Convert the polar equation to rectangular form and identify the graph. r= 7/ (2cosΘ+5sinΘ)

- anonymous

Convert the polar equation to rectangular form and identify the graph. r= 7/ (2cosΘ+5sinΘ)

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- anonymous

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- anonymous

y = r sin(theta) x = r cos(theta) x^2 + y^2 = r^2 plug and chug

- anonymous

yea so how would i plug it in

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## More answers

- anonymous

cuz i did this last week and i kinda forgot sorry

- myininaya

Try getting your cos and sin with r over there by multiplying both sides by 2cos(theta)+5sin(theta)

- anonymous

okay. i did that

- myininaya

So you have: \[2rcos(\theta)+5rsin(\theta)=7 ?\]

- anonymous

why would you put r w/ each one of the terms isnt there just 1?

- myininaya

\[r=\frac{7}{2\cos(\theta)+5 \sin(\theta)} => r(2\cos(\theta)+5 \sin(\theta))=7\] Distributive property!

- anonymous

oh ok i gotchu

- myininaya

you should have it from here. Use the equations sour gave you.

- anonymous

thannk you(:

- anonymous

what if the question's like r= -3sinΘ ?

- myininaya

Try multiplying r on both sides recall rsin(theta)=y and r^2=x^2+y^2

- anonymous

hmm ok...

- anonymous

so r^2= -3rsinΘ?

- myininaya

right so now you are able to replace rsin(theta) with y and r^2 with x^2+y^2

- anonymous

okay

- anonymous

then youd have?

- anonymous

the square root of that right?

- myininaya

just do exactly what I said replace r sin(theta) with y replace r^2 with x^2+y^2

- anonymous

i did tht

- myininaya

you are done what else do you want?

- anonymous

is tht my final answer

- anonymous

oh ok

- myininaya

putting an equation in terms of just x and y ( or just x or just y) is called a cartesian equation if that was the directions to just right as a cartesian equation then you are done

- myininaya

write*

- anonymous

right, sorry. i missed a problem on my homework it was r= 2sinΘ-4cosΘ

- myininaya

multiply both sides by r

- anonymous

do you square both sides 1st

- anonymous

thts what i emant

- myininaya

replace r^2 with x^2+y^2 replace r sin(theta) with y replace r cos(theta) with x

- anonymous

ok i did that: x^2+y^2= 2y-4x

- myininaya

yep. :) that is right you wrote the poloar equation as a cartesian equation if that is all the directions said to do then you are done.

- anonymous

do you stop there

- anonymous

ok

- anonymous

aand i missed r= 3+3cosΘ

- anonymous

so i multiplied both sides by r 1st

- anonymous

and got x^2+y^2=3r +3x but dont i need to get rid of the r?

- myininaya

you could subtract both sides by 3x giving you \[x^2+y^2-3x=3r \] then square both sides and replace the r^2 with x^2+y^2

- anonymous

why square both sides?

- myininaya

I just prefer my answer without radicals and me squaring both sides will allow me to write my answer without radicals

- anonymous

so what would be final answer then

- myininaya

what if you with what I was saying take my equation and then square both sides and then replace the r^2 with x^2+y^2 you will have your final answer

- anonymous

yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?

- myininaya

\[(x^2+y^2-3x)^2=(3r)^2 \] the us what you get when you square both sides your equation is not equivalent to that you squared each term sorta but not really it is whatever you do to one side of equation you can do to the other and the equation still holds

- myininaya

\[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]

- anonymous

so..

- anonymous

im not following are you saying keep left side just squared in parentheses?

- myininaya

It is easier than multiplying it out, don't you think?

- anonymous

yes. so only change right side. like substitue

- myininaya

You can multiply it out if you want. Just do it correctly, but the if the directions just say write as a rectangular (aka cartesian equation) than you are done. If the left hand side was already in terms of x and y. r^2 was the only thing that needed to be converted.

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