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MaplePrimes Activity

These are replies submitted by Kitonum

@AHSAN  The line

p:=unapply(eval(p(x),%), x);

gives a standard way to define  p  as a function of  x  for later use.

To find  lambda  we only leave one condition  p(0)=0  and then express  lambda  through  k  using the second condition  p(1)=0 :

h := k - (k - 1)*x;
DE := diff(p(x), x) = 6/h^2 - 12*lambda/h^3;     
BC := p(0) = 0;                    
dsolve({BC, DE}, p(x));


@AHSAN  See update to my answer.

@Carl Love  Interestingly, plot3d also works for plotting spatial curves, but it doesn't respond to color input (everything is drawn in black):

f := (x, y) -> x^2 + y^2 - 12*x + 16*y;
display(plot3d(f(x, y), x = -9 .. 9, y = -9 .. 9), pointplot3d([[6, -8, f(6, -8)]], color = red, symbol = solidcircle, symbolsize = 18), view = [-4.2 .. 8.2, -8.2 .. 4.2, -100 .. 100], plot3d([cos(t), sin(t), 1 - 12*cos(t) + 15*sin(t)+1], t = 0 .. 2*Pi, color=red, orientation = [-15, 68, 5]));


@Carl Love  I just raised this ellipse up a little (by 1) so that it is entirely above the surface f(x,y)

@vv  I wrote   ln(2+5*t)  for  t>=0 . It is equivalent to  ln(piecewise(t = 0, 2, 0 < t, 2 + 5*t))  for  t>=0 .  Probably OP simply does not know that by default a piecewise function on unspecified intervals is considered equal to 0.  

@mbirkner  Yes you are right. This approach is only suitable for small numbers. Large numbers require a special custom procedure. I have no idea how this can be done yet.

Use the prefix notation for multiplication:


M := Matrix(4, 4); for m from 0 to 4 do for n from 0 to 4 do M(m+1, n+1) := %factorial(2*m+3*n)/`%*`(%factorial(m+2*n+1), %factorial(m), %factorial(n)) end do end do; M

Matrix(%id = 18446745872818692814)


`~`[InertForm:-Display](M, inert = false)

Matrix(%id = 18446745872857841406)






@acer  Thank you. You are right of course. I did not take into account that initially pairs are specified as sets, and when we get rid of  x  and   sets remain, and Maple automatically sorts them in ascending order, for example  {x=-12, y=-23} => {-23,-12} . I just thought that for a beginner, the approach using  seq  and  rhs is conceptually easier.

@tomleslie  Thanks. I didn't know that Maple accepts systems written in vector form. The help (Maple 2018) does not say about this.

@jalal  Nice! Only the shape of the heart needs to be slightly stretched horizontally.


Anim4 := animate(textplot, [[op(convert(Position(t)+(1.3/4)*Vitesse(t), list)), `#mover(mi("V"),mo("&rarr;"))`], align = [right, below], font = [times, bold, 18]], t = t0 .. t1, frames = 100):


@acer  I thought this trait was just a typo, since in further transformations OP assumes that all symbols are real and positive.


1. Look closely at the picture. The coordinates of several points in your solution do not match what we see. If we assume that point  A  is on the coordinate plane  yOz, then there should be  A(0,6,2), B(-3,0,4), C(5*cos(Pi/3),0,5*sin(Pi/3))

2. Of course, if the coordinates of all points are known, then there is no problem in finding any angles. This can be done using geom3d package or in any other way. The problem is that the condition also contains 2 forces and this obviously should be taken into account in the solution.

3. To include these forces in the solution, I assumed that the OA rod is not rigidly fixed in the wall, but simply rests against it and at the same time can freely tilt to any side. I also assumed that the first coordinate of point A is unknown (it is not entirely clear from the figure whether point  A lies in the plane  yOz  or is shifted to the side). Under these assumptions, we have 3 unknowns and 3 equations for finding them, relying on static equilibrium (see my solution for details).


@yangtheary All answers in the last lines of code. I have added the required notation for clarity.

@Yo  You must provide a list of the main variables:

degree(alpha*x[1]^3+beta*x[1]*x[2]^2, [x[1],x[2]]);


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