[Firam]

The problem is to find the arc length of ln (1-x^2) from 0 to Pi/4.

I got to sqrt [(1 + (4x^2)/(x^2-1)^2]. I thought it was legal to multiply by (x^2-1)^2 still within the sqrt, so that it cancels out. And I get (x^2-1)^2 + 4x^2.

My calculator disagrees. At the end I get (Pi^3)/192 + Pi/4 or roughly .947 but the calc says the answer should be 1.33.

 

[joneeboi]

Did you type that out right? I have no idea why you are finding the square root. Is it because you're finding the arc length?
Did you try

ln(1-x^2) = ln[(1-x)(1+x)] ?

Also, where is that (4x^2)/(x^2-1)^2 coming from?

With all these questions, am I even helping? Lool.

 

[vagarach]

Ok no. I totally misread the definition of arc length. But I confirmed your calculator is correct, and that your 'cancelling' thing is wrong, but it is legal to multiply whereever you please by (x^2-1)^2/(x^2-1)^2.

I would work further, but I have a stats exam to write tomorrow.

 

[Firam]

Quote:

Originally Posted by vagarach Just apply the log law ln(ab) = ln(a) + ln(b). Then apply the property of additivity of integrals, and use integration by parts. Done deal. (this all assumes you have some calculus knowledge, duh )

Joneeboi, no you are not helping. Thanks for trying but I just put in part of the problem first I had to do a derivative and then square it. That then goes into the integral or more specifically the definite integral of sqrt (1+[That squared derivative]).

I'm in Calc II and I know integration by part but I can't do it while I'm stuck under the square root. It seemed too simple for an extra credit problem. I think I might be missing somthing.


2. Find the length of the curve:

y = ln(1-x2) , 0 [less than or equal to] x [less than or equal to] Pi/4.

Edit: You figured out you were talking about the wrong thing.

 

[vagarach]

One way would be to add the 1 and the 4x^2 / (1-x^2)^2 together. Take the root of the top and the bottom. Then you get sqrt( 1 + 2x^2 + x^4) / (1-x^2).
Now you can then do integration by parts, since d/dx [arctanh(x)] = 1/1-x^2.

Hope this helps.

 

[Firam]

Quote:

Originally Posted by vagarach One way would be to add the 1 and the 4x^2 / (1-x^2)^2 together. Take the root of the top and the bottom. Then you get sqrt( 1 + 2x^2 + x^4) / (1-x^2). Now you can then do integration by parts, since d/dx [arctanh(x)] = 1/1-x^2. Hope this helps.

I don't think I ever have somthing that looks like that. I got to the integral of (x^2+1) * |1/(x^2-1)|. Again I'm stuck.

 

[Firam]

I i removed the road block and split it up and came up with an integral I could solve and this one:

Integral of (x^2)/(1-x^2)

It looks better but still not there.

 

[pne]

my calculus is rusty but it does look like it would be easier if you used arctan..

 

[Firam]

Quote:

Originally Posted by pne my calculus is rusty but it does look like it would be easier if you used arctan..

arctan is 1/(1+x^2), which is not what I have unless I don't understand.

 

[digitalcat]

Isn't it the integration of (1+X^2)/(1-X^2) from 0 to pi/4 according to the arc length formula?

Edit: oops, little mistake, that should be the absolute value of (1+X^2)/(1-X^2), although it doesn't matter here. It integrates into -X-log(1-X)+log(1+X). Evaluate it at pi/4 and 0, take the difference and it is 1.3332