Question

given the length of the curve, find the length of the ellipse (in general )
length of the curve given by
\(\color{blue}{L:\int\limits_{a}^{b}\sqrt{r'(t) r'(t) }dt}\)
where r′(t) is the tanget vector

Answer 1

length of the curve givven by

\(L:\int\limits_{a}^{b}\sqrt{r'(t) r'(t) }dt \)
where \( r'(t)\) is the tanget vector

Answer 2

you mean you want to find the arc length of ellipse?

Answer 3

yes

Answer 4

i don't think you will be able to get the arc length ... since the integral involved will be non elementary.

Answer 5

i used this
\(r(t)=<a\ cos t ,b\ sin t>\)
\(r'(t)=<-a\sin t, b\ cos t\)
\(L:\int\limits_{0}^{2\pi}\sqrt{r'(t) r'(t) }dt\)

Answer 6

yes thats what im talking about i the integral i got was XD

Answer 7

although you can approximate it ..

Answer 8

\(L:\int\limits_{0}^{\pi}\sqrt{(-a\sin t, b\ cos t)(-a\sin t, b\ cos t) }dt \)

Answer 9

So should i use numerical approximation ?

Answer 10

hmm ... you can express it as special function. yes ... numerical approximation definitely works.

Answer 11

but for numerical approximation you need ... the values of 'a' and 'b'

Answer 12

well i defined the function , but i was asked to use that L formula
so i would use approximation for the integral :|
i know i can use other function but u know

Answer 13

\(a=0\)
\(p=\pi\)
or i can calculate the half length of it then × by 2

Answer 14

\[ \int_0^{2\pi} \sqrt{a^2 \sin^2 (t) + b^2 \cos^2 (t)}dt = \]

Answer 15

change cos to sin

Answer 16

it wont work :|
u mean using this
\(sin^2 t=1-cos^2 t\)

Answer 17

yes ... and express it as
\[ \int \sqrt{ 1 - k^2 \sin(t)}dt\]

Answer 18

oh !

Answer 19

http://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind

Answer 20

then assume sin x ?

Answer 21

yes ... you will get this series in 'k'
http://upload.wikimedia.org/math/0/a/4/0a471c07d0e9e6e7a6ba81b6cfbe7349.png

the higher power you choose to evaluate the series ... more accurate you get the arc length of ellipse.

Answer 22

i see !
so it need some work !

Answer 23

assume that b>a, atleast it won't hurt to change 'a' to 'b' since ellipse is symmetric.

you will get
\[
\int_0^{2\pi} \sqrt{a^2 \sin^2(t) + b^2 (1 - \sin^2(t))} \; dt = \int_0^{2\pi} b \sqrt {\left( 1 - \frac{b^2 - a^2}{b^2} \sin^2 (t)\right )} dt \]

yes

Answer 24

well thank you !!!
but i guess its enough to show that
it wont work for general cuz i have no idea
about a,b

Answer 25

\[ 4 b \int_0^{\pi/ 2 } \sqrt{1 - k^2 \sin^2 (t)}dt = 4b \cdot \frac{\pi}{2} \sum_{n=0}^\infty \left[ \frac{(2n)!}{2^{2n}(n!)^2} \right ]^2 \cdot \frac{k^{2n}}{2n-1} \]

Answer 26

where \( k^2 = \frac{b^2 - a^2}{b^2} \)

Answer 27

a, b are just parameters for an ellipse that determiners it's features.

Answer 28

yeah i know :)

Answer 29

thank you again !

Answer 30

yw