Finding length along ln(x)

  • #1
113
0
I'm currently trying to find the length along function of ln(x) for the heck of it.

I set up this integral for length

L= int/ sqrt(1+(y')^2)

so y'=1/x

so the integral becomes

int/ sqrt(1+(1/x^2)) = int/ sqrt(x^2+1)/x

So I used trig substitution. I set
tanA=x
secA=sqrt(x^2+1)
dx = sec^2(A) dA

so the integral becomes

int/ secA/tanA * sec^2A dA= int/ sec^2A cscA dA

d/dA cscA = -cotAcscA rearrange this and get -dsecA = cscA

so the integral becomes - int/ sec^2A dsecA

set u=secA

so the integral becomes - int/ u^2 du

so -(u^3/3) evaluated from limit a to b

let the original limits be x=1 and x=5
therefore they become A=arctan1 and A=arctan5
and u=sec(arctan(1)) and u=sec(arctan(5))
where sec(arctan(1))=sqrt(2) and sec(arctan(5)=sqrt(26)

However, if you plug these limits in you get a negative answer. Where did I go wrong?
 

Answers and Replies

  • #2
34
2
d/dA cscA = -cotAcscA rearrange this and get -dsecA = cscA

Sorry, but this is meaningless! You need to learn better about differential forms. In reality, we have

[tex]-d\,\sec A=-\sec A\tan A \,dA.[/tex]
 
  • #3
113
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d/dA cscA = -cotAcscA

so dcscA = -cotAcscA dA

d cscA/-cotA = cscA dA

cscA/cotA = secA

-dsecA = cscA dA

Which step is wrong?
 
  • #4
34
2
d cscA/-cotA = cscA dA

This is step is wrong because [tex]-\cot A[/tex] is not a constant with respect to A. So you cannot divide by [tex]-\cot A[/tex] in the differential form.
 
  • #5
113
0
Oh ok, I see that now. Thanks. I'll try a different way of solving this problem.
 
  • #6
34
2
I think this integral cannot be evaluated at all except by numerical methods. Ask your teacher about it. I guess there's a theorem in Differential Algebra that can decide whether an integral may be evaluated in terms of elementary functions or not.
 
  • #7
113
0
I did manage to get this far

int/ sec^2A cscA
= int/ (tan^2A + 1) cscA
= int/ tanAsecA + int/ cscA
= secA + int/ cscA

is this right?

isn't the integral of cscA equal to ln| tan x/2 | ?
 
  • #8
34
2
I think you've got a point. And you can write

[tex]\int\frac{\sqrt{1+x^2}}{x}dx=\sqrt{1+x^2}-\ln\left|\frac{\sqrt{1+x^2}+1}{x}\right|[/tex]

And, if I haven't done any typo,

[tex]\int_1^5\frac{\sqrt{1+x^2}}{x}dx=4.3674894278899,[/tex]

which is the length you desired. Yeah, two heads do think better than one. Mission accomplished! Now, please check my results, noting also that

[tex]\int\csc A\, dA=-\ln|\csc A+\cot A|=\ln|\tan A/2|.[/tex]
 
  • #9
113
0
Yeah, looks right to me. Thanks for the help.
 

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