# `x=arcsint , y=ln(sqrt(1-t^2)) , 0<=t<=1/2` Find the arc length of the curve on the given interval. Arc length of a curve C described by the parametric equations x=f(t) and y=g(t), `a<=t<=b` where f' and g' are continuous on [a,b] and C is traversed exactly once as t increases from a to b, then the length of the curve is given by,

`L=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt`

We are given:`x=arcsin(t),y=ln(sqrt(1-t^2)), 0<=t<=1/2`

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Arc length of a curve C described by the parametric equations x=f(t) and y=g(t), `a<=t<=b` where f' and g' are continuous on [a,b] and C is traversed exactly once as t increases from a to b, then the length of the curve is given by,

`L=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt`

We are given:`x=arcsin(t),y=ln(sqrt(1-t^2)), 0<=t<=1/2`

`x=arcsin(t)`

`dx/dt=1/sqrt(1-t^2)`

`y=lnsqrt(1-t^2)`

`dy/dt=1/sqrt(1-t^2)d/dt(sqrt(1-t^2))`

`dy/dt=1/sqrt(1-t^2)(1/2)(1-t^2)^(1/2-1)(-2t)`

`dy/dt=-t/(1-t^2)`

Now let's evaluate arc length by using the stated formula,

`L=int_0^(1/2)sqrt((1/sqrt(1-t^2))^2+(-t/(1-t^2))^2)dt`

`L=int_0^(1/2)sqrt(1/(1-t^2)+t^2/(1-t^2)^2)dt`

`L=int_0^(1/2)sqrt((1-t^2+t^2)/(1-t^2)^2)dt`

`L=int_0^(1/2)sqrt(1/(1-t^2)^2)dt`

`L=int_0^(1/2)1/(1-t^2)dt`

`L=int_0^(1/2)1/((1+t)(1-t))dt`

Using partial fractions integrand can be written as :

`L=int_0^(1/2)1/2(1/(1+t)+1/(1-t))dt`

Take the constant out and use the standard integral:`int1/xdx=ln|x|+C`

`L=1/2int_0^(1/2)(1/(1+t)+1/(1-t))dt`

`L=1/2[ln|1+t|+ln|1-t|]_0^(1/2)`

`L=1/2{[ln|1+1/2|+ln|1-1/2|]-[ln1+ln1]}`

`L=1/2[ln|3/2|+ln|1/2|]`

`L=1/2[0.4054651081+0.69314718056]`

`L=1/2[1.09861228867]`

`L=0.54930614433`

Arc length of the curve on the given interval is `~~0.549`

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