musicdramalife00
write each polar equation in rectangular form:
r=3sec(Theta)
r=costheta+sintheta
r=5/3costheta+8sin



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musicdramalife00
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^theta after the last sin

myininaya
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You need to use these equations:
\[r^2=x^2+y^2 ; rcos(\theta)=x ; rsin(\theta)=y \]

musicdramalife00
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i did, but like it ended up like x^2+y^2= 25+9x^2r^230xr/64y^2 for the last one

myininaya
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We want the equations just in terms of x and y instead of r and theta.

musicdramalife00
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i don't know what i did wrong

myininaya
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so the last one is:
\[r=\frac{5}{3}\cos(\theta)+8\sin(\theta) \]?

musicdramalife00
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no uh (5)/(3costheta+8sintheta)

myininaya
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oh so you meant 5/(3costheta)+8sin(theta)) and not 5/3cos(theta)+8sin(theta)

musicdramalife00
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mhm
5 / (3cos+8sin)

myininaya
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\[r= \frac{5}{3\cos(\theta)+8\sin(\theta)}\]
Multiply both sides by that one fraction's denominator

myininaya
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\[r(3\cos(\theta)+8\sin(\theta))=5 \]
Distribute then substitute the equations I gave you

musicdramalife00
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3xr+8ry=5 but then you'll need to put r on one side cause you would need to square it. and the answer comes out all weird

myininaya
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well rcos(theta) is x not just cos(theta
same thing with the y deal
rsin(theta) is y not just sin(theta)

musicdramalife00
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whoops how about the secant one

myininaya
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put it in terms of cos by using an identity

myininaya
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then put it in some form where you can use the equations I gave you

musicdramalife00
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sosoososososo it would be x^3+y^2x=3 AND uh the other one x^2+y^2=x+y?

myininaya
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for the first one? no...

myininaya
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I haven't looked at the second one...
You are making this way too hard it looks like.

myininaya
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first one
you did this as first step right?:
\[r=\frac{3}{\cos(\theta)}\]


myininaya
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Notice this only has a cos in it...
how do you get cos with the r over there so you can use the equation rcos(theta)=x

musicdramalife00
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multiply both sides with r

myininaya
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no

myininaya
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try again

myininaya
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How do you undo multiplication?

myininaya
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Or division?

musicdramalife00
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oh omg

myininaya
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we have division by cos
to undo that you ?


myininaya
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yes

musicdramalife00
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thanks

myininaya
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so for that second one...

myininaya
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you actually got it right

myininaya
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but how did you go about it
you didn't choose a complicated way right?

musicdramalife00
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yep uh
multiplied both side with r and then substituted the stuff in

myininaya
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oh good

myininaya
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do you have any other questions? do you think you got this better?

musicdramalife00
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uh yea i have a few more

musicdramalife00
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for standard forms of polar equation uh how would graph r=2cos3theta

myininaya
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Plug in values of theta and see what the r output is

myininaya
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then remember how you graph using polar coordinates
r is the distance from the origin and direction is given by r to from the angle
theta is the angle

musicdramalife00
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like how would you do this with a graphing calculator

myininaya
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So for example if I plug in theta=0
\[r=2 \cos (3 \cdot 0)=2\cos(0)=2(1)=2 => \text{ graph the point } (2,0^o)\]
dw:1388269865874:dw

myininaya
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find 0 degrees and then go to the circle with radius 2

musicdramalife00
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cause wksht says not to use a table

myininaya
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dw:1388269927627:dw
then plug in more points

musicdramalife00
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oh i know how to graph but not without plugging in points

myininaya
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I haven't looked at a calculator in ages

myininaya
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did you try looking up how to input polar equations into whatever calculator type you have

myininaya
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some calculators require a difference procedure in order to do certain things


myininaya
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find mode
and choose polar or pol or polargc

myininaya
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then just type your equation in

musicdramalife00
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u what's the website


myininaya
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What ?