Divergence of a variable vector

[gtfitzpatrick]

Homework Statement

v = (a.r)r where r=xi+yj+zk and a is a constant vector
show $$\nabla$$.v = 4(a.r)

I let a= ai+bJ+ck

then (a.r) = ax+by+cz

then this (a.r)r = ax$$^{2}$$i+by$$^{2}$$j+cz$$^{2}$$k

$$\nabla$$.v = da1/dx+da2/dz+da3/dz =2ax+2by+2cz

which is equal to 2(a.r)

am i wrong or the book?

 

[Hootenanny]

then this (a.r)r = ax$$^{2}$$i+by$$^{2}$$j+cz$$^{2}$$k

This is not true. For example, consider the scalar $\lambda$,

$$\lambda\left(x\hat{i}+y\hat{j}+z\hat{k}\right) = \left(\lambda x\hat{i}+ \lambda y\hat{j}+\lambda z\hat{k}\right)$$

And in your case,

$$\lambda = ax+by+cz$$

 

[gtfitzpatrick]

got ya, thanks a million