Find the arc length of the curve . y=arcsin(x)+sqrt(1-x^2)   0<=x<= 1. show steps.

Asked on by ryghukil

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neela | High School Teacher | (Level 3) Valedictorian

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Arc length = Int (1+y'^2)dx.

y' = 1/sqrt(1-x^2) - x/(sqrt(1-x^2).

1+y'^2  = 1/(1-x^2)+x^2/(1-x^2)-2x/(1-x^2) =(1-x)^2/(1-x^2)

Arc length = Integral sqrt[(1-x)^2/(1-x^2)]dx, x=0 to x=1

=Integral(1-x)/sqrt(1-x^2) dx

=Integral {1/sqrt(1-x^2)} - integral [xdx/sqrt(1-x^2)} bet x=0 to x=1

=arcsinx - Integral [(-1/2)dt/sqrt t], where t = 1-x^2

=arcsinx +(1/2)(-2/1)t^(1/2), between x =0 to x=1.

={[arc sinx - (1-x^2)^(1/2)]x=1}- {arcsimx-(1-x^2)^1/2]at x=0}

={arc sin 1- 0} - {arcsin0-1}

arcsin1 +1 Or



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