Question

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)?

Answer 1

\[x=2\cos ~t\]\[y=2\sin~t\]
\[0 \le t < \pi\]

Answer 2

You taking A levels?

Answer 3

Yes. How did you know?

Answer 4

got the feel of it

Answer 5

ok, so just differentiate x and y

Answer 6

x=2cost
dx/dt=-2sint
--------------------
y=2sint
dy/dt=2cost

Answer 7

then \[\frac{dy}{dx}=\frac{2cost}{-2sint}\]
\[\frac{dy}{dx}=-\cot~t\]

Answer 8

\[y-y_1=m(x-x_1)\]
\[y-2=-\cot~t(x-0)\]
\[y=-xcot ~t +2\]

Answer 9

Try Substitute (0,2) into x=2cos t and y=2sin t

Answer 10

t should be pi/2

Answer 11

Ok, that's the answer. but why?

Answer 12

Oh, I know why. Thank you! :D

Answer 13

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