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Equation of the circumference of an ellipse parametric equations

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mickellowery
#1
Jul30-10, 03:08 PM
P: 69
1. The problem statement, all variables and given/known data
Consider the ellipse given by the parametric equation x=3cos(t) y=sin(t) 0[tex]\leq[/tex]t[tex]\leq[/tex]2[tex]\Pi[/tex]. Set up an integral that gives the circumference of the ellipse. Also find the area enclosed by the ellipse.


2. Relevant equations
[tex]\int[/tex][tex]\sqrt{1+(dy/dx)^2}[/tex]dt


3. The attempt at a solution
[tex]\int[/tex][tex]\sqrt{1+(-2/3 cot(t))^2}[/tex]dt It should also be the integral from 0 to 2[tex]\Pi[/tex] I'm not sure what I did wrong but I know that -2/3 cot(t) is not right.
area: A=2[tex]\int[/tex]1/2 (2/3 tan(t))dt
=[tex]\int2/3 tan(t)dt[/tex]
=-2/3ln(lcos(t)l) evaluated from [tex]\Pi[/tex] to 0
I know that I got something wrong here too, and I assume it is the 2/3 tan(t) but I'm not sure what I did wrong again.
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Raskolnikov
#2
Jul30-10, 05:56 PM
P: 193
Your formula for arc length is wrong. It's

[tex]
\int \sqrt{1 + (\frac{dy}{dx})^2} \ dx
[/tex]

[tex]
= \int \sqrt{1 + (\frac{\frac{dy}{dt}}{\frac{dx}{dt}})^2} \ \frac{dx}{dt}dt
[/tex]


EDIT: Pathetic lapse of judgement on my part >_>...ignore this post.
Mark44
#3
Jul30-10, 06:23 PM
Mentor
P: 21,216
Arc length for a parametrized curve can also be written this way:
[tex]\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt[/tex]

gomunkul51
#4
Jul30-10, 06:28 PM
P: 276
Equation of the circumference of an ellipse parametric equations

@Raskolnikov:

[tex]

= \int \sqrt{1 + (\frac{\frac{dy}{dt}}{\frac{dx}{dt}})^2} \ \frac{dx}{dt}dt

[/tex]

how did you get that and how do you plan to integrate it? :)


But as far as I know the ds element for parametrization: f(x(t),y(t)) = x(t) + y(t) is:

[tex]
\int f(x,y) ds = \int f(x(t),y(t)) \sqrt{\left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2}dt
[/tex]
mickellowery
#5
Jul30-10, 07:02 PM
P: 69
OK the way I did it I actually used [tex]\int[/tex][tex]\sqrt{1+dy/dt/dx/dt}[/tex] and that's how I came up with the -[tex]\frac{2}{3}[/tex]cot(t)2 I had -[tex]\frac{2}{3}[/tex][tex]\frac{cos(t)}{sin(t)}[/tex] and I simplified it to cot(t). Would this even be the right formula to use for the circumference? I thought that arc length would be the right choice as long as I evaluated it from 0 to 2[tex]\Pi[/tex].
Mark44
#6
Jul30-10, 07:55 PM
Mentor
P: 21,216
Quote Quote by mickellowery View Post
OK the way I did it I actually used [tex]\int[/tex][tex]\sqrt{1+dy/dt/dx/dt}[/tex]
This isn't the right formula for arc length. If you simplify Raskolnikov's formula in post 2, you get the one I showed in the next post.
Quote Quote by mickellowery View Post
and that's how I came up with the -[tex]\frac{2}{3}[/tex]cot(t)2
I had -[tex]\frac{2}{3}[/tex][tex]\frac{cos(t)}{sin(t)}[/tex] and I simplified it to cot(t). Would this even be the right formula to use for the circumference? I thought that arc length would be the right choice as long as I evaluated it from 0 to 2[tex]\Pi[/tex].
mickellowery
#7
Jul30-10, 08:10 PM
P: 69
Oh geez I just noticed a typo in the original problem. It should be x=3cos(t) y=2sin(t) not y=sin(t) sorry about that.
mickellowery
#8
Jul30-10, 08:29 PM
P: 69
Alright so with the correct equations would the proper integral for the circumference be:

[tex]\int[/tex][tex]\sqrt{(-3sin(t))^2 +(2cos(t))^2}[/tex]dt

And then for the area enclosed by the ellipse would I use [tex]\int[/tex](3cos(t)-2sin(t))2 evaluated from 0 to [tex]\Pi[/tex]?
Mark44
#9
Jul30-10, 08:46 PM
Mentor
P: 21,216
You should simplify the integrand. Also, the limits are from 0 to [itex]2\pi[/itex].

Tip: Put all your LaTeX code inside one pair of tex tags.
Instead of this:
[tex]\int[/tex][tex]\sqrt{(-3sin(t))^2 +(2cos(t))^2}[/tex]

do this:
[tex]\int \sqrt{(-3sin(t))^2 +(2cos(t))^2}dt[/tex]
Dick
#10
Jul30-10, 09:32 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
You should probably also notice that they only asked you so set up the circumference integral, not solve it. It's an elliptic integral. It's not elementary. But you should be able to solve the area integral.


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