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AmTran_Bus
 one year ago
Find the arc length of ellipse
x=4sintheta and y=3costheta
AmTran_Bus
 one year ago
Find the arc length of ellipse x=4sintheta and y=3costheta

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AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0So I know you can say\[\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2}\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0and also \[\frac{ x }{ 4 }=\sin \theta , \frac{ y }{ 3 }= \cos \theta\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0so (x/4)^2 +(y/3)^2 =1

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Taking the derivative, you got 4 cos theta and 3 sin theta

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1id use: \[\int \sqrt{(x')^2+(y')^2}~dt\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0That isnt the formula in my book? How did you get it and why does it work? Just from the trig identity?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1its the general setup for the arclength "s", which is just the pythag thrm s, the section, squared; is equal to sum of the square of the parts dw:1438546179447:dw s^2 = x^2 + y^2, or simply s = sqrt(x^2 + y^2) as s and x and y get small ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438546232411:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ds = sqrt(dx^2 + dy^2) now when all of these are "with respect to x", then dx=1 ds = sqrt(1 + (y')^2) is the initial introducion

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if it is with respect to "t", dx = x'

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Ok, well thank you then. So all I have is\[\int\limits_{0}^{2\pi}\sqrt{16\cos^2 \theta+9\sin^2 \theta }\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Ok. And that is an answer choice.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1lets hope memory serves us right then lol

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Yea, for real. I apparently do not get this real well, cause its still fuzzy why every other example they have said use the (dy/dx)^2 thing.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0That is, the very first one I posted.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we could define y in terms of x if you are working with parameters of course .. but its just extra work

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the thing about arclength is, we dont want to be confined to a coordinant system ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1anyways, thats my perspective :)
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