MHB Graph $r(t)$ for t = $\pi/4$: Position and Tangent Vectors

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The discussion centers on graphing the position vector $r(t) = \sin(t)i + 2\cos(t)j$ and its tangent vector at $t = \pi/4$. The position vector at this point is $r(\pi/4) = \frac{\sqrt{2}}{2}i + \sqrt{2}j$, while the tangent vector is $r'(\pi/4) = \frac{\sqrt{2}}{2}i - \sqrt{2}j$. Participants highlight that the original equation represents an ellipse, with the graph extending from -1 to 1 on the x-axis and -2 to 2 on the y-axis. The conversation also notes successful contributions from users in helping to visualize these vectors. Overall, the thread emphasizes the importance of understanding both the position and tangent vectors in relation to the graph of the ellipse.
ineedhelpnow
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$r(t)=sin(t)$i $+ 2cos(t)$j
$t= \pi/4$

sketch the position vector and the tangent vector

$r'(t)=cos(t)$i $- 2sin(t)$j

$r(\pi/4)= \frac{\sqrt{2}}{2}$i $+ \sqrt{2}$j

$r'(\pi/4)= \frac{\sqrt{2}}{2}$i $- \sqrt{2}$j

$\left\langle \frac{\sqrt{2}}{2}, \sqrt{2} \right\rangle$
$\left\langle \frac{\sqrt{2}}{2}, - \sqrt{2} \right\rangle$

can someone help me graph these. the original equation is the graph of a ellipse/circle from -1 to 1 on the x-axis and -2 to 2 on the y axis
 
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ineedhelpnow said:
$r(t)=sin(t)$i $+ 2cos(t)$j
$t= \pi/4$

sketch the position vector and the tangent vector

$r'(t)=cos(t)$i $- 2sin(t)$j

$r(\pi/4)= \frac{\sqrt{2}}{2}$i $+ \sqrt{2}$j

$r'(\pi/4)= \frac{\sqrt{2}}{2}$i $- \sqrt{2}$j

$\left\langle \frac{\sqrt{2}}{2}, \sqrt{2} \right\rangle$
$\left\langle \frac{\sqrt{2}}{2}, - \sqrt{2} \right\rangle$

can someone help me graph these. the original equation is the graph of a ellipse/circle from -1 to 1 on the x-axis and -2 to 2 on the y axis

Let's see...

[desmos="-4,4,-3,3"]x^2+y^2/4=1;y=2x\left\{0<x<\frac{\sqrt{2}}{2}\right\};y=2\sqrt{2}-2x\left\{\frac{\sqrt{2}}{2}<x<\sqrt{2}\right\}[/desmos]
 
ok so i literally posted this question on a bunch of different sites and once again MHB was the first to succeed :) thanks ILS!
 
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