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3D Vector Feilds

Anonymous Thu 22 Apr 04:29:47 2021 No.13013055 View Reply Original Report

Quoted By: >>13013121 >>13013137 >>13013159 >>13013188 >>13013292 >>13014418

How do I find where the vectors in a vector field are pointing (towards origin, away from it, etc.) given a parametric equation (<x+y+z,0,0> or something) in 3D space?

Anonymous

Anonymous Thu 22 Apr 2021 04:53:07 No.13013094 Report

Quoted By:

>>12997581
/sqt/

Anonymous

Anonymous Thu 22 Apr 2021 05:01:24 No.13013121 Report

Quoted By: >>13013137

>>13013055
The field is given by F(x,y,z). The path is given by r(t)=x,y,z(t) where r and F are vector valued. The angle between these vectors at a point in time are given by r(t)•F(r(t))=|r(t)|*|F(x,y,z(t))|*cos?

Anonymous

Anonymous Thu 22 Apr 2021 05:08:16 No.13013137 Report

Quoted By:

>>13013121
>>13013055
i just realized this is probably not qhat you want. The vector along the parametric path at any point in time is v(t)=r'(t)=dr/dt. The angle you want is given by v(t)•F(r(t))

Anonymous

Anonymous Thu 22 Apr 2021 05:17:57 No.13013159 Report

Quoted By: >>13013180 >>13013185

>>13013055
Take the gradient.

Anonymous

Anonymous Thu 22 Apr 2021 05:24:29 No.13013180 Report

Quoted By: >>13013271

>>13013159
And then?

Anonymous

Anonymous Thu 22 Apr 2021 05:25:18 No.13013185 Report

Quoted By: >>13013271

>>13013159
gradient is a change over space, not a measure of angle between vectors

Anonymous

Anonymous Thu 22 Apr 2021 05:27:36 No.13013188 Report

Quoted By:

>>13013055
pro tip, use $\mathbf{x} [x + y + z] + \mathbf{y} + \mathbf{z}$ notation

Anonymous

Anonymous Thu 22 Apr 2021 06:06:20 No.13013271 Report

Quoted By:

>>13013180
>>13013185
Sorry, misread the OP and thought he was going for normal vectors for a parameterized surface (which you do find with gradient)

Anonymous

Anonymous Thu 22 Apr 2021 06:17:36 No.13013292 Report

Quoted By:

>>13013055
by looking at the vector components??

Anonymous

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Anonymous Thu 22 Apr 2021 12:50:03 No.13014403 Report

Quoted By:

>How do I find where the vectors in a vector field

Anonymous

Anonymous Thu 22 Apr 2021 12:55:56 No.13014418 Report

Quoted By:

>>13013055
Normalize the vector. In this case it is always pointing in the x direction. You can think of it like
(x + y + z) e_x + 0 e_y + 0 e_z where e_i are the basis vectors for euclidean space.

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