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Gradient of vector

aplrt

New member
Feb 12, 2013
3
I'm a bit confused here. If I have Y(x2,x3,x4)=(sqrt(1-x2^2-x3^2-x4^2),x2,x3,x4), how do I find the magnitude of the gradient? I know that for Y(s)=(sqrt(1-s^2),s) the gradient is (-s/sqrt(1-s^2),s) and the magnitude of the gradient is 1/sqrt(1-s^2), and I'm supposed to get an expression similar to this. If I put r=(x2,x3,x4) then it would be Y(r)=(sqrt(1-r.r),r), but I'm not sure how to proceed here. Thankful for any help.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I'm a bit confused here. If I have Y(x2,x3,x4)=(sqrt(1-x2^2-x3^2-x4^2),x2,x3,x4), how do I find the magnitude of the gradient? I know that for Y(s)=(sqrt(1-s^2),s) the gradient is (-s/sqrt(1-s^2),s) and the magnitude of the gradient is 1/sqrt(1-s^2), and I'm supposed to get an expression similar to this. If I put r=(x2,x3,x4) then it would be Y(r)=(sqrt(1-r.r),r), but I'm not sure how to proceed here. Thankful for any help.
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.
 

aplrt

New member
Feb 12, 2013
3
Thank you for your reply! I'm still confused though, how will it look like, will it be
\[\nabla Y = (\frac{-x_{2}}{\sqrt{1-...}}+1,\frac{-x_{3}}{\sqrt{1-...}}+1,\frac{-x_{4}}{\sqrt{1-...}}+1)\] 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 :/
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Thank you for your reply! I'm still confused though, how will it look like, will it be
\[\nabla Y = (\frac{-x_{2}}{\sqrt{1-...}}+1,\frac{-x_{3}}{\sqrt{1-...}}+1,\frac{-x_{4}}{\sqrt{1-...}}+1)\] 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 :/
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?
 

aplrt

New member
Feb 12, 2013
3
Basically, it is a mapping from a disk to a 3-sphere (hypersphere) - a point (x2,x3,x4) in R3 is mapped to the 3-sphere by (sqrt(1-x2^2-x3^3-x4^),x2,x3,x4)=(x1,x2,x3,x4). I need the magnitude of the gradient to see how evenly or how far apart the points are mapped, but using the first expression I get a very clumsy equation that doesn't look right (the result should be something similar to 1/sqrt(1-...)). It would be fine to take the magnitude of the partial derivative with respect to x1, since this is the direction I'm interested in, if that's easier.