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- Thread starter aplrt
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- Feb 5, 2012

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Hi aplrt,

The gradient of a scalar field \(f(x_1,\,x_2,\,\cdots,\,x_n)\) on a rectangular coordinate system is given by,

\[\nabla f = \left(\frac{\partial f}{\partial x_1 },\,\frac{\partial f}{\partial x_2},\,\cdots ,\,\frac{\partial f}{\partial x_n }\right)\]

So in your case,

\[\nabla Y = \left(\frac{\partial Y}{\partial x_2 },\,\frac{\partial Y}{\partial x_3 },\,\frac{\partial Y}{\partial x_4}\right)\]

Hope you can calculate the partial derivatives and continue the problem.

Kind Regards,

Sudharaka.

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\[\nabla Y = (\frac{-x_{2}}{\sqrt{1-...}}+1,\frac{-x_{3}}{\sqrt{1-...}}+1,\frac{-x_{4}}{\sqrt{1-...}}+1)\] or will it look like \[\nabla Y = ((\frac{-x_{2}}{\sqrt{1-...}},1,0,0),(\frac{-x_{3}}{\sqrt{1-...}},0,1,0),(\frac{-x_{4}}{\sqrt{1-...}},0,0,1))\] The second expression doesn't even make sense to me and the first one doesn't look right (and doesn't give the right magnitude). What confuses me is that it's the partial derivatives of a mapping and not a function, I'm not sure how to deal with mappings :/

- Feb 5, 2012

- 1,621

Your first expression is correct. I suppose your second way of writing indicates the same thing. Can you please tell me what is the context of this problem? And where did you find the notation that you have used in your second expression?

\[\nabla Y = (\frac{-x_{2}}{\sqrt{1-...}}+1,\frac{-x_{3}}{\sqrt{1-...}}+1,\frac{-x_{4}}{\sqrt{1-...}}+1)\] or will it look like \[\nabla Y = ((\frac{-x_{2}}{\sqrt{1-...}},1,0,0),(\frac{-x_{3}}{\sqrt{1-...}},0,1,0),(\frac{-x_{4}}{\sqrt{1-...}},0,0,1))\] The second expression doesn't even make sense to me and the first one doesn't look right (and doesn't give the right magnitude). What confuses me is that it's the partial derivatives of a mapping and not a function, I'm not sure how to deal with mappings :/

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