find arc legnth of curve
x=arcsint , y=ln sqrt(1-t^x)
- formula S b,a sqrt(f'(t)^2+g'(t)^2
- i got down to S0,1/2 1/x^2-1 if that is right but dont know what to do next.
Write integral for area of surface about the x-axis
x=(1/4)t^2 , y=t+2 , 0<=t<=4
- formula 2pi S a,b g(t) sqrt(dx/dt)^2+(dy/dt)^2
- about about half way and am stuck.
Please help fast as I will have to leave soon.
2007-06-22 03:13:00 · 1 answers · asked by fcb10121 1 in Science & Mathematics ➔ Mathematics
In your first problem are you sure you meant to write y=ln(sqrt(1-t^x)) instead of y = ln(sqrt(1-t^2))
because that makes it godawful complicated.
On the other hand, if it's squared rather than to the power of x, I've got sqrt(2) int(0,1) 1/((1-t^2)^1/4) dt
which is not too tough to figure.
(why did you integrate from 0 to 1/2? the curve exists on t in [0,1). you didn't list any bounds)
2) What is g(t) in this formula meant to represent?
2007-06-25 20:19:03 · answer #1 · answered by kozzm0 7 · 0⤊ 0⤋