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Taking derivative along the radial line....

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Hi, I am working on the fault diagnosis of the substation grounding grid. I want to calculate the derivative of magnetic flux density along the radial line...Could anyone help how to take this derivative. Like we have x and y-axis if i draw a line y=x and then take derivative with respect to that line. What i did is r=sqrt(x^2+y^2) and the find d/dr but this is wrong......

2 Replies Last Post 26 apr 2015, 00:03 GMT-4

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Posted: 10 years ago 25 apr 2015, 13:07 GMT-4
Right -- you need to know what direction you want. So if I define c = the cosine of an angle around the origin and s is the sine of the angle, then you want:

d/dr = d/dx ∂x/∂r + d/dy ∂y/∂r

by the chain rule, where r is along a radial direction.

But ∂x/∂r = c, and ∂y/∂r = s, so then you get:

d/dr = c d/dx + s d/dy

That's it. No squares or square roots.
Right -- you need to know what direction you want. So if I define c = the cosine of an angle around the origin and s is the sine of the angle, then you want: d/dr = d/dx ∂x/∂r + d/dy ∂y/∂r by the chain rule, where r is along a radial direction. But ∂x/∂r = c, and ∂y/∂r = s, so then you get: d/dr = c d/dx + s d/dy That's it. No squares or square roots.

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Posted: 10 years ago 26 apr 2015, 00:03 GMT-4
Thanks Daniel, I got your point so in this case derivative with respect to phi will be:
d/dphi=-r*s(phi)*d/dx + r*c(phi)*d/dy....
right?
Thanks Daniel, I got your point so in this case derivative with respect to phi will be: d/dphi=-r*s(phi)*d/dx + r*c(phi)*d/dy.... right?

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