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# Calculus/Calculus

Question
Find arc length of curve y=log sec x for x=0 to x=pi/3.

I proceeded thus-
dy/dx=1/sec x*sec x tanx= tan x; (dy/dx)^2=tan^2 x
Arc length=Int Sqrt(1+tan^2 x)dx=Int sec x dx from 0 to pi/3
=log(sec x+ tan x)from 0 to pi/3
=log(sec 60 +tan 60)- log(sec 0 + tan 0)
=log(2+1.732)- log 1=log 3.732 - 0 =1.317.

The answer in the book is 1.732. Am I wrong somewhere or there is a misprint in the book? Please advise

Questioner:MS
Country:Haryana, India
Category:Calculus
Private:No
Subject:Calculus

Question: Find arc length of curve y=log sec x for x=0 to x=pi/3.

I proceeded thus-
dy/dx=1/sec x*sec x tanx= tan x; (dy/dx)^2=tan^2 x
Arc length=Int Sqrt(1+tan^2 x)dx=Int sec x dx from 0 to pi/3
=log(sec x+ tan x)from 0 to pi/3
=log(sec 60 +tan 60)- log(sec 0 + tan 0)
=log(2+1.732)- log 1=log 3.732 - 0 =1.317.

The answer in the book is 1.732. Am I wrong somewhere or there is a misprint in the book? Please advise.
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My advice:  Your computation certainly looks correct to me.  I worked this out and got exactly what you did.

I would have written  ln(sec x) = - ln(cos x) and gone from there, but it makes no difference.

Calculus

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#### Paul Klarreich

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