Question

Find the arc length of ellipse

x=4sintheta and y=3costheta

Answer 1

So I know you can say\[\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2}\]

Answer 2

and also \[\frac{ x }{ 4 }=\sin \theta , \frac{ y }{ 3 }= \cos \theta\]

Answer 3

so (x/4)^2 +(y/3)^2 =1

Answer 4

Taking the derivative, you got 4 cos theta and -3 sin theta

Answer 5

@amistre64

Answer 6

id use:
\[\int \sqrt{(x')^2+(y')^2}~dt\]

Answer 7

That isnt the formula in my book? How did you get it and why does it work? Just from the trig identity?

Answer 8

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



s^2 = x^2 + y^2, or simply

s = sqrt(x^2 + y^2)

as s and x and y get small ...