Mark's Calc III - Individual question feedhttp://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Tue, 01 Jul 2014 16:22:31 -0500From Fundamental circular motion how do we get the tangent line when given t?http://calc3.askbot.com/question/30/from-fundamental-circular-motion-how-do-we-get-the-tangent-line-when-given-t/So I know fundamental circular motion is $\vec{p} = \langle{\cos(t)},{\sin(t)}\rangle$ and we are given the equation: $\vec{p}(t) = \langle{t} + \cos(4t), {− \sin(4t)}\rangle$ and $t=\pi/3$. I want to know how from fundamental motion we can write out a parametrization equation and with that find a line tangent to that point. *Comment*: Does this refer to a particular problem?Mon, 30 Jun 2014 17:57:28 -0500http://calc3.askbot.com/question/30/from-fundamental-circular-motion-how-do-we-get-the-tangent-line-when-given-t/Answer by Mark for From Fundamental circular motion how do we get the tangent line when given t? http://calc3.askbot.com/question/30/from-fundamental-circular-motion-how-do-we-get-the-tangent-line-when-given-t/?answer=32#post-id-32I'm not sure that I understand the relationship between your two $\vec{p}$'s. Can you clarify?Mon, 30 Jun 2014 20:38:05 -0500http://calc3.askbot.com/question/30/from-fundamental-circular-motion-how-do-we-get-the-tangent-line-when-given-t/?answer=32#post-id-32Answer by Anna for From Fundamental circular motion how do we get the tangent line when given t? http://calc3.askbot.com/question/30/from-fundamental-circular-motion-how-do-we-get-the-tangent-line-when-given-t/?answer=42#post-id-42This 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! Tue, 01 Jul 2014 16:22:31 -0500http://calc3.askbot.com/question/30/from-fundamental-circular-motion-how-do-we-get-the-tangent-line-when-given-t/?answer=42#post-id-42