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Question: A potential function in 2D Input mode

Question:A potential function in 2D Input mode

janhardo 335 Maple
2dmath context-menu vectorcalculus
September 24 2020
0

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 

-----------------------------------------------------------------------


 

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

NULL

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]

(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]

 (7)

"(->)"

F

 (8)

F

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

 (9)

NULL

NULL

M := exp(x)*cos(y)+y*z; N := x*z-exp(x)*sin(y); P := x*y+z

x*y+z

 (10)

"(->)"

x

 (11)

M1 := exp(x)*cos(y)+y*z

exp(x)*cos(y)+y*z

 (12)

 =

exp(x)*cos(y)+y*z

 (13)

"(->)"y   ( to z )

NULLNULL 

M2 := exp(x)*cos(y)+y*z

exp(x)*cos(y)+y*z

 (14)

 =

exp(x)*cos(y)+y*z

 (15)

"(->)"-exp(x)*sin(y)+z (to y) NULL

N1 := x*z-exp(x)*sin(y)

x*z-exp(x)*sin(y)

 (16)

 =

x*z-exp(x)*sin(y)

 (17)

"(->)"x  (to z)

N2 := x*z-exp(x)*sin(y)

x*z-exp(x)*sin(y)

 (18)

 =

x*z-exp(x)*sin(y)

 (19)

"(->)"-exp(x)*sin(y)+z (to x)

P1 := x*y+z

x*y+z

 (20)

 =

x*y+z

 (21)

"(->)"x  (to y )

P2 := x*y+z

x*y+z

 (22)

 =

x*y+z

 (23)

"(->)"y  ( to x )

 

 

Test for to be a conservative field

P1=N1 ,  M1=P2, N2=M2

 

int(F, x).... integrating F  with respect to x ,holding  y and z fixed
Its a vector form F , probably F must be converted to scalar form? (command?)

exp(x)*cos(y)+y*z+x*z-exp(x)*sin(y)+x*y+z

exp(x)*cos(y)+y*z+x*z-exp(x)*sin(y)+x*y+z

 (24)

"(->)"

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

 (25)

Ok, can integrate to 3 variables now

Now the partial diretives from f

M1 = exp(x)*cos(y)+y*z"(->)"y*z*x+exp(x)*cos(y) + g(y,z) constant of integration of f (the potential function)NULL

So : "f(x,y,z)= "y*z*x+exp(x)*cos(y)NULL + g(y,z) (1) # a total mess as notation here done by me

 

The logic from this all not yet completely understood and lost oversight , but this equation (1) must be further  integrated/differentiated  to get

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

Step by step following the text example in Thomas Calculus (page 1075, example 2) in 2 D mode is not going well.

 

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