- #1
don_anon25
- 36
- 0
The problem I am working on asks me to find the curve on the surface z=x^(3/2) which minimizes arc length and connects the points (0,0,0) and (1,1,1).
Here's what I did:
Integral [sqrt(dx^2+dy^2+dz^2)]
Integral [dx sqrt (1+(dy/dx)^2 +(dz/dx)^2]
Integral [dx sqrt (1 + (dy/dx)^2 + 9x/4)] since dz = 3/2 x^(1/2) dx
Thus the "functional" is sqrt (1 + (dy/dx)^2 + 9x/4).
Can I now take derivatives and substitute directly into the Euler-Lagrange equation and solve for y? Where/how do I apply the initial conditions -- that the endpoints are (0,0,0) and (1,1,1)?
Am I on the right track with this one?
Here's what I did:
Integral [sqrt(dx^2+dy^2+dz^2)]
Integral [dx sqrt (1+(dy/dx)^2 +(dz/dx)^2]
Integral [dx sqrt (1 + (dy/dx)^2 + 9x/4)] since dz = 3/2 x^(1/2) dx
Thus the "functional" is sqrt (1 + (dy/dx)^2 + 9x/4).
Can I now take derivatives and substitute directly into the Euler-Lagrange equation and solve for y? Where/how do I apply the initial conditions -- that the endpoints are (0,0,0) and (1,1,1)?
Am I on the right track with this one?