@janhardo 

Some things are left:

- return display  : when using this ?

- role of [ ] in 2 examples ?

yv:=unapply( f, indets(f, 'name') [ ] )~(xv);   

plot
                            ( f,
                              indets(f, 'name') [ ] =a..b,
                              color=red
                            ),
- p,L:=approxL( f, a, b, N):    p, L  ?  

@tomleslie 

Looking again to the code.

I aspected that yv  -values of f  should be stored in a array, but that happened here.

yv:=unapply( f, indets(f, 'name')[])~(xv);

But there is not Array made for this at forehand , well perhaps it has to do with the  ~ (xv) operator
It depend then from what datacontainer type( source container)  is used by ~(xv) ?

In this case it was a Array ,but was it a list then you get probabably a target datacontainer  list?   

Its s strange symbol [ ], it can also used as a range ? [ ]= a..b

plot
                            ( f,
                              indets(f, 'name')[]=a..b,
                              color=red
                            );

Than there is also return display   why is this ?

The procedure is p, L: = approxL( f, a, b, N):  were stands p, L for ?

So i am unsure about the [ ] used to free a variable from a set  and used a range [ ]=a..b

And unsure about the ~ operator about source datatcontainer to target datacontainer

@tomleslie 
Thanks

Very helpful.

Seems that [ ] is used for {x} to get into x  for the right function definition
Its a good idea/approach  to strip elements of the commando and to see what it actually does with the outcome.  

 

@Carl Love 

This basic derivation for a chordal segment there is used arcsin why not  sin ? 
Makes it difference in a calculation.

Its is more userfriendler to use in calculation degrees , and the central angle is measured in degrees in the basic derivation. 

@Carl Love 

I am trying to figure out further this command

So the command yv:=unapply( f, indets(f, 'name')[])~(xv) basically means

1. figure out the name of the idependent variable in the expression 'f'
2. Using 'f' and the name of its independent variable, convert these to an appliable function
3. Apply this function to every element in the container 'xv'

yv is now  a Array filled with y-values of f  , its construction of yv to a Array is by using [ ]  
Using this  [ ]  how ?  

@dharr 

The plot is showing the right result, good.

Takes a lot of time to proof geometrical results in Maple, but its no problem   

@dharr 

Thanks

Using the geometry package for confirming results in geometric expression
Why there is not something in Maple like geometric Expression ? 

@Carl Love 

Thanks

Interesting, because maybe you has used a instant formulae for the chordal segment, but that is not the case
Highlight is the formulae, but what is so special on this formulae ?

@Carl Love 

Thanks

Its clever too , more then the other example

Using a circle segment seems to be the fastest way., but the code is more advanced then the other example.
That makes it for now more difficult for me, but i am continuing to try to imagine : could i come up with solution this by myself ? 

You have used first the formulae for the 1/2  of a circle segment as i have seen here. ( how do you get this ..from a handbook?) 

http://www.pandd.demon.nl/numwis/cirkelsect.htm#een

 
Has Maple a inbuilt formulae repository for handling euclidian geometry ? 

@Ronan 

Thanks

Did not account for it  radians or degrees in GX, but now i do. 
Removing the absolute stripes could be better , because Maple get be freed from this calculation

If the answer is - then make it +  ..or make this no sense ? 

There are more approaches for solving of course and using export in Geometric expressions gives this pic
But doing this with Maple..

Note : those absolute (modulus) |  is for a positieve area meant ?, so leaving them out is easier for Maple to handle the equation ?

@Ronan 

Thanks

You do it better then my, because my equation seems to still be correct but another one unsolvable.

The question was about two farmers : http://www.davdata.nl/math/grazinggoat.html

R = 1.1587 L  ( L is radius smallest circle ) 
If R has this length : half of smallest circle can be grazed.
  

Geometric expression can perhaps figure out the solution for the overlapping circles , but i did not yet figure out how to to this. 
 

@Ronan 

Hello,

It is a Curvilinear Polygon

You draw the two circles and add the two intersections points ( via contstruct panel: intersection) and these two points are used for the definition of the overlapping area of the two circles.

Choose the arc- drawing  in the drawing panel , then click on a intersection point and drag a curvelinear line to the other intersection point.

Now this a half part and back again for the other part.

Then select the area and then calculate symbolic area ( see panels ).

@Rouben Rostamian  

Thanks

Clever done this by you and experienced user i think.

It started with a (big) circle c1 with his midpoint in the center of a xoy  coordinate system.
c2 is a circle with his midpoint on x-axis ( R, 0 )

We can calculate half of the blue area with two integrals for the upperparts of the two circles
For circle c1 there can be a area integrated from X (intersection ) to X= L along the x-axis. 

For circle c2 there can be a area intergrated from XOZ (x=0) to X (intersection )  along the x-axis
 

This two values of the integrated areas, doubled gives the blue area  sketched in Geometric Expression. = A
We get a equation.

Note: never have seen this %[1] : if % can have multiple output ? ( depends on input : here on one line with mutiple input )

Only the integration with ( integration interval , ? , ? ).

That's it.