well this one is only 2 dimensional, but the method is not the problem. I know how to do the problem.
What I need to know is this :
The original curve was r(t)= 3cos(t), 15sin(t). Given this curve, to get the length from 0 to 2pi i'd have to take 3 times the integral of the following
How do I integrate: (1+24cos(t))^(1/2)
It is a definite integral from 0 to 2pi. I have been whacking at it for a long time now, but keep running into the problems caused by the fact that
0 and 2 pi are the same angle. I also get sine in the denominator sometimes which of course evaluated at 0 and 2pi is equal to 0, thus giving an undefined answer.
I've tried messing around with half angle laws, trig identities, etc.

Normally I get most problems right away, but I don't think any of my calculus teachers ever taught me how to do this one.
I assume it's a combination of some u substituion and trig identities? idk


Anyway, if anybody could outline the method (Don't have to do it for me, but either way I'd just do different similar exercises for practice if you did) that would be AMAZING

THANKS

i don't think it's expected for anyone to find the indefinite integral for the majority of the arc length integrations in regular calculus
use your calculator to integrate it

t -> (1+24cos(t))^(1/2) is not defined on ]Arccos(-1/24),2*pi - Arccos(-1/24)[ which is a interval of measure =/= 0, hence your function is not integrable on [0,2pi]
Either I understood wrong what you wrote or you made a mistake somewhere

I wrote it down properly, I think you made a mistake. It definitely is integrable, I've tested it with calculators.
If it is solvable (which I'm fairly certain that it is), there must be a way that you can rearrange it to make it work.
Though it might be one of those ones that just can't be done by a person, only a machine that is capable of an approximate answer, which is how I did it.


We don't get to use calculators to integrate them generally in my class. If this one is possible to do in your head, I'll find a way. I already got the answer right by chugging it into the calculator, but I'm one of those people who can't sleep when something like this is unanswered.
I am able to do many of them very similar to this simply in my head, but this one I just can't do without the aid of a calculator, and I figured it must be possible since it wasn't marked differently.
But who knows maybe they expect us to do that. I just hate that they'd assign us to plug it into a calculator. It just isn't learning to me

e: this is a definite integral by the way. I think I listed the limits of integration in my original post.

This post was edited by ringo794 on Jun 17 2014 11:29pm

Your problem is to find the perimeter of an ellipse.

There is no solution using classical functions (algebraic, exponential, trigonometric or other).
This is an example of elliptic integral.

Since the 17th century (Kepler), approximate values have been given.
Some examples : (take a = 3 and b = 15 for your problem)

(Euler) : L = pi * sqrt( 2(a² + b²) )

(Ramanujan) : L = pi * (a+b) * ( 3 - sqrt( 4 - h) )
with h = ( (a-b) / (a+b) )²

Wolframalpha.com gives ~ 63.03

I talked to a lot of people and this was essentially what we came to (yours has more detail ofc) Thanks!
I just plugged the integral into my calculator (something I just hate doing) after I was told by a professor that this could not be solved by means of calc 1-3

i don't think you quite understand what he was getting at
your function is imaginary for a part of that domain, close to when cosine becomes negative

I understand what they are saying, but the integral is able to be solved, and gives a very real answer, which is what I was posting for.

This post was edited by ringo794 on Jun 19 2014 12:59am