AmTran_Bus
  • AmTran_Bus
Find the arc length of ellipse x=4sintheta and y=3costheta
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AmTran_Bus
  • AmTran_Bus
So I know you can say\[\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2}\]
AmTran_Bus
  • AmTran_Bus
and also \[\frac{ x }{ 4 }=\sin \theta , \frac{ y }{ 3 }= \cos \theta\]
AmTran_Bus
  • AmTran_Bus
so (x/4)^2 +(y/3)^2 =1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

AmTran_Bus
  • AmTran_Bus
Taking the derivative, you got 4 cos theta and -3 sin theta
AmTran_Bus
  • AmTran_Bus
@amistre64
amistre64
  • amistre64
id use: \[\int \sqrt{(x')^2+(y')^2}~dt\]
AmTran_Bus
  • AmTran_Bus
That isnt the formula in my book? How did you get it and why does it work? Just from the trig identity?
amistre64
  • amistre64
its 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
  • amistre64
|dw:1438546232411:dw|
amistre64
  • amistre64
ds = 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
  • amistre64
if it is with respect to "t", dx = x'
AmTran_Bus
  • AmTran_Bus
Ok, 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
  • AmTran_Bus
*d theta
amistre64
  • amistre64
yep
AmTran_Bus
  • AmTran_Bus
Ok. And that is an answer choice.
AmTran_Bus
  • AmTran_Bus
Thanks.
amistre64
  • amistre64
lets hope memory serves us right then lol
AmTran_Bus
  • AmTran_Bus
Yea, 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
  • AmTran_Bus
That is, the very first one I posted.
amistre64
  • amistre64
we could define y in terms of x if you are working with parameters of course .. but its just extra work
AmTran_Bus
  • AmTran_Bus
Ok.
amistre64
  • amistre64
the thing about arclength is, we dont want to be confined to a coordinant system ...
amistre64
  • amistre64
anyways, thats my perspective :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.