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Algebra 8 Online
OpenStudy (mollykenney):

Convert the rectangular equation to a polar equation that expresses r in terms of theta x=8

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

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

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

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

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

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.

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

OpenStudy (puppernupper):

how do i post a question

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