@Preben Alsholm 

Thanks!

This illustrate your reasoning fine and also my geometrical idea of it, but it is more then a plane alone , because y can be any function.
That's what i ike with Maple to see directly a plot to support a idea
The plot shows that partiel directive of x =0. ( f[x]=0)   
For u = f(x,y,z) if one of the partial  directive =0 , you could plot this too then...no its beyond 3D
 

Its clear now with a function of 2 or 3 variables if one of the part directives = 0 in those functions , the remaining 2D or 3D functions are the ones with only those variables not mentioned as partial directive variable. (the same idea for integrating for the intergation constant )
This was the case with calculation made for a potential function for a vectorfield. 

Download een_part_afgeleide_is_0_voor_functie_van_twee_var.mw

@Preben Alsholm 
Thanks!
That's not that easy to follow your reasoning.
But is was thinking more on a geometrical explanation 
Suppose z =f(x,y)  and a partial derative to x and this derative = 0 
What kind of surface you get ? ..it must be a plane   parallel to the x-axis     

The part derative of y could be any number and if it is 0 then you can get a x-y plane or planes above or below  
Make this sense too ?
 

@Preben Alsholm 
Thanks!

You are right for Maple, it has already proven algorithms built in what represents a enormous amount of scientific knowledge. 
But with pdsolve it gives the result what i am looking for namely  h(z) !    

pdsolve(diff(g(y, z), y) = 0);
                        g(y, z) = _F1(z)

 The reasoning why pdsolve comes with this result h(z) , there must be some one?.i am curious what it is.

@vv 

Thanks 
Amazing procedure and if correct it shows de constants for integrating in 1 to 3 variables 
For one variable gives

F := exp(x);
f := IntWithConst(F, x);
                       f := exp(x) + _F()
But that is a number of course and for two and 3 variables it are functions as constants

The derive of a scalarfunction  out of a vectorfunction is for some points not clear 
But studying some vids makes it more clear !

See for ff : Scalar_Potential.mw

@janhardo 
Unfortanely could not follow complete the handling of the partial deratives  and integrating

In general integrating in 1,2 and 3 variables and their constants?
What says that one partial derative is zero in u=f(x,y,z) ?

Could this figure out in Maple symbolically ?..yes it can, but how
With pdsolve( two or more variables) and dsolve (one variable)

example

A := exp(x)*cos(y) + y*z

f := int(A, x) + g(y, z);
              f := y z x + exp(x) cos(y) + g(y, z)

The constant is here g(y,z) 

Now diff(f,y) should be B. Obtain that derivative and compare to B.

diff(y*z*x + exp(x)*cos(y) + g(y, z), y) - B = 0;
                         d             
                        --- g(y, z) = 0
                         dy            

This says g(y,z) is independent of y, so g(y,z) = h(z), a function of z only. That makes f become

ff := eval(f, g(y, z) = h(z));
               ff := y z x + exp(x) cos(y) + h(z)

@rlopez 

Thanks

Well, if you have reached all goals in life, then this person is a satisfied (enlighted) person i belief.
( maslow pyramid)

I noticed for your worksheet that it is maple 1d input in a 2D document and this is not how a regular student tackle his math
i think.
But i like the direct style, but for that is needed  a some more working knowledge of Maple commands     
 

@rlopez 
Thanks!

Interesting see another solving method and helpful.
I figured out how to do it in 2D mode too 
Only try to follow a bookexample and there is no curl  there mentioned
Picked up :  it is conservative field and those filelds are not rotating, so curl = 0 then    

That's a long time ago 50 years back, at that time i was a teenager of 16 years old

thomas_calculus-vb2-blz_1075_-vraag_forum_deel2.mw
   

@tomleslie 

Thanks!

With pdsolve you get the complete general function and with ScalarPotential you must add constant to the equation
Did not studied yet the outcome of a PDE to a serie of ODE 
But the idea is to this do it by first principles in 2D mode, like a student must do 
 

With the documentmode is it possible to get quick idea of a pattern, but beware.
For example i integrated  for f(x,y,z)= x+y+z with respect to x 
The pattern is that x ( whatever function x is ) is integrated and  y and  z are multiplied with only x  
x + y + z  -> int = 1/2*x^2 + y*x + z*x  
2*x + y + z-> int =x^2 + x*y + x*z
x*sin(x) + y + z -> int= y*x + z*x + sin(x) - x*cos(x)

That's brings the question can i see in documentmode to respect what variable is diferentiate/integrate , because there is no notation of it

You can do this for integration/differentation for the 3 variables x,y,z to get a idea how the final function is derived from the vectorfunction 

Or the idea is 
F is vectorfunction (given) => gradF is gradientfield  = dF/dx, dF/dy,dF/dz (round d) 
So integrate for the 3 partial deratives gives a scalar function back ( no need to look for pattern  for integrate/differentate)  

@9009134 

If you click on plot ( in document mode) , you get a context menu. There is a tickbox  with scaling constrained option too.
You can experiment with other plot options there. 

@vv 
Thanks 
For the good advice.
Understanding a math concept is the first step and then practicing by hand 
To do it on your own makes it even harder and probably for many math topics impossible without a teacher
But these day internet can be source of  studymaterial. 
Must be possible for me think as B Ed in math to study some calculus topics, if am taking the time for it of course.
 

@vv 

It is a sparringpartner Maple for understanding more  calculus, but it follows not always a textbook for integration for not adding a constant
 

@vv 

Thanks

There are two functions involved f(x) and f(y) for  z= f(x,y)  for partiel differentation
(If you consider it as two intersections curves from a surface as defining a partiele directive)
So then f[x](x,y) =0    ..y(x) is left over as constant

It s getting more complicated in another example with v(x,y,z) for a vectorfield:.has it a potential function?

@tomleslie 

Thanks

Its solved as you mentioned : there is a display command needed 

I execute a animate command without display in the table 

All is working fine i think 

@tomleslie 

Thanks

Indeed

The first worksheet attached if you take this one or take the one  from dr Lopes the output of a intermediate animate plot gives this message.

But the final animate works.   

@tomleslie 

What should i could  aspect  for the arttached worksheet ?
Most all of the commands are suppresed by a colon and focus is on the animation of the positionvector   

When i do a iniline with Ctrl+=  for P1 there is message : 

          [Length of output exceeds limit of 1000000]

So this animation cannot be seen , but the final anmation works