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I have a function f in two variables y and theta, where y=z^2+x^2. There are numerical values for x,z,and theta ( x and z are the dimensions),I have to do calculations to obtain the velocity, so I need a way to express the vector of velocity in two components x and z separately.The following is a simple example of my task:

f[y_,θ_] = 2 y Cos θ + 4 y^2 (1 - 3 Cos θ)

gradf[y_, θ_] = Grad[f[y, θ], {y, θ}]

v[y_, θ_] = f[y, θ]/gradf[y, θ]

I made the below way, but it is not correct.

v[(z^2 + x^2) _, θ_] = f[y, θ]/gradf[y, θ]

I need separation of variables as components of vector, where:

z = {0.026239934196048018`, 0.031239934196048022`, 
 0.03623993419604802`, 0.04123993419604802`, 0.04623993419604802`, 
 0.05123993419604802`};
x = {-0.0065135092956195365`, -0.0065135092956195365`, 
-0.0065135092956195365`, -0.0065135092956195365`, 
-0.0065135092956195365`, -0.0065135092956195365`};

k = Sqrt[y]= Sqrt[z^2 + x^2]
θ = ArcCos[z/k]

Thank you.

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    $\begingroup$ Can you clarify what you want to do exactly? Do you just want to replace y by z^2+x^2 ? $\endgroup$
    – banone
    Commented Jan 27, 2020 at 11:30
  • $\begingroup$ Hi Banone, My goal is to get the velocity in two variables z and x only. In other words, it is to obtain v [z_,x_] $\endgroup$
    – Fuad
    Commented Jan 27, 2020 at 20:07
  • $\begingroup$ and what about θ? $\endgroup$
    – banone
    Commented Jan 28, 2020 at 7:18
  • $\begingroup$ I already have numerical data for z,x, and θ. I don't know if I can evaluate the function using the values of θ and then the function will be v[(z^2 + x^2)] , so I'll get two parts (because of gradient) both of them in (z^2 + x^2). Then how can I obtain v[x_,z_]? $\endgroup$
    – Fuad
    Commented Jan 28, 2020 at 14:00
  • $\begingroup$ Ok, I understand. I suggest making an interpolation function for θ(x,z) and include it in your definition. Can you provide your data or equivalent example data? $\endgroup$
    – banone
    Commented Jan 28, 2020 at 14:43

1 Answer 1

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I was wrong in the comments, since you have an analytical expression for θ, there is no need for an InterpolationFunction.

f[y_, θ_] = 2 y Cos[θ] + 4 y^2 (1 - 3 Cos[θ]);
gradf[y_, θ_] = Grad[f[y, θ], {y, θ}];
v0 = f[y, θ]/gradf[y, θ];

There was a typo, the Cos didn't have brackets in the first line, also I added v0 to save the evaluated expression before adding the other definitions:

y = z^2 + x^2;
k = Sqrt[z^2 + x^2];
θ = ArcCos[z/k];

Then you can just:

v[x_, z_] = v0

to evaluate the function you just map it to your x,z-data:

mesh = Transpose[{x, z}]
v[Sequence @@ #] & /@ mesh

or like this:

v[#[[1]], #[[2]]] & /@ mesh

Let me know if you have any questions.

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1
  • $\begingroup$ Thank you very much! $\endgroup$
    – Fuad
    Commented Jan 30, 2020 at 20:29

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