Write the following equation in polar coordinates (using t for ): (x2 + y2)2 = x2 - y2 becomes ____?____ - cos(2t) = 0 I know that r2=x2+y2 but i'm not sure where everything else leads me.
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Originally Posted by mickles Write the following equation in polar coordinates (using t for ): (x2 + y2)2 = x2 - y2 becomes ____?____ - cos(2t) = 0 I know that r2=x2+y2 but i'm not sure where everything else leads me. $\displaystyle x^2+y^2=r^2$ $\displaystyle x=r\cos(t)$ $\displaystyle y=r\sin(t)$ Try this subs and see what happens
Originally Posted by TheEmptySet $\displaystyle x^2+y^2=r^2$ $\displaystyle x=r\cos(t)$ $\displaystyle y=r\sin(t)$ Try this subs and see what happens Okay i did this: r^2 = r^2cos^2(x)+r^2sin^2(x) 1 = cos^2(x) +sin^2(x) So 1 = cosx + sinx
I also inserted x and y into the equation in my first post and i got this: cos^2(x)-sin^2(x) = (r^2cos^4(x)+2r^2cos^2(x)sin^2(x)+r^2sin^4(x))
Originally Posted by mickles Write the following equation in polar coordinates (using t for ): (x2 + y2)2 = x2 - y2 becomes ____?____ - cos(2t) = 0 I know that r2=x2+y2 but i'm not sure where everything else leads me. Remember that $\displaystyle \cos^2(t)-\sin^2(t)=\cos(2t)$ so we get $\displaystyle (r^2)^2=r^2\cos^2(t)-r^2\sin(t) \iff r^4=r^2(\cos^2(t)-\sin^2(t))$ $\displaystyle r^4=r^2(\cos(2t)) \iff r^2=\cos(2t) \iff r^2-\cos(2t)=0$
Originally Posted by TheEmptySet Remember that $\displaystyle \cos^2(t)-\sin^2(t)=\cos(2t)$ so we get $\displaystyle (r^2)^2=r^2\cos^2(t)-r^2\sin(t) \iff r^4=r^2(\cos^2(t)-\sin^2(t))$ $\displaystyle r^4=r^2(\cos(2t)) \iff r^2=\cos(2t) \iff r^2-\cos(2t)=0$ Oh okay i understand now. I forgot the identity where cos^2(2t)-sin^2(2t) = cos(2t) Thank you very much i understand now!
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