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Mathematics 14 Online
OpenStudy (lukecrayonz):

Eliminate the parameter and write the correspnding rectangular equation whose graph represents the curve. x=sqrt(t) y=1-t

OpenStudy (lukecrayonz):

I got the x, t=x^2, but the y..

OpenStudy (anonymous):

substitute that into the second equation... y=1-x^2

OpenStudy (lukecrayonz):

Oh.. really? lmfao

OpenStudy (lukecrayonz):

I swear I look at things and put WAY too much thought into them.

OpenStudy (lukecrayonz):

@dpaInc how about for x=4+2cos(theta) and y=-1+2sin(theta)?

OpenStudy (anonymous):

these are polar equations... change back using x^2+y^2=r^2 tan(theta)=y/x x=rcostheta y=rsintheta

OpenStudy (lukecrayonz):

x=4+2cos(theta)-->(x-4/2)^2=cos^2theta

OpenStudy (lukecrayonz):

((x-4)/2)^2+((y-1)/2))^2=1

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