danielatha4
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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?
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?