Length of Curve: Find Radius Around (0,0)

[soofjan]

Homework Statement

Find the length of the curve on the sufrace z=xy, whose projection on the xy plane is a 1 radius circle around (0,0).

Homework Equations

The Attempt at a Solution

Length of a curve - usually linear integration of the first kind, when f(x,y,z)=1. So I tried:
x=cos(t), y=sin(t), 0<t<2*pi for the circle, z=cos(t)*sin(t). I got the integrand: sqrt(1+cos(2t)^2), which is not integrable.
Am I missing something, or could this be a mistake in the question..?

Thanks!

 

[jegues]

Homework Statement

Find the length of the curve on the sufrace z=xy, whose projection on the xy plane is a 1 radius circle around (0,0).

Homework Equations

The Attempt at a Solution

Length of a curve - usually linear integration of the first kind, when f(x,y,z)=1. So I tried:
x=cos(t), y=sin(t), 0<t<2*pi for the circle, z=cos(t)*sin(t). I got the integrand: sqrt(1+cos(2t)^2), which is not integrable.
Am I missing something, or could this be a mistake in the question..?

Thanks!

How did you get that integral? The formula for length of a curve as I know it is,

\int^{t_{2}}_{t_{1}} \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2} + (\frac{dz}{dt})^{2}}dt

 

[soofjan]

x' = -sin(t)
y' = cos(t)
z = cos(t)*sin(t) = sin(2t)/2 --> z' = cos(2t)

(x')^2+(y')^2+(z')^2 = [-sin(t)]^2+cos(t)^2+cos(2t)^2 = 1+cos(2t)^2

hence, sqrt[1+cos(2t)^2]

 

[LCKurtz]

Homework Statement

Find the length of the curve on the sufrace z=xy, whose projection on the xy plane is a 1 radius circle around (0,0).

Homework Equations

The Attempt at a Solution

Length of a curve - usually linear integration of the first kind, when f(x,y,z)=1. So I tried:
x=cos(t), y=sin(t), 0<t<2*pi for the circle, z=cos(t)*sin(t). I got the integrand: sqrt(1+cos(2t)^2), which is not integrable.
Am I missing something, or could this be a mistake in the question..?

Thanks!

I get the same thing and you are correct that it is not an elementary integral.

 

[soofjan]

Ok, thanks.

 

[Char. Limit]

If it matters, and it probably doesn't, you can express the integral as an incomplete elliptic integral function of the second kind. It's actually rather interesting.