From Fundamental circular motion how do we get the tangent line when given t?

answered 2014-07-01 16:22:31 -0600

Anna
101 ●3 ●5

updated 2014-07-01 21:28:23 -0600

Mark
166 ●1 ●5 ●12 https://sites.google.com/...

This is what I did for that problem, I'm not sure about it though:

So, we need to fill out the equation $ \vec{p}(t)= \vec{p_0} + \vec{d}t $

I got $ \vec{p_0} $ by plugging the given value of $ t= \pi/3 $ into the equation $ \vec{p}(t) = \langle t+cos(4t), -sin(4t) \rangle $ to get $ \vec{p_0} = \langle \pi/3-1/2, \sqrt{3}/2 \rangle $

Then I got $ \vec{d} $ by taking the derivative of $ \vec{p}(t) $ and the plugging in $ t= \pi/3 $

$ \vec{p}(t)= \langle t +cos(4t), -sin(4t) \rangle $

$ \vec{v}(t)= \langle 1-4sin(4t), -4cos(4t) \rangle $

$ \vec{d}= \langle 1-4sin(4\pi/3),-4cos(4\pi/3 \rangle $

$ \vec{d}= \langle 1+2\sqrt{3},1/2 \rangle $

Then plugging everything into the final equation I got, $$ \vec{p}(t)= \langle \pi/3-1/2,\sqrt{3}/2\rangle + \langle 1+2\sqrt{3},1/2 \rangle t $$

Comment: Looks good!