Plot a function based on Derivative - Gradient field

Question

I'm trying to plot the gradient field of a function in such a way that is possible to change it easily, only editing the function.

Consider the code:

xmin := -2; xmax := -xmin; ymin := -2; ymax := -ymin;
f[x_, y_] := x^2 + y^2
Plot3D[f[x, y], {x, xmin, xmax}, {y, ymin, ymax}]

So, the following code produces a very fancy result:

VectorPlot[{2 x, 2 y}, {x, xmin, xmax}, {y, ymin, ymax}, 
 StreamPoints -> Coarse, StreamColorFunction -> Hue]

Since {D[f[x, y], x], D[f[x, y], y]} produces {2 x, 2 y} I tried to define

Gradf := {D[f[x, y], x], D[f[x, y], y]}

and use

VectorPlot[Gradf, {x, xmin, xmax}, {y, ymin, ymax}, 
 StreamPoints -> Coarse, StreamColorFunction -> Hue]

but I got a lot of errors as

General::ivar: -1.99971 is not a valid variable. >>

General::stop: Further output of General::ivar will be suppressed during this calculation. >>

I really don't understand why it is not working.

Any idea how to solve this so I could only change the definition of f and plot again?

Answer 1

The problem is that by defining Gradf as fun := D[x^2,x] you are saying "Ok, you will find the Derivative of x^2 in terms of x later on." by using SetDelayed (:=).

Thus, when fun appears in your VerctorPlot or more generally Plot, Mathematica sees it as: "Ok, now I'm going to find the Derivative of fun in terms of x, for each x = {0, 10}":

Plot[fun, {x, 0, 10}]

General::ivar: 0.0002042857142857143` is not a valid variable.

meaning that Mathematica, as you commanded, is trying to find the Derivative of x^2 in terms of x with x equal to 0.0002042857142857143 (case n°1 bellow). Which is obviously not possible since x is not a Variable anymore.

A part of the excellent answer from rm-rf here says:

• If you're plotting a function, whose definition depends on the output of another possibly expensive computation (such as Integrate, DSolve, Sum, etc. and their numerical equivalents) use = or use an Evaluate with :=. Failure to do so will redo the computation for every plot point! This is the #1 reason for "slow plotting".

Then all you have to do is to follow what has been said above: defining your function with Set (=) so it Evaluates the function to x before plotting it (case n°2 bellow), or Evaluate fun inside Plot so it Evaluates the fun to x before plotting (case n°3 bellow). What is internally happening can be seen with Trace:

Trace@With[{fun := D[x^2, x]}, Plot[fun, {x, 0, 10}]]

Trace@With[{fun = D[x^2, x]}, Plot[fun, {x, 0, 10}]]

Trace@With[{fun := D[x^2, x]}, Plot[Evaluate@fun, {x, 0, 10}]]

---

Now to answer the question on your specific case you have two possibilities:

xmin := -2; xmax := -xmin; ymin := -2; ymax := -ymin;
f[x_, y_] := x^2 + y^2

With[{Gradf = {D[f[x, y], x], D[f[x, y], y]}}, 
  VectorPlot[Gradf, {x, xmin, xmax}, {y, ymin, ymax}, 
  StreamPoints -> Coarse, StreamColorFunction -> Hue]]

(* or *)

With[{Gradf := {D[f[x, y], x], D[f[x, y], y]}}, 
  VectorPlot[Evaluate@Gradf, {x, xmin, xmax}, {y, ymin, ymax}, 
  StreamPoints -> Coarse, StreamColorFunction -> Hue]]