music101
Convert the polar equation to rectangular form and identify the graph.
r= 7/ (2cosΘ+5sinΘ)
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music101
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sourwing
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y = r sin(theta)
x = r cos(theta)
x^2 + y^2 = r^2
plug and chug
music101
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yea so how would i plug it in
music101
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cuz i did this last week and i kinda forgot sorry
myininaya
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Try getting your cos and sin with r over there by multiplying both sides by 2cos(theta)+5sin(theta)
music101
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okay. i did that
myininaya
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So you have:
\[2rcos(\theta)+5rsin(\theta)=7 ?\]
music101
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why would you put r w/ each one of the terms isnt there just 1?
myininaya
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\[r=\frac{7}{2\cos(\theta)+5 \sin(\theta)} => r(2\cos(\theta)+5 \sin(\theta))=7\]
Distributive property!
music101
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oh ok i gotchu
myininaya
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you should have it from here. Use the equations sour gave you.
music101
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thannk you(:
music101
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what if the question's like r= -3sinΘ ?
myininaya
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Try multiplying r on both sides
recall rsin(theta)=y
and r^2=x^2+y^2
music101
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hmm ok...
music101
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so r^2= -3rsinΘ?
myininaya
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right so now you are able to replace rsin(theta) with y and r^2 with x^2+y^2
music101
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okay
music101
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then youd have?
music101
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the square root of that right?
myininaya
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just do exactly what I said replace r sin(theta) with y
replace r^2 with x^2+y^2
music101
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i did tht
myininaya
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you are done
what else do you want?
music101
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is tht my final answer
music101
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oh ok
myininaya
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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
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write*
music101
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right, sorry. i missed a problem on my homework it was r= 2sinΘ-4cosΘ
myininaya
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multiply both sides by r
music101
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do you square both sides 1st
music101
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thts what i emant
myininaya
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replace r^2 with x^2+y^2
replace r sin(theta) with y
replace r cos(theta) with x
music101
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ok i did that: x^2+y^2= 2y-4x
myininaya
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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.
music101
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do you stop there
music101
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ok
music101
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aand i missed r= 3+3cosΘ
music101
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so i multiplied both sides by r 1st
music101
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and got x^2+y^2=3r +3x but dont i need to get rid of the r?
myininaya
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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
music101
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why square both sides?
myininaya
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I just prefer my answer without radicals and me squaring both sides will allow me to write my answer without radicals
music101
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so what would be final answer then
myininaya
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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
music101
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yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?
myininaya
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\[(x^2+y^2-3x)^2=(3r)^2 \]
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
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\[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]
music101
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so..
music101
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im not following are you saying keep left side just squared in parentheses?
myininaya
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It is easier than multiplying it out, don't you think?
music101
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yes. so only change right side. like substitue
myininaya
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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.