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The parametric equations of a curve are x=2cos t and y=2sin t, for 0 <= t <2 pi. What is the value of t at the point (0,2)?
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\[x=2\cos ~t\]\[y=2\sin~t\] \[0 \le t < \pi\]
You taking A levels?
Yes. How did you know?
got the feel of it
ok, so just differentiate x and y
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x=2cost dx/dt=-2sint -------------------- y=2sint dy/dt=2cost
then \[\frac{dy}{dx}=\frac{2cost}{-2sint}\] \[\frac{dy}{dx}=-\cot~t\]
\[y-y_1=m(x-x_1)\] \[y-2=-\cot~t(x-0)\] \[y=-xcot ~t +2\]
Try Substitute (0,2) into x=2cos t and y=2sin t
t should be pi/2
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Ok, that's the answer. but why?
Oh, I know why. Thank you! :D
Because x=2cos t and y=2sin t is a part of the curve, and y=−xcot t+2 also part of a curve, So, substitute (0,2) into both equation will get the same value
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