Question
the path of a particle in the xy-plane is vector r = (cos2t, sint) for t for all [-pi/2, pi] where t represents time. Sketch the path. Is it a smooth curve?
How do i sketch the path? do i just plug in random points between [-pi/2, pi]? how would i connect it together? and how do i know if it's a smooth curve? Thank you.
How do i sketch the path? do i just plug in random points between [-pi/2, pi]? how would i connect it together? and how do i know if it's a smooth curve? Thank you.
Answers
If that pair of coordinates represents (X,Y), then the easiest way is to do pretty much what you already suggested, but do it systematically in small increments. That is, work out a set of points as follows. -pi/2 is approximately -1.57, so
t=-1.57, X=cos(2t)=-1, Y=sin(t)=-1
t=-1.4, X=cos(2t)=-0.942, Y=sin(t)=-0.985
t=-1.3, X=cos(2t)=-0.856, Y=sin(t)=-0.963
etc etc. Then just plot X vs Y. If you've got access to a spreadsheet you'll find it very quick to do that.
Just one question though: are you sure that interval isn't [-pi/2, +pi/2]? If so, then yes, it's a smooth curve between those limits. But if you're correct about the upper limit for t being pi, then the curve is rather more complicated: it runs smoothly from (-1,1) through (0,0), up to (-1,1) and then bounces back on itself again to (0,0), overwriting itself in the process. If I had to bet, I'd guess that the question has been copied down wrong.
t=-1.57, X=cos(2t)=-1, Y=sin(t)=-1
t=-1.4, X=cos(2t)=-0.942, Y=sin(t)=-0.985
t=-1.3, X=cos(2t)=-0.856, Y=sin(t)=-0.963
etc etc. Then just plot X vs Y. If you've got access to a spreadsheet you'll find it very quick to do that.
Just one question though: are you sure that interval isn't [-pi/2, +pi/2]? If so, then yes, it's a smooth curve between those limits. But if you're correct about the upper limit for t being pi, then the curve is rather more complicated: it runs smoothly from (-1,1) through (0,0), up to (-1,1) and then bounces back on itself again to (0,0), overwriting itself in the process. If I had to bet, I'd guess that the question has been copied down wrong.
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