A potential function in 2D Input mode

Its easy to calculate a potential function for a vectorfield
But now by hand in 2D Input mode ..a struggle

Doing your homework in 2D Input mode is something totally different then the structurized explanations done in Maple input for lessons by an instructor.

Partiel differentation with respect to what variable, is not notated in 2D mode, so a serie of them showing up in the document says nothing.and reexecuting them a couple of times doubles the output if you have used a inline notation.
Basic idea here f-> f ' -> F(f ') = f and handling variables x,y, and z for integrating/differentation

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restart

with(Student:-VectorCalculus)

BasisFormat(true)

v := VectorField(`<,>`(x, y, z))

Vector(3, {(1) = x, (2) = y, (3) = z})

(1)

Student:-VectorCalculus:-ScalarPotential(v); "_noterminate"

(1/2)*x^2+(1/2)*y^2+(1/2)*z^2

(2)

Thats easy to use the ScalarPotential for a manual exercise

Example 2: finding a Potential function

Show that F = [exp(x)*cos(y)+y*z, x*z-exp(x)*sin(y), x*y+z] is conversative and find a potential function for it.

F := VectorField(`<,>`(exp(x)*cos(y)+y*z, x*z-exp(x)*sin(y), x*y+z))

Vector(3, {(1) = exp(x)*cos(y)+y*z, (2) = x*z-exp(x)*sin(y), (3) = x*y+z})

(3)

"(->)"

y*z*x+exp(x)*cos(y)+(1/2)*z^2

(4)

NULL

ScalarPotential(F)

y*z*x+exp(x)*cos(y)+(1/2)*z^2

(5)

So "f(x,y,z):=y z x+(e)^x cos(y)+(z^2)/2 + C :"

Now manual

'F' = value(F)

F = (Vector(3, {(1) = exp(x)*cos(y)+y*z, (2) = x*z-exp(x)*sin(y), (3) = x*y+z}))

(6)

(exp(x)*cos(y)+y*z)*`#mover(mi("e"),mo("&lowbar;"))`[x]+(x*z-exp(x)*sin(y))*`#mover(mi("e"),mo("&lowbar;"))`[y]+(x*y+z)*`#mover(mi("e"),mo("&lowbar;"))`[z]