Question #3e769

3 Answers
Nov 11, 2016

#rho=(-sintheta+root2(sin^2theta+80cos^2theta))/(2cos^2theta)#
for #0< theta < 2pi#

Explanation:

Starting from the equations of passage from cartesian (rectangular) to polar coordinates
#x=rhocostheta#, #y=rhosintheta# and #rho=root2(x^2+y^2)#

we can replace them inside the given equation and get

#rho^2cos^2theta+rhosintheta-20=0# this can be seen as a second degree equation in the unknown #rho# that can be solved like this

#rho=(-sintheta+-root2(sin^2theta+80cos^2theta))/(2cos^2theta)#
that describes our function for #0< theta < 2pi#.

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 < 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

#rho=(-sintheta+root2(sin^2theta+80cos^2theta))/(2cos^2theta)#
for #0< theta < 2pi#

Nov 11, 2016

Assuming that intended equation is #x^2+(y-4)^2=4^2#, # r =8 sin theta, theta in [0, pi]#. .

Explanation:

#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, 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=OP=(CP+CO)cos(pi/2-theta)=8 sin theta.#

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

and radius.

#OP^2=OC^2+CP^2-2(OC)(CP)cos angle OCP#

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 [0, pi]#, for the whole circle.

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

not used #(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

Nov 11, 2016

Assuming the equation was #x^2+(y-4)^4=16#

#r=8sintheta#

Explanation:

#x=rcostheta#
#y=rsintheta#

#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#