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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you take the (dx/dt)^2+ (dy/dt)^2 amd then take the ontegral ove rthe inerval is this idea correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A differential element of a curve in the xy 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,\[\frac{ds}{dt}=\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} \rightarrow ds =\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integrating,\[s=\int\limits_{t_1}^{t_2}\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x=e^t \sin t \rightarrow \frac{dx}{dt}=e^t(\cos t  \sin t )\]\[y=e^t \sin t \rightarrow x=e^t \sin t \rightarrow \frac{dy}{dt}=e^t(\sin t + \cos t )\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Squaring and adding each, you have\[\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2=2e^2t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your arc length is\[s=\int\limits_{0}^{2}\sqrt{2}e^tdt=\sqrt{2}(e^21)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*should be 2e^(2t) in secondlast part of derivation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you've got some errors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x= e^t cos t , not e^t sin t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i got that :D pretty neat calculus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, transcribing from notes to this site hardly ever goes to plan.
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