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I'm trying to calculate the curvlenght for y=ln(cosx) , 0<=x<=pi/6. But I can not figure out what I'm doing wrong.
Using sec(x) is not allowed (because we don't learn that here in Sweden).
Using formula ds = squareroot(1 + (dy/dx)^2) dx. dy/dx=-sinx/cosx
squareroot( 1 + (dy/dx)^2 ) dx = squareroot( 1 + (-sinx/cosx)^2 ) dx = squareroot( 1 + (tanx)^2 ) dx =
// 1+tan^2(x) = 1/cos^2(x) //
= squareroot( 1/cos^2(x) ) dx = ( squareroot(1) ) / ( squareroot(cos^2(x) ) dx = 1/cosx dx =
// 1/cosx = cosx/cos^2(x) //
= cosx/cos^2(x) dx =
//cos^2(x) = 1- sin^2(x) //
= cosx/( 1 - sin^2(x) ) dx = 1/( 1 - sin^2(x) * cosx dx
// Using u = sinx , du = cosx dx //
= 1/( 1 - u^2 ) du = 1/( (1 - u)(1+u) ) du =
// 1/( (1 - u)(1+u) ) = ( A/(1-u) ) + ( B/(1+u) ) => A+Au+B-Bu = 1 => A+B=1 and A-B=0 <=> A=1/2 , B=1/2 //
1/( 1 - u^2 ) du = (1/2) * ( (1/(1+u) + 1/(1-u) ) du = (1/2) * ( ln|1+u| + ln|1-u| ) =
// ln(A) + ln(B) = ln(A*B) , (1+u)(1-u)=1-u^2 //
= (1/2) * ln|1-u^2| =
// changing back u=sinx //
= (1/2) * ln| 1-sin^2(x) | = (1/2) * ln| cos^2(x) | = ln|cosx|
ds = squareroot(1 + (dy/dx)^2) dx , 0<=x<=pi/6 <=> ln| cosx | 0<=x<=pi/6
But when calculating ln| cos(pi/6) | - ln|1| I get -0,14... The correct answer should be (1/2)*ln(3) which is about 0,5
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Hi!
What I see is that:
Hope it helps!
Jose
Make everything as simple as possible, but not simpler. -- Albert Einstein
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. . . Really?
. . . . .
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