🚀 go-pugleaf

Kevin (Kat32181@aol.com/kevin@webcom.com) wrote in can.schoolnet.math.sr:
> In reviewing for a final that is on Thursday the 21 I have come across
> 4 problems that I cannot solve.  Any help would be greatly apreciated.
This group gets very light traffic, so it gets checked only occasionally.
Looks like four days was not enough to get a reply.  You should have
cross-posted to sci.math & alt.algebra.help.  (One post directed to three
groups, not three separate posts.  Then people who read multiple groups
will see your post only once.)

> 1) Show that:
>    grad (ln (determinant of the vector r)) = (vector r)/(r squared)
I think you should have "length" instead of "determinant".
If I'm wrong, what IS your definition of determinant of a vector?
If I'm right, then it is just do the derivatives:
  (d/dx,d/dy,d/dz)Ln(Sqrt(x^2+y^2+z^2) = (x,y,z)/(x^2+y^2+z^2)

> 2) The symbol (E_ijk) is the permutation symbol (or Levi-Civita
> density) and has the following properties (For i, j, k = 1,2,3):
>        (E_ijk) = 0 if any index is equal to any other index
>        (E_ijk) = +1 if ijk form an even permutation of 123
>        (E_ijk) = -1 if ijk form an odd permutation of 123
> Show that the cross product:
>        (vector C) = (vector A) cross (vector B)
> can be expressed as:
>        (C subscript i) = (Sigma_jk)(E_ijk)(A subscript j)(B subscript k)
What is your textbook's DEFINITION of the cross product?
Aside from the redundant Sigma (0 exactly when E_ijk=0, 1 otherwise),
this IS what I usually use as definition for the cross product!
(You ARE using the Einstein summation convention? - implied sums on
 repeated indices.)

> 3) Let V1, V2, V3, V4 be outward vectors perpendicular to the four faces
> of a tetrahedron with magnitures equal to the areas of those faces.
> Show that:  V1 + V2 + V3 + V4 = 0
Let A, B, and C be the edge vectors from one vertex of the tetrhedron,
choosen so that A, B, & C form a positively oriented basis.
Then:  V1 = B x A /2;  V2 = C x B /2;  V3 = A x C /2;
Two edges of the fourth face are parallel to (B-A) and (C-A) so that
V4 = (B-A) x (C-A) /2.  (Since A,(B-A),&(C-A) are positively oriented.)
Multiply out V4 and add the four expressions.
(Use B x C = - C x B and A x A = 0)

> 4) If R is the position of a particle of mass m and vector F is the
> external force on the particle, then the torque (relative to the origin)
> is:  T = R x F.  Show that:  T = d(L)/dt  where L = R x mV (V = velocity).
Just calculate the derivative d(L)/dt using the product rule.
You need to know some high school physics:  dR/dr = V, dV/dt=A and F=mA.

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