Convert the rectangular equation to a polar equation that expresses r in terms of theta
x=8
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OpenStudy (astrophysics):
\[x=r \cos \theta \]
OpenStudy (astrophysics):
and \[r^2=x^2+y^2\]
OpenStudy (mollykenney):
8=rcostheta
and r^2= 8^2 + y^2 ??
OpenStudy (astrophysics):
You want r in terms of theta
OpenStudy (astrophysics):
So good, you have \[8 = r \cos \theta \] what's next
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OpenStudy (mollykenney):
find y so we have sin theta ?
OpenStudy (astrophysics):
No, why would you need y, all you need is x = rcos theta, maybe I shouldn't have put the second equation.
OpenStudy (mollykenney):
or it's just 8/costheta
OpenStudy (astrophysics):
Yes, that's it
OpenStudy (mollykenney):
wow! thank you haha
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OpenStudy (astrophysics):
Np
OpenStudy (mollykenney):
would you mind helping me on one more problem or at least explaining the ending to it?
OpenStudy (mollykenney):
The rectangular coordinates of a point are given. Find polar coordinates of the point such that r>0 and 0<(or equal to) theta < 2pi
(-10,10)
OpenStudy (nubeer):
First you need to find R.. you can do that by
r2=x2+y2
OpenStudy (mollykenney):
ok and i get 0
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OpenStudy (nubeer):
:3 no
r2 = (-10)^2 + (10)^2
r^2 = 100 +100
do you see where u went wrong?
OpenStudy (mollykenney):
I didn't put parenthesis
OpenStudy (mollykenney):
r = 14.1?
OpenStudy (nubeer):
no.. you kept the negative sign out when you were about to square..
so you got -100.. thats why you got 0.. parenthesis just reduces chance of a mistake.
OpenStudy (nubeer):
yes
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OpenStudy (mollykenney):
and that is the final answer?
OpenStudy (mollykenney):
nevermind
OpenStudy (nubeer):
We now find tan t using the formula tan theta = y / x
OpenStudy (mollykenney):
-1
OpenStudy (mollykenney):
14.1,-1
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OpenStudy (nubeer):
you have to find the value of theta .. whatever is theta..
that would be your other coordinate..
OpenStudy (mollykenney):
-1 is not the coordinate for theta?
OpenStudy (nubeer):
no.. its not theta..
its tan theta =-1
take inverse of tang.. and keep in mind that in which quadrant tan is negative.
OpenStudy (mollykenney):
-45
OpenStudy (nubeer):
tan is negative in second quadrant.. and the angle there is between 90 and 180 so theta would be in between here.
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OpenStudy (nubeer):
|dw:1459407701367:dw|
OpenStudy (nubeer):
|dw:1459407738872:dw|
this would be your theta
OpenStudy (mollykenney):
150 degrees
OpenStudy (nubeer):
how did u got 150?
OpenStudy (nubeer):
well you got -45.. so subtract it from 180
=180-45
change it into radian.. that would be your other coordinate
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