How do you find the arc length of the curve y=lnx over the interval [1,2]?
2 Answers
Apply the arc length formula.
Explanation:
y=lnx
y'=1/x
Arc length is given by:
L=int_1^2sqrt(1+1/x^2)dx
Rearrange:
L=int_1^2sqrt(x^2+1)/xdx
Multiply numerator and denominator by
L=int_1^2(x^2+1)/(xsqrt(x^2+1))dx
Integration is distributive:
L=int_1^2x/sqrt(x^2+1)dx+int_1^2 1/(xsqrt(x^2+1))dx
Apply the substitution
L=[sqrt(x^2+1)]_ 1^2+intsectheta/tanthetad theta
Simplify:
L=sqrt5-sqrt2+intcscthetad theta
Integrate directly:
L=sqrt5-sqrt2-[ln|csctheta+cottheta|]
Rewrite in terms of
L=sqrt5-sqrt2-[ln|(1+sectheta)/tantheta|]
Reverse the substitution:
L=sqrt5-sqrt2-[ln((1+sqrt(x^2+1))/x)]_1^2
Insert the limits of integration:
L=sqrt5-sqrt2-ln((1+sqrt5)/(2(1+sqrt2)))
Rearrange for clarity:
L=sqrt5-sqrt2+ln2-ln((1+sqrt5)/(1+sqrt2))
Length =
Explanation:
The arc length of a function
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