## anonymous one year ago STuck? if cartesian equation of x= sint, y=cos^2t is = x^2 + y = 1... how does the graph in the range of -2pi <t<2pi end up being an upside down parabola with values that go beyond -1 and 1?

is you r question, the graph being a parabola or having a range beyond -1 and 1?

2. anonymous

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3. anonymous

So am I trying to graph with incorrect points or is my graph correct?

yes its clear that x can only take values between -1 and 1 since $$-1\leq \sin(t) \leq 1$$ so your graph is correct

5. anonymous

so if my professor asked me to identify the behavior of the graph according to the cartesian equation I derived from these x and y functions... I could tell him that a particle traveling along the path of this graph would actually travel clockwise then bounce back and travel counterclockwise between the -1 and 1 on the x-axis..???? does that make sense?

6. anonymous

because thats' mainly my thing, Im trying to understand the behavior of how a particle would travel along this path. starting at -2pi and ending at 2pi.. and since my graph starts and ends at -1 and 1... would my conclusion be correct? (about the bouncing back and forth)

7. anonymous

are you still there?

yes the idea of travelling is derived from the time - t So if we take the distance on x direction, the relationship of that with the time is given as $$x = \sin(t)$$ so there itself it says that the x change in a oscillatory way

9. anonymous

ok. fantastic!!!!!!! thank you. I just wanted to make sure I was on the right track. And you seem pretty confident. :) Thank you very much.

you are welcome