- #1
DarkSamurai
- 7
- 0
Homework Statement
[tex]x = \frac{u^{2} + v^{2}}{2}[/tex]
[tex]y = uv[/tex]
[tex]z = z[/tex]
Find the arc length given:
[tex]u(t) = cos(t), v(t) = sin(t), z = \frac{2t^{\frac{3}{2}}}{3}[/tex]
Homework Equations
[tex]ds^{2} = dx^{2} + dy^{2} + dz^{2}[/tex]
In curvilinear coordinates thhis becomes
[tex]ds = \sqrt{h^{2}_{1}du^{2}_{1} + h^{2}_{2}du^{2}_{2} + h^{2}_{3}du^{2}_{3}}[/tex]
The Attempt at a Solution
First I need to get the scale factors, so I took the derivative of each x, y, z component.
I came up with:
[tex]dx = udu - vdv[/tex]
[tex]dy = vdu + udv[/tex]
[tex]dz = dz[/tex]
I then found the scale factors,
[tex]h_{1} = h_{u} = \sqrt{u^{2} + v^{2}}[/tex]
[tex]h_{2} = h_{v} = \sqrt{u^{2} + v^{2}}[/tex]
[tex]h_{3} = h_{z} = 1[/tex]
Then we inject the scale factors into the element arc length formula.
[tex]ds = \sqrt{h^{2}_{1}du^{2}_{1} + h^{2}_{2}du^{2}_{2} + h^{2}_{3}du^{2}_{3}}[/tex]
I'm not sure what to do about du1, du2, and du3. Are they just dx, dy and dz? And if so, would this mean I have to integrate 3 times to get the arc length?