Find the length of the parametric curve curve x = et cos t, y = et sin t for 0 <= t <= 2.

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Find the length of the parametric curve curve x = et cos t, y = et sin t for 0 <= t <= 2.

Mathematics
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you take the (dx/dt)^2+ (dy/dt)^2 amd then take the ontegral ove rthe inerval is this idea correct
A differential element of a curve in the x-y plane, ds, is given in terms of the differentials of x and y by Pythagoras' Theorem. \[(ds)^2=(dx)^2+(dy)^2\]Dividing both sides of the equation by (dt)^2, you have\[\left( \frac{ds}{dt} \right)^2=\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2\]and from this,\[\frac{ds}{dt}=\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} \rightarrow ds =\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]
Integrating,\[s=\int\limits_{t_1}^{t_2}\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]

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yup :)
\[x=e^t \sin t \rightarrow \frac{dx}{dt}=e^t(\cos t - \sin t )\]\[y=e^t \sin t \rightarrow x=e^t \sin t \rightarrow \frac{dy}{dt}=e^t(\sin t + \cos t )\]
Squaring and adding each, you have\[\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2=2e^2t\]
So your arc length is\[s=\int\limits_{0}^{2}\sqrt{2}e^tdt=\sqrt{2}(e^2-1)\]
*should be 2e^(2t) in second-last part of derivation.
you've got some errors
x= e^t cos t , not e^t sin t
otherwise its great
yeah i got that :D pretty neat calculus
well, transcribing from notes to this site hardly ever goes to plan.
:)

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