Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

 

I have the first and second fundamental forms of a surface. The Bonnet theorem tells me that this is enough to determine a surface in Euclidean space. Has this theorem been implemented in Maple?

1. Write a procedure that enables to compute the volume of a cone. Remember that a cone volume can be calculated using the following equation:

𝑉𝑜𝑙𝑐𝑜𝑛𝑒 =1/3𝜋𝑟2ℎ
with r and h generated randomly, both in the range from -100 to 100.

If Maple returns negative figures for r and h, ask it to replace these values with their absolute values.

If Maple returns 0 for r and h, Maple asks directly the user to input desired values.


2) With reference to point 2, you have the following matrix:

[
−65 1 32
1 −12 3

12 −2 2
]
Write a procedure that chooses randomly two numbers from the above matrix and assigns them to r and h. Then, it compares the two cone volumes and returns the largest value.

I was plotting some 3d stuff and wanted to animate it. I used orientation in the plot3d to get the orientation I wanted. When trying to animate them the orientation is ignored.

Specifically though I'm suing Pk = plot3d(...);
then animate(display, [P1, ..., Pn]);
 

(more or less, my real code is more complex but that is the idea).

Where can I tell animate to use a specific orientation? I tried adding it to the options of display but it didn't work either. Seems animate and display completely ignore orientation.

 

 

 

I am trying to solve a differential equation numerically. But I am facing some difficulties. It is saying that ODE system has a removable singularity at r = 0. How can I remove the singularity?  Please check the attached figure and try to help.

Version: Maple 2020.1

When i set the color for the gridlines it only seems to be applied for the major-tick gridlines as the following trivial example shows:

plot(sin(t), t = 0 .. Pi, axes = frame, background = "#303030", color = "Orange", axis = [gridlines = [color = "#707070", linestyle = dot]])

I assume it must also be possible to also specify the color for the minor tick-marks gridlines?

The obvious (?) variant "axis=[gridlines = [color = ["#707070", "#707070"] , ... " just seems to crash maple (nothing happens when the plot() expression is evaluated).

I'm unable to find anything in the documentation regarding this and it only seems to imply that the color should be applied to both major & minor gridlines which is not the case.

?

 

 

 

 

Hi everyone! Currently I'm studying magnetism, and I was thinking that maybe seeing represented the helix movement of an atom with a vector v parallel to the Magnetic Field B subject to the Force of Lorentz and the Hysteresis cycle created in the magnetazation of a ferromagnetic material could help me understand it more. I tried to create the plots in Maple 2015 but I couldn't.. anyone can help me by creating those two plots?

I notice that when I enter ?convert or ?convertininto the maple prompt, the help window opens once and then ceases to function.

Update. It seems that the problem is not restricted to a specific command, this occurs for any command. To replicate, open help using F1, exit maple help, then open help again using F1. Maple help won't open again. I have to do a hard exit to get it to work again.

Platform details:

Hello,

Recently I switched from Mathematica to Maple. I have the following code 

solve({pf'(e)- psi'(e) - tau - mu*pi(tau)/(1 - pi(tau)) = 0}, {tau}) assuming 0 < p, 0 < tau, 0 < mu, 0 <= pi< 1

and I am getting a result of 

RootOf(-D(pf)(e)*pi(_Z) - mu*pi(_Z) + _Z*pi(_Z) + 2*pi(_Z) + D(pf)(e) - _Z - 2)

I have to questions: (a) what does that _Z means? (b) is it possible to express the above solution in a more "readable" way?

Thanks in advance

I want to plot some oscillating function (function involve sin and cos) as a function to time (t). And when I try a large interval like t=0..1000, it is going to take forever to plot the graph and the system becomes a little laggy. Is there any way I can do to make it compute quickly?

Thank you all

I am currently trying to solve the following ODE using numerical methods:

diff(U(x), x$2) + [(z+ I*y)2/k12 -k22 + ((z+I*y)/k3)*sech(x)^2U(x)

Where the complex value (in this case, omega, has been written as z+Iy). I believe dsolve has capabilities for solving this as an initial value problem with complex values and thus to solve this as a boundary value problem I aim to use fsolve to find a zero other a function which is the (IVP solution) - (the non-initial boundary). This has worked very well for the case where y=0 however does not work for values of y>0, and it seems the problem is with fsolve. Any advice on how to deal with this problem, perhaps alternatives to fsolve? 


 

restart;

M__h := 0.352e-1;

0.352e-1

 

0.34e-1

 

0.8354e-1

 

0.96e-2

 

.123

 

0.7258e-1

 

0.214e-1

 

0.219e-1

 

.123

 

.7902

 

.11

 

0.136e-3

 

0.5e-1

 

0.8910e-1

 

0.45e-1

 

.7

 

.7214

 

1.354

 

0.235e-1

(1)

pdes := [diff(B(t, x), t) = M__h-beta__1*B(t, x)*G(t, x)/N__h+beta__2*B(t, x)*G(t, x)/N__h-mu__h*B(t, x)+sigma__h*E(t, x)*(diff(B(t, x), x, x)), diff(C(t, x), t) = beta__1*B(t, x)*G(t, x)/N__h-u[1]*C(t, x)/(1+C(t, x))-mu__h*C(t, x)*(diff(C(t, x), x, x)), diff(DD(t, x), t) = beta__2*DD(t, x)*G(t, x)/N__h-u[1]*DD(t, x)/(1+DD(t, x))-mu__h*DD(t, x)-delta__1*DD(t, x)*(diff(DD(t, x), x, x)), diff(E(t, x), t) = u[1]*C(t, x)/(1+C(t, x))+u[1]*DD(t, x)/(1+DD(t, x))-(mu__h+sigma__h)*E(t, x)*(diff(E(t, x), x, x)), diff(F(t, x), t) = M__b-beta__3*F(t, x)*C(t, x)/N__b+beta__4*F(t, x)*DD(t, x)/N__b-mu__b*F(t, x)*(diff(F(t, x), x, x)), diff(G(t, x), t) = beta__3*F(t, x)*C(t, x)/N__b+beta__4*F(t, x)*DD(t, x)/N__b-mu__b*G(t, x)*(diff(G(t, x), x, x))];

[diff(B(t, x), t) = 0.352e-1-0.891056911e-1*B(t, x)*G(t, x)-0.96e-2*B(t, x)+0.8910e-1*E(t, x)*(diff(diff(B(t, x), x), x)), diff(C(t, x), t) = .6791869919*B(t, x)*G(t, x)-0.45e-1*C(t, x)/(1+C(t, x))-0.96e-2*C(t, x)*(diff(diff(C(t, x), x), x)), diff(DD(t, x), t) = .5900813008*DD(t, x)*G(t, x)-0.45e-1*DD(t, x)/(1+DD(t, x))-0.96e-2*DD(t, x)-0.235e-1*DD(t, x)*(diff(diff(DD(t, x), x), x)), diff(E(t, x), t) = 0.45e-1*C(t, x)/(1+C(t, x))+0.45e-1*DD(t, x)/(1+DD(t, x))-0.9870e-1*E(t, x)*(diff(diff(E(t, x), x), x)), diff(F(t, x), t) = .7214-.1739837398*F(t, x)*C(t, x)+.1780487805*F(t, x)*DD(t, x)-1.354*F(t, x)*(diff(diff(F(t, x), x), x)), diff(G(t, x), t) = .1739837398*F(t, x)*C(t, x)+.1780487805*F(t, x)*DD(t, x)-1.354*G(t, x)*(diff(diff(G(t, x), x), x))]

(2)

bcs := [(D[2](B))(t, 0) = 0, (D[2](B))(t, 1) = 0, (D[2](C))(t, 0) = 0, (D[2](C))(t, 1) = 0, (D[2](DD))(t, 0) = 0, (D[2](DD))(t, 1) = 0, (D[2](E))(t, 0) = 0, (D[2](E))(t, 1) = 0, (D[2](F))(t, 0) = 0, (D[2](F))(t, 1) = 0, (D[2](G))(t, 0) = 0, (D[2](G))(t, 1) = 0, B(0, x) = 100, C(0, x) = 70, DD(0, x) = 50, E(0, x) = 70, F(0, x) = 100, G(0, x) = 70]

[(D[2](B))(t, 0) = 0, (D[2](B))(t, 1) = 0, (D[2](C))(t, 0) = 0, (D[2](C))(t, 1) = 0, (D[2](DD))(t, 0) = 0, (D[2](DD))(t, 1) = 0, (D[2](E))(t, 0) = 0, (D[2](E))(t, 1) = 0, (D[2](F))(t, 0) = 0, (D[2](F))(t, 1) = 0, (D[2](G))(t, 0) = 0, (D[2](G))(t, 1) = 0, B(0, x) = .100, C(0, x) = .70, DD(0, x) = .50, E(0, x) = .70, F(0, x) = .100, G(0, x) = .70]

(3)

sol := pdsolve(pdes, bcs, numeric);

module () local INFO; export plot, plot3d, animate, value, settings; option `Copyright (c) 2001 by Waterloo Maple Inc. All rights reserved.`; end module

(4)

sol:-plot3d([B(t, x), C(t, x)], t = 0 .. 20, x = 0 .. 20)

Error, (in pdsolve/numeric/plot3d) unable to compute solution for t>HFloat(0.25):
Newton iteration is not converging

 

``


 

Download spatial_1.mw

Hello

I have no choice but use Grid:-Map and Grid:-Seq in my calculations due to the size of them.  Here is a very small example that is puzzling me (Perhaps I did something really silly and did not realize). 

ansa:=CodeTools:-Usage(Grid:-Map(w->CondswithOnesolutionTest(w,eqns,vars,newvars,tlim),conds5s)):

with the following result:

ansa:=set([{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}], [{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}], [{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}], [{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}])

The same thing but now using only map

ansb:=CodeTools:-Usage(map(w->CondswithOnesolutionTest(w,eqns,vars,newvars,tlim),conds5s)):
ansb:={[{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}], [{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}], [{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}], [{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}]}

(This is what I expected as the result).

 

Why did Grid:-Map add set to the answer?  What am I missing?  

 

Many thanks

 

Here is the problem. I start Maple 2020 on windows 10. Run a script which takes 1-2 days to complete. 

During this time, I can't use that Maple at all, since it is busy. 

I could start Maple 2019, and that runs as completely separate process. But I want to use Maple 2020 since some things in my scripts do not work on Maple 2019 that work on Maple 2020.

If I start a new instance of Maple 2020, by doing Start->Maple 2020. it does seem to start it OK, but I noticed it seems to be somehow still connected to the one running somehow.  May be they are sharing the same interface?

I can use the new instance now and open new worksheet and use it. But it seems to become very slow, as if it is sharing something with the other Maple 2020 running the long script which uses lots of resources. It is not RAM issue, I have 64 GB RAM, and there is plenty of free RAM left. 

When I close the new Maple 2020 workseet I started, I get a message asking if I want to save the worksheet that I have open from the earlier instance which is still running ! 

I say no ofcourse, as I do not want to terminate that instance, I want to keep it running until the script is completed.

My question is: Could someone may be explain exactly what happens when one starts new Maple 2020, while one is allready running? Why it seems they are sharing either the interface or something else.  How to start completely separate Maple 2020 instance on same PC while one is allready running?

With Mathematica, this issue does not happen. I can start two instances of same version on same PC, and there is nothing shared between them at all.  This does not seem to be the case with Maple.

Maple 2020.1 on windows 10.

 

Is it possible to determine an analytic solution to the following system of two differential equations for $A$ and $B$ using Maple.  My suspicion is that trial and error would find an analytic solution in theory and so that Maple could find the solution.  M is a constant and \sigma is some arbitrary function of t and the spatial coordinates. 

\[ \Bigg( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{2} \Bigg( 1 + \frac{M}{2 \sqrt{x^2 + y^2 + z^2}} \Bigg) \Bigg( \frac{\partial \sigma}{\partial x }\frac{\partial}{\partial x} +\frac{\partial \sigma}{\partial y}\frac{\partial}{\partial y} +\frac{\partial \sigma}{\partial z}\frac{\partial}{\partial z} \Bigg) \Bigg)B=0, \]

\[\frac{d A}{dt} = AB.\]

Furthermore, the boundary conditions are 

\[B \rightarrow -1  \: \text{as}  \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty,\]

\[A \rightarrow e^{-t} \: \text{as} \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty \]

System_of_Equations.pdf

 

I do not quite understand why prof asks this question. Or I am doing right? Where can I improve? Or I understand this question completely wrong. To be honest, I did not get the point 

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