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nokia8650
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x = 2cot t
y = (sin t)^2
t is greater than 0 but less than or equal to pi/2
The cartesian can be found using trig identities to be:
y = 8/ (4+ x^2)
What would be the range of the cartesian equation? I think it would be x is greater than or equal to 0, since when t = pi/2, x = 0, and as t tends to 0, x tends to infinity.
Am I correct?
Thank you.
y = (sin t)^2
t is greater than 0 but less than or equal to pi/2
The cartesian can be found using trig identities to be:
y = 8/ (4+ x^2)
What would be the range of the cartesian equation? I think it would be x is greater than or equal to 0, since when t = pi/2, x = 0, and as t tends to 0, x tends to infinity.
Am I correct?
Thank you.