Divergence of a variable vector

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The discussion centers on the divergence of the vector field v = (a.r)r, where r is defined as xi + yj + zk and a is a constant vector. The user attempts to show that ∇·v = 4(a.r) but arrives at ∇·v = 2(a.r) instead. They question whether their calculations or the textbook are incorrect. A counterexample is provided to clarify the misunderstanding regarding the scalar multiplication of vectors. Ultimately, the user acknowledges the correction and expresses gratitude for the clarification.
gtfitzpatrick
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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?
 
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gtfitzpatrick said:
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
 
got ya, thanks a million
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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