`x=arcsint , y=ln(sqrt(1-t^2)) , 0<=t<=1/2` Find the arc length of the curve on the given interval.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial