in general , arc length = ∫ √(1 + (dy/dx)^2) dx from x=a to x=b
for yours:
y = ln(cosx)
dy/dx = -sinx/cosx = -tanx
arc length = ∫ (1 + tan^2 x)^(1/2) dx from 0 to π/6
= ∫ (sec^2 x)^(1/2) dx from 0 to π/6
= ∫ secx dx from 0 to π/6
= | [ ln( (tanx) + secx) ] | from 0 to π/6
= | ln(tan(π/6) + sec π/6) - (ln((tan0) + sec 0) ) |
= | ln (1/√3 + 2/√3) - ln( 0 + 1) |
= | ln(3/√3) - 0 |
= l ln(3) - ln(√3) |
= ln 3 -(1/2)ln 3
= (1/2) ln 3
the integral of secx should be part of your repertoire of basic integrals, if not,
here is the derivation of ∫secx dx = ln(tanx) + secx
https://www.youtube.com/watch?v=d7bAxU5CDtk
Find the arc length of the curve
y=ln(cosx) from
x=0 to
x=π/6.
Length:
1 answer
1 answer