A particle travels in a path such that the position vector
)
at time
is given by
=3\cos{(t)}\mathbf{i}+2\sin{(t)}\mathbf{j}, t\ge 0)
If the positive y axis points north and the positive x axis point east, find correct to two decimal places, the bearing of point
, the position of the particle at
from:
The originAlright I was able to find this one out fairly easily, I found the point and the quadrant the point was in, 4th quadrant point
)
. Then I knew that the angle I find needs to add
to it, so I went..
}&=\frac{\sqrt{2}}{\frac{3}{\sqrt{2}}}\\<br />\theta&=\tan^{-1}{\left ( \frac{2}{3} \right )}\\<br />&=33.69^o<br />\end{align*})
I then added
to it and I got
which is the answer.
Now here is the part I was having trouble with
The initial positionThe initial position vector was
=3\mathbf{i})
which I then converted to cartesian form to find the bearing and I got
)
. Now my assumption was because it was on the x-axis I just subtract
from my previous answer to get the bearing but that gave me a different answer.
Appreciate any help
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