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anonymous
 2 years ago
Can anyone explain what r stands for in PSet 8, problems 4A1, 2, and 3? For instance, problem 4A2(b) is:
 find the gradient field of w=ln(r).
Is r the position vector <x,y>? In that case, wouldn't the function w really be a vector function with component ln(x) and ln(y)?
Yet, the answer given in the solutions is that the gradient field of w = (xi+yj)/r^2
anonymous
 2 years ago
Can anyone explain what r stands for in PSet 8, problems 4A1, 2, and 3? For instance, problem 4A2(b) is:  find the gradient field of w=ln(r). Is r the position vector <x,y>? In that case, wouldn't the function w really be a vector function with component ln(x) and ln(y)? Yet, the answer given in the solutions is that the gradient field of w = (xi+yj)/r^2

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phi
 2 years ago
Best ResponseYou've already chosen the best response.0r is the distance from the origin (the "r" component of polar coordinates ( r, \(\theta\) ) \[ r= \sqrt{x^2+y^2}\] \[ w= \ln r \\ \nabla w=\left< \frac{\partial}{\partial x} \ln r, \frac{\partial}{\partial y} \ln r\right>\\ \nabla w= \left<\frac{1}{r} \frac{\partial }{\partial x} \ln (x^2+y^2)^{\frac{1}{2}},\frac{1}{r} \frac{\partial }{\partial y} \ln (x^2+y^2)^{\frac{1}{2}}\right>\] finish taking the partial derivative with respect to x (and y) to get the result

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks. I had thought of that, but made a silly mistake on differentiating and got the wrong result. Problem solved!
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