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Divergence of a variable vector

  • #1

Homework Statement



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

I let a= ai+bJ+ck

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

then this (a.r)r = ax[tex]^{2}[/tex]i+by[tex]^{2}[/tex]j+cz[tex]^{2}[/tex]k

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

which is equal to 2(a.r)

am i wrong or the book?
 

Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
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then this (a.r)r = ax[tex]^{2}[/tex]i+by[tex]^{2}[/tex]j+cz[tex]^{2}[/tex]k
This is not true. For example, consider the scalar [itex]\lambda[/itex],

[tex]\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)[/tex]

And in your case,

[tex]\lambda = ax+by+cz[/tex]
 
  • #3
got ya, thanks a million
 

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