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anonymous
 one year ago
help please :(((( supposed to find cartesian equation for x = sin t and y = cos^2 t ... in the range 2pi<t<2pi... I don't understand if this is supposed to be a semi circle or circle?... and what direction the travel is?..
anonymous
 one year ago
help please :(((( supposed to find cartesian equation for x = sin t and y = cos^2 t ... in the range 2pi<t<2pi... I don't understand if this is supposed to be a semi circle or circle?... and what direction the travel is?..

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0x = sin t y = cos^2 t x^2 + y = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now I do understand that equation and when I try to plot points on my graph I end up with a semi circle in quadrant 1 and 2 along the xaxis, but I don't understand what the behavior would be if it's a semi circle, would a particle only be traveling along the half circle from 1 to 1 and then make another loop ? or is the direction assuming two passes along the semi circle.?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0do you really think \(x^2+y=1\) is a circle ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, but Im confused because the range is supposed to be from 2pi to 2pi?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Maybe forget about range for now what do you know about the graph of \(x^2+y=1\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that x^2 is a parabola and y would be a horizontal shift

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0right, try sketching the graph

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0if it helps, split the given range of \(t\) into multiple intervals like : (0, pi/2) (pi/2, pi) (pi, 3pi/2) (3pi/2, 2pi) and see what part of the graph gets traced in each interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry to bother you but I am still having a bit of trouble, I thought I figured it out, but Im still stuck. :(
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