1. 1. In spherical coordinates, the position vector has the following Cartesian components:

      

      where we have used the spherical coordinate unit vectors in Cartesian components

      

      These unit vectors have time derivatives

      

      Thus particle velocities are given by

      

      In components, then , and .

   2. Dividing the infinitesimal displacement vector

      

      by dt reproduces the velocity above

      

2. 
   
   
   1. We find at the point (3,2):

      

      The rate of change of V in the direction is given by the unit vector in this direction dotted into the gradient (this gives the projection of the gradient onto the -direction). Here normalizing our unit vector

      

      so

      

      Since this is negative, the potential is decreasing, at rate (in whatever units are presumed given). The force component , so that .

3. We prove that is always zero by calculating it in Cartesian coordinates:

   

   Each component of is thus the difference of mixed partial derivatives, for example

   

   Since the order of partial differentiation does not matter,

   

   and the mixed partial derivatives cancel, leaving .

   We are to check this manually for . Here

   

   so

   

4. For ,

   1. 

   2. This is a nonconservative force, so we must find a loop C for which the work integral is nonzero, and evaluate the work integral. The gradient varies in a complicated way, so let's try finding a loop only in the xy-plane (fixing z). We note that for z=0, vanishes, so loops in the xy-plane at z=0 will give vanishing loop integral. But let's try z=1. For z=1, . Since this has nonzero z-component, we should get a nonzero loop integral in the xy-plane, perpendicular to the z-axis. (Note: you are unlikely to have picked the same loop as I do, but the reasoning establishing a loop with nonzero loop integral and calculating that loop integral correctly are what matters.)

      

      This gives the loop integral

      

      Since , this gives

      

      The two x-integrals cancel (since doesn't depend on the different values of y), leaving us with

      

   For ,

   1. 

   2. This is a nonconservative force, so we must again find a loop C for which the work integral is nonzero, and evaluate the work integral. Again let's try finding a loop only in the xy-plane (fixing z). For z=0, . Since this has nonzero z-component, we should get a nonzero loop integral in the xy-plane, perpendicular to the z-axis.

      

      This gives the loop integral

      

      Since , which gives in our z=0 plane, this gives

      

   Lastly, for ,

   1. 

   2. Since this force is conservative, we must find its potential. We could do this via a work integral as done in Review problem 6b. We could also guess by trying to integrate: we know that

      

      Integrating the x-partial derivative,

      

      Then integrating the y-partial derivative adds a term

      

      (Note that the 2xy term in came from the first term in -V, which we'd already deduced from the x-integration.) Finally integrating the z-partial derivative shows that g(z) can only be a constant:

      

5. 1. This is equal to in problem 4. Thus, from problem 4,

      

   2. 
      

      

      The only tricky leg is the diagonal one, for which

      

      Since along this leg z= 2-y, we have

      

      so along this leg

      

      Having stated all variables on this leg in terms of y only, we're prepared to write our loop integral

      

      Since , which equals in our x=0 plane, this gives

      

• About this document ...