Imagine a brachistochrone shaped path made of a frictionless flexible metal strip which reacts to the force of a weighty sliding object.

Depending on its flexibility and the object's mass, what would be the strip's initial shape for fastest descent between its top and bottom? How would its shape change during the object's descent?

Perhaps an aircraft emergency escape slide or the fastest path for a slalom skier exemplify this kind of situation.

I want to solve x'(t)=A(t)x(t)+b(t) for given A, b, and intital conditions x(0).

I want to use matrixDE from the package DEtools. (I know there are probably other ways but this would be the most convenient way for further tasks).

My question is: How can I specify the initial values using matrixDE?

A:=Matrix(2,2,[1,-2,4,-5])

b:=Matrix(2,1,[3,7])

sol:=matrixDE(A,b,t)

For dsolve we have

dsolve({ode,ics})

But what about matrixDE?

Hello, I have a problem with solving this DE:

y'' - y = -4sin^3x + 9sinx where y(0)=y(2pi). I have solved it but i dont know how to do it in Maple. Thanks in advace.

Dear all, 

I did these commands in Maple:

restart:
x:=Matrix([[x1],[x2]])

W:=<<w11 | w12>, <w21 | w22>>

v:= (W.x)

<seq(diff(v,t),t=W)>;

Then, the answer was this:

Vector[column](8, [x1, 0, 0, x1, x2, 0, 0, x2])

What I expected was this:

Would you tell me how to map or reorder the calculation result from Maple?

Thank you

question.mw

Hello,

One of the features I find most useful is the ability to use Explore and plot3d(in Worksheet mode) to visualize how an expression changes for given parameters.  Unfortunately, when I use Explore(plot3d...) the default output is a very narrow box with a small 3d plot and sliders underneath.  In other programs like matlab or mathematica, resizing a plot is simply a matter of resizing the plot box/window, but I do not understand why it seems almost like an impossibility in Maple, I must be missing something:

• The resulting 3d plot in the Explore box does not have resize handles
• Using the 'size' parameter does not work with plot3d, like it does with plot
• If I under Tools->Options->Display I change Plot Display to 'Window', that seems to have no effect.  I expected it to open the plot in its own resizeable window but that does not seem to be the case.
• There are resizeable 'handles' on only left and right side of the Explore bounding box, but that does not resize the plot
• If I click on the plot there are is a 'table' and 'cell' tab that shows up in the right pane:
  • If I increase the cell width and height in pixels, the plot goes blank. I have to rerun the script to get the (small) plot to come back
• What is even more strange, is that a few times I have stumbled across some 'magical' sequence of resizing the Explore box horizontally, changing table scale settings and the width/height of the plot which eventually does give me a bigger plot.
• I haven't been able to nail down what the magical sequence is, and 90% of the times I can't get it to work.  Not to mention that the second I run the script again it all resets itself and at best I have to redo the complex 'magical' sequence of adjustments every time.

It can't be this difficult to resize a plot, can it?  I love Maple, I think there is nothing else like it, but this is a rather embarrasing state of things for such a simple, basic functionality.

I did search and found a script for doing it programmatically on a plot but it relies on having a handle(foo:=plot...) to the plot which doesn't seem to work for Explore.

Thanks for any help.

Hi everyone,

I'm a student studying pde's and I was trying to find a tool for me to understand it a lot better.

The heat equation is given by: Ut = a^2*Uxx

Take the example of a rod that is insulated at both ends (establishing BC's) of Ux(0,t) = 0 and Ux(L,t) = 0. Let's define the intial condition for any temperature at point x as x*(L-X).  We know that if we try to solve the steady-state solution, setting Ut = 0, we get Uxx = 0 which implies the general solution is U(x) = C1x + C2

From our boundary conditions we can see that a rod insulated at both ends -- and I should say laterally also -- should have a graph that turns from a quadratic to a horizontal line that is defined as the average of the initial conditon function from 0 to L.

In summary: Does maple have a feature to animate the function turning from a quadratic to a horizontal line? I think it would be beneficial in the long term for learning about BC's and visualizing them in my head after I play around with it.

restart:
N:=3:
for i from 1 to N do
x[i]:=evalf((2*i-1)/(N)) :
end do:
for i from 1 to N do
t[i]:=evalf((2*i-1)/(N)) :
end do:
f:=unapply(x*t,x,t);
g:=unapply(x^2*t,x,t);

Question 1: We have two functions f and g as above. How can we create a table which contains merged cells as follows by Maple? 

Question 2:  How to convert or export the table to LATEX?

I want to solve the following ode: 

ode := diff(x(t), t) = k*(a - x(t))*(b - 2*x(t))^2

k, a and b are positive real numbers and I want to add the condition that a-x(t) >= 0 and b-2x(t) >=0, in addition to x(0)=0

how to I do that?

Hi

I have been making a 3d graph of the level sets of a function. Here is the code for the 3d graph:

display(seq(seq(plot3d([i/sin(u), u, j], u = 0 .. 3/2, t = 0 .. 10, view = [0 .. 10, 0 .. Pi/2, 0 .. 10], color = i*j), i = 1 .. 10), j = 1 .. 10))

Each curve is a different level set- and I'd liketo colour them all individually -so people can tell tham appart. Any variation of the code I've made makes each of the curves black.

 

Hi everybody

I have a problem with finding the interpolation function from a data list.

In Mathematica I can find the interpolation function from a set of data simply as below:

data := {{13, -2}, {12, -1}, {11, 0.0}, {10, 1}, {9, 2}, {8, 3}, {7, 
   4}, {6, 5}, {5, 6}, {4, 7}, {3, 8}, {2, 9}, {1, 10}}

g := Interpolation[dat]

Plot[g[x], {x, 1, 9}] (see the attached file for the diagram)

The mathematica software gives simply the interpolating function and then we can work with it.

However, I cannot do the same procedure with maple.

Would you be so kind as to let me know how can I make an interpolating function for these set of data in maple?

Thank you very much and best regards.

Hadi

How would one go about and solve a boundary value problems i.e.:

y'' + a* y = 0 under dirilecht boundary y(0) = 0 and y(L) = 0; I know this shouldn't yield a trivial solution

likewise if i wanted to do neuman or mixed conditions, how should i approach that? 

thanks

Dear all, 

Would you let me know how to run a matrix differentiation?

restart;
with(LinearAlgebra):

with(VectorCalculus):

x:=<<x1>,<x2>>

W:=<<w11, w12>|<w21, w22>>
Cal:=Transpose(W.x).(W.x)

## Up to this point, there was no problem 

map(diff,Cal,x) ## Then, this operation caused an error

What I wanted to do is 

diff(VectorCalculus:-Norm(W.x)^2, x)

Thank you, 

In Kwon Park 

L := proc(N) N; end:
plot([seq(L(min(5, max(-5, 5^(x*10) + 5^(x*k))),k=0..10))], x = 0..3);

Warning, expecting only range variable x in expression max(-5,5^(10*x)+5^(x*k)) to be plotted but found name k

Removing L and everythign works. I want maple to first compute the inner expression before calling L. But even thenn it shouldn't matter in this case, it is just an identity function.

There should be no reason why this can't be done. I experience this *type* of problem with maple a lot. Where it seems to try to compute everything symbolically and breaks in some cases for unknown reasons.

I want to find the numbers a, b, c, d, t, m, n of this equation. I tried
 

restart:
 k := 0:
 for a to 10 do
for b to 10 do 
for c to 10 do 
for d to 10 do 
for t to 2 do 
for m to 10 do
for n to 10 do 
if a > c and igcd(a, b, c, d, t, m, n) = 1 and abs(b)+abs(d)-n <> 0 then X := [solve(abs(a*x+b)+abs(c*x+d)-t*x^2+m*x-n = 0)]; if nops(X) = 6 and type(X[1], integer) and type(X[2], integer) and type(X[3], integer) and type(X[4], integer) and type(X[5], integer) and type(X[6], integer) then k := k+1; L[k] := [a, b, c, d, t, m, n, X[]] 
end if end if
end do end do end do end do end do end do end do; 
L := convert(L, list); 
k; 
L;

I can not get the result for along time. How can I get the result and reduce the time?

Is it possible to create the operator command : |x| so that it passes x to the VectorCalculus Norm function so that if I write:

>  | < 3, 0, 4> |

the result is 5?