7 years, 2 days

## How to prove that a partial derative is ...

Maple

Got here a  function g(y,z) and if the partial directive with respect to y is 0 , then  there is left over a function h(z)
Can this ber proven in Maple, can't figure it out by reasoning.
Perhaps a geometrical explanation makes more sense ?

## A potential function in 2D Input mode...

Maple

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

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

 (1)

 (2)

Thats easy to use the ScalarPotential for a manual exercise

Example 2: finding a Potential function

Show that  is conversative and find a potential function for it.

 (3)

 (4)

 (5)

So

Now manual

 (6)

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 (8)

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 (10)

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 (12)

=

 (13)

( to z )

 (14)

=

 (15)

(to y)

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=

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(to z)

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=

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(to x)

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=

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(to y )

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=

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( to x )

Test for to be a conservative field

P1=N1 ,  M1=P2, N2=M2

.... 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?)

 (24)

 (25)

Ok, can integrate to 3 variables now

Now the partial diretives from f

=  + g(y,z) constant of integration of f (the potential function)

So :  + 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

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

## Indefinite integral for two variables ...

Maple

Maple is not adding a function g(y) to the indefinite integral
Maybe i do miss here something ?

 (1)

textbook

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looks like the y is constant  in f(x,y)
Maple is not putting a constant with the integral   , the same for
Think then only the integrand 2x is integrated and y is a constant, but the notation of g(y) is not C

Integral  and then  partieel differentation gives  2x.y

Its about finding a potential function for a vectorfield F

## Positionvector curve...

Maple

I studiy a classroom  with 3 examples of a positionvector(radiusvector).
The last one a positionvector curve defined by a procedure  in table 3 is not yet working

• see enclosed worksheet : animate(F, [z], z = 0 .. 5, frames = 101) is not working yet

## Working with the Student[VectorCalculus ...

Maple

I must say that using the help for this package is not intuiative. ( in general for all )
Once on a page for command i can't  go back to the home page where the mainpage is for the package
Example i was doing something with gradient command and try figure out how to make a function out of it

So i want to look at the eval command in the student vectorcalculus package
This could be easy if there was a button to see all commands ( a home button) on the gradient page

Now i must type in help again ..
evalVF gives a column vector notation , but mixing both notations ? ( as vector , as column vector)

```gf := Gradient(exp(x*y^3*z^2))
```

This i can evaluate via contectpanel for  f(x,y,z)  in a point and as output a colomn vector

Making a vectorfunction for gf is not possible via contectpanel ? , or do i miss something

Note: perhaps showing the package commands in the working document is a idea

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