Partial Solutions to Brainfood #10

These are partial answers so that you can check to see if you are on target. We expect your actual answers to be "fleshed out" a bit more.
  1. dH/dy = y and -dH/dx=x, so the Hamiltonian vector field generated by H is the same as given by F.

    Clearly g(0)=(1,0), so the curve parametrized by g passes through (1,0). A solution curve has the property that the velocity vector at t equals the vector field at g(t). In symbols: g'(t) = F(g(t)). Compute g'(t) and evaluate F at g(t) to show the equality.

  2. g'(0)=(0,1).
  3. DH(x,y) = (-x,y) and g(0)=(1,0), so DH(g(0))=DH(1,0)=(-1,0).
  4. The dot product is zero. It says that the instantaneous rate of change of H is zero when traveling along the curve parametrized by g (at t=0). Another interpretation is that a unit change in t near t=0 gives rise to an expected change in the xy-plane of g'(0). But an expected change of g'(0) gives rise to an expected change of DH(g(0)).g'(0) in the range of H.
  5. The composition is a constant value of 1/2, so the graph is a horizontal line. Note that this is the same value as H(1,0).
  6. The derivative of a constant is zero. Thus H does not change AT ALL along the image of g. Said another way, g parametrizes the level set of H of height 1/2.

Last modified: Fri Nov 22 10:06:06 1996