SOLUTION: Find the polar coordinates of the point whose rectangular coordinates are (4sqrt 3, -4) Find the rectangular coordinates of the point whose polar coordinates are (-4, π/6).

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Question 865281: Find the polar coordinates of the point whose rectangular coordinates are (4sqrt 3, -4)
Find the rectangular coordinates of the point whose polar coordinates are (-4, π/6).
Find a rectangular form of the equation r = 5 cos θ

Found 2 solutions by Edwin McCravy, stanbon:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Find the polar coordinates of the point whose rectangular coordinates are
(4sqrt%283%29, -4)
We will indicate the rectangular coordinates of point P in 
black as P%28x%2Cy%29 and the polar coordinates of point P
in red as red%28P%28r%2Ctheta%29%29

Plot P using its rectangular coordinates P%28x%2Cy%29 = (P%284sqrt%283%29%2C0%29:



The x-coordinate of point P is x=4sqrt%283%29 and the y-coordinate
of point P is -4.

Draw a right triangle with this point P and the origin as
vertices and the right angle on the x-axis.  The legs of this right
triangle x and y are the RECTANGULAR coordinates of point P. The 
hypotenuse r of this right triangle is the first POLAR coordinate 
of P.  The angle theta indicated by the counter-clockwise red 
arc is the second POLAR coordinate of the point red%28P%28r%2Ctheta%29%29 



We only need to calculate red%28r%29 and +theta

red%28r%29%5E2=x%5E2%2By%5E2
red%28r%29%5E2=%284sqrt%283%29%29%5E2%2B%28-4%29%5E2
red%28r%29%5E2=16%2A3%2B16
red%28r%29%5E2=48%2B16
red%28r%29%5E2=64
red%28r%29=8

tan%28red%28theta%29%29=y%2Fx=%28-4%29%2F%284sqrt%283%29%29=-1%2Fsqrt%283%29

Therefore red%28theta%29 in the 4th quadrant is red%2811pi%2F6%29, and

P%284sqrt%283%29%2C0%29 = red%28P%288%2C11pi%2F6%29%29



Find the rectangular coordinates of the point whose polar coordinates are
red%28P%28-4%2Cpi%2F6%29%29.
Let's draw the point red%28P%28-4%2Cpi%2F6%29%29 using its polar coordinates.

First we draw the angle red%28theta%29 with a dotted line through
the origin:



Next we locate the value of r on the x-axis, then we swing an
arc from that point on the x-axis around to the dotted line,
like the green arc below swinging from -4 on the x-axis to
the dotted line, and mar that point red%28P%28-4%2Cpi%2F6%29%29.



Then we erase the green arc and draw a right triangle with 
this point P and the origin as vertices and the right angle 
on the x-axis, and indicate the RECTANGULAR coordinates x,
y, of the point P.  Since we swung the point from the point
x=-4, the hypotenuse red%28r=-4%29.

  

Now we calculate x and y.

cos%28theta%29=x%2Fred%28r%29
cos%28pi%2F6%29=x%2Fred%28-4%29
sqrt%283%29%2F2=x%2Fred%28-4%29
2x=red%28-4%29sqrt%283%29
x=-2sqrt%283%29

sin%28theta%29=y%2Fred%28r%29
sin%28pi%2F6%29=y%2Fred%28-4%29
y=red%28-4%29sin%28pi%2F6%29
y=red%28-4%29%281%2F2%29
y=-2

So red%28P%28-4%2Cpi%2F6%29%29 = P%28-2sqrt%283%29%2C-2%29



Find a rectangular form of the equation red%28r+=+5+cos%28theta%29%29
Always substitute trig functions first, 

sin%28red%28theta%29%29=y%2Fred%28r%29, cos%28red%28theta%29%29=x%2Fred%28r%29, tan%28red%28theta%29%29=y%2Fx

and always wait until last to replace red%28r%29 by red%28r%29=sqrt%28x%5E2%2By%5E2%29

red%28r%29=+5cos%28theta%29
red%28r%29+=+5%28x%2Fred%28r%29%29
red%28r%29+=+5x%2Fred%28r%29
red%28r%5E2%29+=+5x
%28sqrt%28x%5E2%2By%5E2%29%29%5E2+=+5x
x%5E2%2By%5E2=5x

You can then put it in standard form of a circle,

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

getting:

%28x-5%2F2%29%5E2%2B%28y-0%29%5E2=%285%2F2%29%5E2

whose graph is a circle with center (h,k) = (5%2F2,0) 

and radius r=5%2F2: 



Edwin

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the polar coordinates of the point whose
rectangular coordinates are (4sqrt 3, -4)
Note:: the point is in QIV so the angle is in QIV
-----------------
r = sqrt[(4sqrt(3))^2 + 4^2] = sqrt[64] = 8
angle = t = arctan(-4/4sqrt(3)) = arctan(-1/sqrt(3)) = (11/6)pi
Ans: (8,(11/6)pi)
====================================
Find the rectangular coordinates of the point
whose polar coordinates are (-4, π/6).
x = -4(cos(pi/6)) = -4(sqrt(3)/2) = -2sqrt(3)
y = -4(sin(pi/6)) = -4(1/2) = -2
=====================================
Find a rectangular form of the equation
r = 5 cos θ
----
r = sqrt(x^2+y^2)
theta = arctan(y/x)
----
Therefore cos(theta) = x/sqrt(x^2+y^2)
----------------------
Substituting you get:
=================
sqrt(x^2+y^2) = 5*x/sqrt(x^2+y^2)
--------
x^2 + y"^2 = 5x
----
Cheers,
Stan H.
=================