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### Vector Calculus

```
Date: 02/10/99 at 07:42:48
From: D. Denduyver
Subject: Vector Calculus

Can you help me and my son prove the following identity:

a . (grad(a . v) - curl (v x a)) = div v,

where a = constant unit vector
. = scalar product
x = vector product
v = vector

We've found that:

a . curl (v x a) = a .(( a . V) v - a (V . v)) = a. (a . V)v - V . v

but a . (grad(a . v)) - a . (a . V)v + V . v is not equal to V . v

with V = del or nabla operator.

Can you also give us a title of a textbook where we can find the proof
of de (p,r) equation of an ellipse with (p, r) = pedal coordinates?

Thank you very much,

Dany Denduyver
```

```
Date: 02/10/99 at 09:51:12
From: Doctor Mitteldorf
Subject: Re: Vector Calculus

My general approach for vector calculus identities is to write them
out as coordinates, using the Einstein summation convention. So A.B
becomes AiBi and V x F becomes EijkVjFk. For the cross product, use
epsilon-ijk, which I've written Eijk.

E123 = E312 = E231 = 1 and E321 = E213 = E132 = -1, and all their
combinations, like E112 and E333, are zero.

The particular identity you're working with is really much easier than
it looks, because a is a unit vector.  You can take a to be the unit
vector in the 3 direction, which I'll call z, without loss of
generality.

a.v = v3, and (a.V) is just V3 (Notation:  I'm using V for del and V3
means "del sub 3", the derivative in the z direction.)

So a . grad(a . v) = V3(v3), i.e., the derivative in the z direction
of the z component of v.

Now for the second term: (v x a) has a component in the 1 direction
equal to v2, and its component in the 2 direction is -v1.

We need the 3 component of the curl of this (because we're taking the
dot product of the curl with a). This is V1 (v x a)2 - V2 (vxa)1, or,
using the above, -V1(v1) - V2(v2).

So, collecting all terms, we have V3(v3) + V1(v1) + V2(v2), which is
indeed the divergence of v.

I learned vector calculus years ago from E. M. Purcell, _Electricity
and Magnetism_, Vol. II of the Berkeley Physics Course.

I'm not familiar with pedal coordinates, so I'm leaving this message
up where another doctor might be able to direct you to an appropriate
reference.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 02/10/99 at 18:29:22
From: Doctor Anthony
Subject: Re: Vector Calculus

I believe you have had a reply on the first question so will confine
my remarks to the pedal equation of the ellipse.

If phi = angle between the tangent at a point on the curve and the
radius vector to that point (polar coordinates) then a simple diagram
shows

tan(phi) = r.d(theta)/dr

sin(phi) = r.d(theta)/ds

cos(phi) = dr/ds

If p = perpendicular from the origin to the tangent then

p = r.sin(phi)

To express p in terms of r we can write

1/p^2 = 1/r^2 cosec^2(phi) = (1/r^2)[1 + cot^2(phi)]

= (1/r^2)[1 + (1/r^2)(dr/d(theta))^2]

1/p^2 = 1/r^2 + (1/r^2)[dr/d(theta)]^2   .........(1)

By eliminating theta between this equation and the polar equation of
the curve we get the equation between p and r, the pedal equation.

Example: Consider the conic with the given equation.

For the curve  2a/r = 1 - cos(theta)  find the (p,r) equation.

Differentiating  -2a/r^2 [dr/d(theta)] = sin(theta)

So,  r.d(theta)/dr = -2a/(r.sin(theta))
and from eqn. of curve,
= -(1 - cos(theta))/sin(theta)
and so,

tan(phi) = -2sin^(theta/2)/[2sin(theta/2).cos(theta/2)]

tan(phi) = -tan(theta/2)  = tan(pi-theta/2)

and                   phi = pi - theta/2

Then,

p = r.sin(phi)
= r.sin(theta/2)

and from eqn. of curve

=  2a.sin(theta/2)/[1-cos(theta)]

=  2a.sin(theta/2)/[2.sin^2(theta/2)]

=  a/sin(theta/2)

=  a.cosec(theta/2)

p^2 = a^2.cosec^2(theta/2)  ..............(2)

but,

r = 2a/(1-cos(theta)) = a/sin^2(theta/2)

therefore,   cosec^2(theta/2) = r/a

and from (2),     p^2 = a^2.r/a

p^2 = a.r

which is the pedal equation of the conic.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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