With the package VectorCalculus we can study the speed and acceleration to their respective components. Considering the visualizaccion and algebraic calculations and to check with their respective commands. Both 2D and 3D.

Velocidad-Aceleració.mw     (in spanish)

Lenin Araujo Castillo

Physics Pure

Computer Science

The attached presentation is the last one of a sequence of three on Quantum Mechanics using Computer Algebra, covering the field equation for a quantum system of identical particles, its stationary solutions and the equations for small perturbations around them and, in this third presentation, the conditions for superfluidity of such a system of identical particles at low temperature. The novelty is again in how to tackle these problems in a computer algebra worksheet.

Download QuantumMechanics3.mw   QuantumMechanics3.pdf

Edgardo S. Cheb-Terrab
Physics, Maplesoft

Funciones.mw ...............................................................

Addition, subtraction, scalar product, vector, projections and graphs with physics packages and plots. With this you can begin to start the physics course for engineering.

Operaciones_con_Vect.mw   (in spanish)

Lenin Araujo Castillo

Physics Pure

Computer Science

This is a 5-days mini-course I gave in Brazil last week, at the CBPF (Brazilian Center for Physics Research). The material will still receive polishment and improvements, towards evolving into a sort of manual, but it is also interesting to see it exactly as it was presented to people during the course. This material uses the update of Physics available at the Maplesoft Physics R&D webpage.

BrasilComputacaoAlgebrica.mw.zip

BrasilComputacaoAlgebrica.pdf 

Edgardo S. Cheb-Terrab
Physics, Maplesoft

We regularly host live webinars on a variety of topics for our customers, and we wanted to make this information available to the MaplePrimes community as well. We will be posting information about new webinars we think will be of interest approximately once per month.

Partnering with the MAA to Revolutionize Placement Testing

In this webinar, we will demonstrate how the Maple T.A. MAA Placement Test Suite can be used to ease the problem of placement testing and how it can benefit your campus in general.

Other topics include:

• How placement testing contributes to student success

• How the MAA placement tests are created, and rigorously validated

• How valid and reliable the MAA placement tests are for entry level mathematics courses

• How you can use the Maple T.A. Placement Test Suite for easy administration, flexible delivery, and fast results

To join us for the live presentation, please click here to register.

We regularly host live webinars on a variety of topics for our customers, and we wanted to make this information available to the MaplePrimes community as well. We will be posting information about new webinars we think will be of interest approximately once per month.

Partnering with the MAA to Revolutionize Placement Testing

In this webinar, we will demonstrate how the Maple T.A. MAA Placement Test Suite can be used to ease the problem of placement testing and how it can benefit your campus in general.

Other topics include:

• How placement testing contributes to student success

• How the MAA placement tests are created, and rigorously validated

• How valid and reliable the MAA placement tests are for entry level mathematics courses

• How you can use the Maple T.A. Placement Test Suite for easy administration, flexible delivery, and fast results

To join us for the live presentation, please click here to register.

In this post we present the solution with Maple to the logical problem of "Gardens Puzzle"

http://www.mathsisfun.com/puzzles/gardens-solution.html

The Puzzle:

Five friends have their gardens next to one another, where they grow three kinds of crops: fruits (apple, pear, nut, cherry), vegetables (carrot, parsley, gourd, onion) and flowers (aster, rose, tulip, lily).

1. They grow 12 different varieties.
2. Everybody grows exactly 4 different varieties
3. Each variety is at least in one garden.
4. Only one variety is in 4 gardens.
5. Only in one garden are all 3 kinds of crops.
6. Only in one garden are all 4 varieties of one kind of crops.
7. Pear is only in the two border gardens.
8. Paul's garden is in the middle with no lily.
9. Aster grower doesn't grow vegetables.
10. Rose growers don't grow parsley.
11. Nuts grower has also gourd and parsley.
12. In the first garden are apples and cherries.
13. Only in two gardens are cherries.
14. Sam has onions and cherries.
15. Luke grows exactly two kinds of fruit.
16. Tulip is only in two gardens.
17. Apple is in a single garden.
18. Only in one garden next to Zick's is parsley.
19. Sam's garden is not on the border.
20. Hank grows neither vegetables nor asters.
21. Paul has exactly three kinds of vegetable.

Who has which garden and what is grown where?

About methods of solution. At first I just wanted to generate all variations and using conditions 1 .. 21 to find all solutions. But even if we use the condition that everybody grows exactly 4 different varieties then the total number variants equals  5!^2*binomial(12,4)^5=427945522455000000

So from the very beginning using some of the conditions 1 .. 21 we maximally reduce the number of possible variants. For example from the conditions 11, 18 and 6 implies that only in one garden are all 4 varieties of flowers. Next we pass through these variants and using conditions 1 .. 21 and finally come to a unique solution:

restart;

Fruits:={apple, pear, nut, cherry}:

Vegetables:={carrot, parsley, gourd, onion}:

Flowers:={aster, rose, tulip, lily}:

Set1:=Flowers:

Garden1:={Set1}:

Set2:=Fruits union Vegetables union Flowers minus {nut,gourd,parsley} minus {apple,cherry,rose}:

Garden2:={seq({nut,gourd,parsley} union {Set2[i]}, i=1..nops(Set2))}:

Set3:=Vegetables union Flowers minus {parsley}:

Garden3:={seq({apple,cherry,pear} union {Set3[i]}, i=1..nops(Set3))}:

Set4:=combinat[choose](Fruits union Vegetables union Flowers minus {onion,cherry} minus {apple,parsley,pear,nut}, 2):

Garden4:={seq({onion,cherry} union Set4[i], i=1..nops(Set4))}:

Set5:=Fruits union Vegetables union Flowers minus {apple, cherry,nut,parsley}:

Garden5:=combinat[choose](Set5, 4):

S:=[]:

for s1 in Garden1 do

for s2 in Garden2 do

for s3 in Garden3 do

for s4 in Garden4 do

for s5 in Garden5 do

s:=[s1,s2,s3,s4,s5]: s_4:=combinat[choose](s,4): m:=0: n:=0: k:=0: p:=0: q:=0:

for i in s do

if `intersect`(i,Fruits)<>{} and `intersect`(i,Vegetables)<>{} and `intersect`(i,Flowers)<>{} then m:=m+1: fi:

if pear in i then n:=n+1: fi:

if tulip in i then k:=k+1: fi:

if aster in i and `intersect`(i,Vegetables)<>{} then p:=p+1: fi:

if i=Fruits or i=Vegetables or i=Flowers then q:=q+1: fi:

od:

if nops(`union`(op(s)))=12 and nops(`union`(seq(`intersect`(op(s_4[j])), j=1..nops(s_4))))=1 and m=1 and n=2 and k=2 and p=0 and q=1 then S:=[op(S),[s1,s2,s3,s4,s5]]: fi:

od: od: od: od: od:

L1:=[seq([[3,Paul],[2,Sam],seq([combinat[permute]([1,4,5])[i,j],[Luke,Zick,Hank][j]],j=1..3)],i=1..6)]:

L2:=[seq([[3,Paul],[4,Sam],seq([combinat[permute]([1,2,5])[i,j],[Luke,Zick,Hank][j]],j=1..3)],i=1..6)]:

L0:=[op(L1),op(L2)]:

L:=[seq(op(combinat[permute](L0[i])),i=1..nops(L0))]:

Sol:=[]:

for l in L do

for s in S do

sol:=[seq([op(l[i]),s[i]], i=1..5)]:

gad1:=op(select(has,sol,parsley)): gad2:=op(select(has,sol,Zick)):

if abs(gad1[1]-gad2[1])=1 and convert([seq(((pear in sol[i][3] implies (sol[i][1]=1 or sol[i][1]=5)) and (sol[i][2]=Paul implies (not lily in sol[i][3])) and (sol[i][1]=1 implies (apple in sol[i][3] and cherry in sol[i][3])) and (sol[i][2]=Sam implies (onion in sol[i][3] and cherry in sol[i][3])) and (sol[i][2]=Luke implies nops(`intersect`(sol[i][3],Fruits))=2) and (sol[i][2]=Hank implies (`intersect`(sol[i][3],Vegetables)={} and not aster in sol[i][3])) and (sol[i][2]=Paul implies nops(`intersect`(sol[i][3],Vegetables))=3)), i=1..5)], `and`) then Sol:=[op(Sol), sol]: fi:

od: od:

for i in Sol do

Matrix(sort(i,(x,y)->x[1]<y[1]));

od;

In this post we present another compact proof of this remarkable theorem without using  geometry package.
The proof uses a procedure called  Cc , which for three points returns a list of the coordinates of the center and the radius of the circumscribed circle.

restart;

Cc:=proc(A,B,C)

local x1, y1, x2, y2, x3, y3, x, y;

x1,y1:=op(A);  x2,y2:=op(B);  x3,y3:=op(C);

solve({(x2-x1)*(x-(x1+x2)/2)+(y2-y1)*(y-(y1+y2)/2)=0, (x2-x3)*(x-(x2+x3)/2)+(y2-y3)*(y-(y2+y3)/2)=0},{x,y});

assign(%);

[simplify([x,y]), simplify(sqrt((x-x1)^2+(y-y1)^2))];

end proc:

Proof for arbitrary triangle:

A, B, C:=[x1,y1], [x2,y2], [x3,y3]:

A1, B1, C1, M:=(B+C)/2, (A+C)/2, (A+B)/2, (A+B+C)/3:

P1:=Cc(A,M,B1)[1]: P2:=Cc(B1,M,C)[1]: P3:=Cc(C,M,A1)[1]:

P4:=Cc(A1,M,B)[1]: P5:=Cc(B,M,C1)[1]: P6:=Cc(C1,M,A)[1]:

Cc1:=Cc(P1,P2,P3):  Cc2:=Cc(P4,P5,P6):

is(Cc1=Cc2);

                                                  true

Let  us consider the general case of symbolic values C(xC,yC). I make use of the idea suggested by edgar in http://www.mapleprimes.com/questions/97743-How-To-Prove-Morleys-Trisector-Theorem : no assumptions.

restart; with(geometry); point(A, 0, 0);
point(B, 1, 0);
point(C, xC, yC);
point(MA, (xC+1)*(1/2), (1/2)*yC);
point(MC, 1/2, 0);
point(MB, (1/2)*xC, (1/2)*yC);
point(E, (0+1+xC)*(1/3), (0+0+yC)*(1/3));# the center of mass
line(l1, x = 1/4, [x, y]);
The coordinates of the center of the first described circle are found as the solutions of the system of the equations of midperpendiculars.

midpoint(ae, A, E); coordinates(ae);


S1 := solve({x = 1/4, ((xC+1)*(1/3))*(x-(xC+1)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

BTW, Maple can't create the midperpendiculars in this case.

point(O1, op(map(rhs, S1)));
                               O1

Simple details are omitted in the above. The coordinates of the centers of the two next described circles are found similarly.
coordinates(midpoint(mce, MC, E));

S2 := solve({x = 3/4, ((-1/2+xC)*(1/3))*(x-5/12-(1/6)*xC)+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O2, op(map(rhs, S2)));

                               O2
coordinates(midpoint(bma, B, MA)); coordinates(midpoint(be, B, E));
  

                

S3 := solve({(xC-1)*(x-(xC+3)*(1/4))+yC*(y-(1/4)*yC) = 0, ((-2+xC)*(1/3))*(x-(4+xC)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O3, op(map(rhs, S3)));

                               O3

Now we find the equation of the circumference which passes through O1, O2, and O3.

eq := a*x+b*y+x^2+y^2+c = 0:
sol := solve({eval(eq, S1), eval(eq, S2), eval(eq, S3)}, {a, b, c});

A long output can be seen in the attached .mw file.

eq1 := eval(eq, sol);

  Now we find (in suspense)  the coordinates of the next center and verify whether it belongs to the sircumference O1O2O3.

coordinates(midpoint(mac, C, MA)); coordinates(midpoint(ec, E, C)); S4 := solve({(xC-1)*(x-(3*xC+1)*(1/4))+yC*(y-3*yC*(1/4)) = 0, ((2*xC-1)*(1/3))*(x-(4*xC+1)*(1/6))+(2*yC*(1/3))*(y-4*yC*(1/6)) = 0}, {x, y});

 point(O4, op(map(rhs, S4)));

                               O4
simplify(eval(eq1, S4));

                             0 = 0

Hope the reader will have a real pleasure to find the two residuary centers and to verify these on his/her own.

geom2.mw

It is well known that the medians of a triangle divide it into 6 triangles.
It is less known that the centers of their circumscribed circles belong to one circumference as drawn below

This remarkable theorem  was proved in the 21st century! Unfortunately, I lost its source.
I can't prove this difficult  theorem by hand. However, I can prove it with Maple.
The aim of this post is to expose these proofs. Everybody knows that it is scarcely possible
to construct a general triangle with help of the geometry package of Maple.
Without loss of generality one may assume that the vertex A is placed at the origin,
the vertex B is placed at (1,0), and the vertex C(xC,yC). We firstly consider the theorem
in the case of concrete values of xC and yC.

restart; with(geometry):with(plots):
point(A, 0, 0);
point(B, 1, 0);
xC := 15*(1/10); yC := sqrt(3); point(C, xC, yC);
triangle(T, [A, B, C]);
median(mA, A, T, MA);
median(mB, B, T, MB);
median(mC, C, T, MC);
line(m1, [A, MA]);
line(m2, [B, MB]);
intersection(E, m1, m2);
triangle(AEMB, [A, E, MB]);
circumcircle(c1, AEMB, 'centername' = C1);
circumcircle(c2, triangle(CEMB, [C, E, MB]), 'centername' = C2);
circumcircle(c3, triangle(CEMA, [C, E, MA]), 'centername' = C3);
circumcircle(c4, triangle(BEMA, [B, E, MA]), 'centername' = C4);
circumcircle(c5, triangle(BEMC, [B, E, MC]), 'centername' = C5);
circumcircle(c6, triangle(AEMC, [A, E, MC]), 'centername' = C6);
circle(CC, [C1, C2, C3]);
IsOnCircle(C4, CC);
                              true

IsOnCircle(C5, CC);
                              true
IsOnCircle(C6, CC);
                              true
display([draw([T(color = black), mA(color = black), mB(color = black), mC(color = black), C1(color = blue), C2(color = blue), C3(color = blue), C4(color = blue), C5(color = blue), C6(color = blue), CC(color = red)], symbol = solidcircle, symbolsize = 15, thickness = 2, scaling = constrained), textplot({[-0.5e-1, 0.5e-1, "A"], [.95, 0.5e-1, "B"], [xC-0.5e-1, yC+0.5e-1, "C"]})], axes = frame, view = [-.1 .. max(1, xC)+.1, 0 .. yC+.1]);

This can be done as a procedure in such a way.

restart; SixPoints := proc (xC, yC) geometry:-point(A, 0, 0); geometry:-point(B, 1, 0); geometry:-point(C, xC, yC); geometry:-triangle(T, [A, B, C]); geometry:-median(mA, A, T, MA); geometry:-median(mB, B, T, MB); geometry:-median(mC, C, T, MC); geometry:-line(m1, [A, MA]); geometry:-line(m2, [B, MB]); geometry:-intersection(E, m1, m2); geometry:-triangle(AEMB, [A, E, MB]); geometry:-circumcircle(c1, AEMB, 'centername' = C1); geometry:-circumcircle(c2, geometry:-triangle(CEMB, [C, E, MB]), 'centername' = C2); geometry:-circumcircle(c3, geometry:-triangle(CEMA, [C, E, MA]), 'centername' = C3); geometry:-circumcircle(c4, geometry:-triangle(BEMA, [B, E, MA]), 'centername' = C4); geometry:-circumcircle(c5, geometry:-triangle(BEMC, [B, E, MC]), 'centername' = C5); geometry:-circumcircle(c6, geometry:-triangle(AEMC, [A, E, MC]), 'centername' = C6); geometry:-circle(CC, [C1, C2, C3]); return geometry:-IsOnCircle(C4, CC), geometry:-IsOnCircle(C5, CC), geometry:-IsOnCircle(C6, CC), geometry:-draw([CC(color = blue), C1(color = red), C2(color = red), C3(color = red), C4(color = red), C5(color = red), C6(color = red), T(color = black), mA(color = black), mB(color = black), mC(color = black), c1(color = green), c4(color = green), c2(color = green), c3(color = green), c5(color = green), c6(color = green)], symbol = solidcircle, symbolsize = 15, thickness = 2) end proc;
SixPoints(1.5, 1.2);

true, true, true, PLOT(...)
 SixPoints(1.5, 1.2)[4];

See geom1.mw

To be continued (The general case will be considered in  part 2http://www.mapleprimes.com/posts/200210-Six-Points-On-Circumference-2 .).

On Thursday, Feb. 27, we are hosting our first-ever Virtual User Summit.   This day provides Maplesoft’s academic community a chance to learn more about the different ways Maplesoft technology is being used in education and research, a chance to interact with Maplesoft employees as well as each other, and a chance to get a glimpse into the future of education.

The virtual nature of this conference is a very tangible example of how much technology has changed our lives.  No less dramatic is the effect of technology on education.  In the keynote presentations at this conference, you will learn about Maplesoft’s vision for the future of education. You’ll also get to see tangible examples of technology that is building towards that vision, including sneak peeks of some things we are working on.

Visit Maplesoft Virtual User Summit for the full agenda and to register.  “Doors open” at 8:30 Eastern Time and the keynote presentations start at 9:00.

We are looking forward to this chance to come together and share our passion for technology and technical education.  Hope to see you there!

Vector using package Physics, LinearAlgebra.

Vectores.mw     (in spanish)

I was recently asked about performing some General Relativity computations from a paper by Plamen Fiziev, posted in the arXiv in 2013. It crossed my mind that this question is also instrumental to illustrate how these General Relativity algebraic computations can be performed using the Physics package. The pdf and mw links at the end show the same contents but with the Sections expanded.

GeneralRelativit.pdf    GeneralRelativity.mw

Edgardo S. Cheb-Terrab 
Physics, Differential Equations and Mathematical Functions, Maplesoft

Code of the animation:

restart;

N := 192:

A := seq(plot([[.85*sin(t)^3-2+1.25*i/N, .85*(13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t)), t = 0 .. Pi*i/N], [-.85*sin(t)^3-2+1.25*i/N, .85*(13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t)), t = 0 .. Pi*i/N], [sin(t)^3+2-1.25*i/N, 13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t), t = 0 .. Pi*i/N], [-sin(t)^3+2-1.25*i/N, 13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t), t = 0 .. Pi*i/N]], color = red, thickness = 5, view = [-3 .. 3, -1.2 .. .9]), i = 1 .. N):

plots[display](A, insequence = true, scaling = constrained, axes = none);