Author Topic: Sketching the Path (Read 585 times) Tweet Share

Gloamglozer · « **on:** April 03, 2011, 03:10:26 pm »

I think I've had a face palm moment. It says:

"Sketch the path of $(e^{-t}\cos(t), e^{-t}\sin(t), 0)$ for $0\leq t\leq 3\pi$

I let $x= e^{-t}\cos(t)$ and $y=e^{-t}\sin(t)$

And I got to $cos^{2}(t) + sin^{2}(t) = e^{2t}$ which doesn't sound right. Can someone please tell me how to do this?

m@tty · « **Reply #1 on:** April 03, 2011, 03:37:19 pm »

$x=e^{-t}\cos(t)$

$y=e^{-t}\sin(t)$

$x^2+y^2 = e^{-2t}(\cos^2(t)+\sin^2(t))=e^{-2t}$

Which is a circle with radius decreasing as t increases.

Now you can use the parametric equations to figure out the start/end point.

Start: $(1, 0, 0)$ sub t = 0

End: $(0, e^{-3\pi}, 0)$ sub t = 3pi

Then to sketch you can ignore the third dimension as the path doesn't move in that direction..

Oh, and the circle goes anti clockwise..

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Author Topic: Sketching the Path (Read 585 times) Tweet Share

Gloamglozer

Sketching the Path

m@tty

Re: Sketching the Path

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