Find Magnitude of Gradient in Vector Y(x2,x3,x4)

In summary: Your understanding is correct. :) I think the first expression should be \[\nabla Y = (\frac{-x_{2}}{\sqrt{1-...}},\frac{-x_{3}}{\sqrt{1-...}},\frac{-x_{4}}{\sqrt{1-...}})\]and the magnitude of the gradient would be \[\sqrt{\left(\frac{-x_{2}}{\sqrt{1-...}}\right)^2+\left(\frac{-x_{3}}{\sqrt{1-...}}\right)^2+\left(\frac{-x_{4}}{\sqrt{1-...}}\right)^2}=1\]since the unit sphere is defined
  • #1
aplrt
3
0
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.
 
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  • #2
aplrt said:
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.
 
  • #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 :/
 
  • #4
aplrt said:
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?
 
  • #5
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.
 

FAQ: Find Magnitude of Gradient in Vector Y(x2,x3,x4)

What is the definition of the magnitude of gradient in vector Y(x2,x3,x4)?

The magnitude of gradient in vector Y(x2,x3,x4) is a measure of the rate of change of a vector field at a specific point. It represents the direction and steepness of the change in the field's values at that point.

How is the magnitude of gradient calculated in vector Y(x2,x3,x4)?

The magnitude of gradient in vector Y(x2,x3,x4) is calculated using the partial derivatives of the vector field with respect to each variable. It is represented by the square root of the sum of the squares of these partial derivatives.

What does a high magnitude of gradient indicate in vector Y(x2,x3,x4)?

A high magnitude of gradient in vector Y(x2,x3,x4) indicates a steep change in the vector field at a specific point. This means that the field's values are changing rapidly and in a specific direction at that point.

How does the magnitude of gradient relate to the direction of change in vector Y(x2,x3,x4)?

The magnitude of gradient in vector Y(x2,x3,x4) is directly proportional to the direction of change in the vector field at a specific point. A higher magnitude of gradient indicates a steeper change in a specific direction, while a lower magnitude indicates a gentler change.

What is the significance of calculating the magnitude of gradient in vector Y(x2,x3,x4)?

Calculating the magnitude of gradient in vector Y(x2,x3,x4) is important in understanding the behavior and characteristics of vector fields. It helps in determining the rate and direction of change in the field at a specific point, which is useful in many scientific and engineering applications.

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