Question

find arclength:
x=arcsint
y=ln square root(1-x^2)
x is between 0 and 1/2

Answer 1

\[\int_{sin(0)}^{sin(1/2)}\sqrt{(x')^2+(y')^2}dt\] maybe?

Answer 2

x = arcsint defines x as an angle

Answer 3

you sure you got this posted right?

Answer 4

ya, oopsss. it should be y= ln(square root of( 1-(t^2))

Answer 5

that does make a little more sense to me :)

Answer 6

arcsin' = 1/sqrt(1-t^2) , if my memory serves me

Answer 7

x = arcsint

sinx = t

cosx x' = 1

x' = 1/cosx = 1/sqrt(1-sin^2x) = 1/sqrt(1-t^2)

Answer 8

y= ln(square root of( 1-(t^2))

y = 1/2 ln (1-t^2)

2y = ln(1-t^2)

2y' = -2t/(1-t^2)

y' = -t/(1-t^2)

Answer 9

squre those up; add them; and sqrt the results :)

Answer 10

thanx