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well if x =sint and y = cost at t = pi/2 u get x = 1, y = 0, at t = pi u get x = 0, y = -1, so it seems like we are going clockwise. u also know that an equation of a circle in parametrics is of some form alike to x = rcos/sin, y = rcos/sin so u always guess something of that form. additionally, we can see here that r = 2, since the magnitude of the distance from the origin to the point (sqrt3,1) is 2. also, we need to get that at t = 0, x =sqrt 3 and y = 1. if we guess x =2 sin(t-alpha) and y = 2cos(t-alpha), we can find that alpha = pi/3. so our paramterics in this case would be x= 2sin(t-pi/3) and y = 2cos(t-pi/3).View attachment 43108
Where do you get the intuition to make x=sint and y=cost for clockwise circles?
yeah thats an option. but slightly more efficient way:Would you do a table of values for most ugly 3d vector graphs like these, cause I can't really find another efficent way to go about it. I just feel like in the exam a table of values would take way to long to do accurately
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