- anonymous

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

- schrodinger

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- anonymous

^theta after the last sin

- myininaya

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

- anonymous

i did, but like it ended up like x^2+y^2= 25+9x^2r^2-30xr/64y^2 for the last one

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## More answers

- myininaya

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

- anonymous

i don't know what i did wrong

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

mhm
5 / (3cos+8sin)

- myininaya

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

- myininaya

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

- anonymous

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

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

- anonymous

whoops how about the secant one

- myininaya

put it in terms of cos by using an identity

- myininaya

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

- anonymous

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

- myininaya

for the first one? no...

- myininaya

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

- myininaya

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

- anonymous

yep

- myininaya

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

- anonymous

multiply both sides with r

- myininaya

no

- myininaya

try again

- myininaya

How do you undo multiplication?

- myininaya

Or division?

- anonymous

oh omg

- myininaya

we have division by cos
to undo that you ?

- anonymous

x=-3

- myininaya

yes

- anonymous

thanks

- myininaya

so for that second one...

- myininaya

you actually got it right

- myininaya

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

- anonymous

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

- myininaya

oh good

- myininaya

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

- anonymous

uh yea i have a few more

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

like how would you do this with a graphing calculator

- myininaya

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

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

- anonymous

cause wksht says not to use a table

- myininaya

|dw:1388269927627:dw|
then plug in more points

- anonymous

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

- myininaya

I haven't looked at a calculator in ages

- myininaya

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

- myininaya

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

- myininaya

http://mathbits.com/MathBits/TISection/PreCalculus/polargraphs.htm

- myininaya

find mode
and choose polar or pol or polargc

- myininaya

then just type your equation in

- anonymous

u what's the website

- anonymous

*uh

- myininaya

What ?

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