positive for any $0 < θ < 2 π$ $0 < theta < 2pi$ .
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

$ρ = \frac{- sin θ + \sqrt[2]{{sin}^{2} θ + 80 {cos}^{2} θ}}{2 {cos}^{2} θ}$ $rho=(-sintheta+root2(sin^2theta+80cos^2theta))/(2cos^2theta)$
for $0 < θ < 2 π$ $0< theta < 2pi$

Answer 2 · 2016-11-11T07:38:50

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

Explanation:

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

radius 4.

Let O be the pole and $P (r, θ)$ $P(r, theta)$ 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 $△$ $triangle$ ,

$r = O P = (C P + C O) cos (\frac{π}{2} - θ) = 8 sin θ .$ $r=OP=(CP+CO)cos(pi/2-theta)=8 sin theta.$

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

and radius.

$O P^{2} = O C^{2} + C P^{2} - 2 (O C) (C P) cos ∠ O C P$ $OP^2=OC^2+CP^2-2(OC)(CP)cos angle OCP$

So,

$r^{2} = 4^{2} + 4^{2} - 2 (4) (4) (2 {cos}^{2} θ - 1)) = 64 (1 - {cos}^{2} θ) = 64 {sin}^{2} θ$ $r^2=4^2+4^2-2(4)(4)(2 cos^2theta-1))=64(1-cos^2theta)=64sin^2theta$

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

Perhaps, some readers might not like this answer, because I have

not used $(x, y) = r (cos θ, sin θ)$ $(x, y)=r(cos theta, sin theta)$ 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

Answer 3 · 2016-11-11T08:06:53

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

$r = 8 sin θ$ $r=8sintheta$

Explanation:

$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

$r^{2} {cos}^{2} θ + {(r sin θ - 4)}^{2} = 16$ $r^2cos^2theta+(rsintheta-4)^2=16$
$r^2cos^2theta+r^2sin^2theta-8rsintheta+cancel(16)=cancel(16)$
$r^2(cos^2theta+sin^2theta)=8rsintheta$
$r^cancel(2)= 8cancel(r)sintheta$
$r=8sintheta$

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