HOW I can find the function that satisfies these boundary conditions?

please help me.

Thanks

FUNCTION.mw

Download FUNCTION.mw

equation of ellipse

eqn := (1/32)*(x-16)^2+(1/2025)*(y+0)^2 = 1

only interested in the positive section

plots[implicitplot]([(1/32)*(x-16)^2+(1/2025)*(y+0)^2 = 1], x = 0 .. 18, y = 0 .. 60, scaling = constrained)

equation for arch length of function

int(sqrt(1+(diff(eqn, x))^2), x = 0 .. 16)

Doesn't seem like the best way to solve this.

Wanting to generate a general fomula I can then use in excel to calculate the arch length of the positive only section of the ellipse

any ideas of what the assumptions could be to have maple solve this?

eqn2 := (x-h)^2/a^2+(y-k)^2/b^2 = 1;
                               2          2    
                        (x - h)    (y - k)     
                eqn2 := -------- + -------- = 1
                            2          2       
                           a          b        
                             "(->)"

     [[                                            (1/2)]  
     [[          / 2  2    2  2      2        2  2\     ]  
     [[    a k + \a  b  - b  h  + 2 b  h x - b  x /     ]  
     [[y = ---------------------------------------------], 
     [[                          a                      ]  

       [                                            (1/2)]]
       [          / 2  2    2  2      2        2  2\     ]]
       [    a k - \a  b  - b  h  + 2 b  h x - b  x /     ]]
       [y = ---------------------------------------------]]
       [                          a                      ]]

eqn3 := (a*k+sqrt(a^2*b^2-b^2*h^2+2*b^2*h*x-b^2*x^2))/a;
                                                     (1/2)
                   / 2  2    2  2      2        2  2\     
             a k + \a  b  - b  h  + 2 b  h x - b  x /     
     eqn3 := ---------------------------------------------
                                   a                      
diff(eqn3, x);
                           2        2                
                        2 b  h - 2 b  x              
          -------------------------------------------
                                              (1/2)  
            / 2  2    2  2      2        2  2\       
          2 \a  b  - b  h  + 2 b  h x - b  x /      a
`assuming`([int(sqrt(1+(diff(eqn3, x))^2), x = c .. d)], [a > 0, b > 0, c >= 0, d > c]);
 

Hello everyone,

I want to solve the matrix equation U^TCU=M, where I know the matrices C and M and I need to find que matrix U.

I already did so many tests but without success.

Could someone help me?

Thanks!

I am unable to do the following multiplication of two vectors(B and C given below), which is possible in Mathematics but not in Maple(Tried all the possible ways)

B := Vector(2, {1 = 2, 2 = 3});
C := Vector(1, {1 = 4});

B . C;
Error, (in LinearAlgebra:-DotProduct) vectors must have the same dimension
B*C;
Error, (in rtable/Product) use *~ for elementwise multiplication of Vectors or Matrices; use . (dot) for Vector/Matrix multiplication
`~`[`*`](B, C);
Error, dimension bounds must be the same for all container objects in an elementwise operation
 

I have the following two questions/issues:

1.) It seems to me that in the Tetrads package, some tetrad is automatically calculated in terms of the metric. Is there any way to reverse the roles, i.e., to have the vierbein determine the metric?, as is anyway a more natural setup, I think. The point is that I would like to work with an arbitrary, non-fixed vierbein, and from it having the metric, Christoffel symbols, etc., calculated 'on the fly', and then ultimatively construct the socalled minimal spin connection from these quantities.

2.) Consider the following expressions:

Simplify(e_[a,mu ]*Christoffel[~mu,nu,rho]);
Simplify(e_[a,nu ]*Christoffel[mu,~nu,rho]);
Simplify(e_[a,rho]*Christoffel[mu,nu,~rho]);

As the Christoffel symbols do not transform as tensors, I find this notation unfortunate, as one cannot simply move the vierbein through a (partial) derivative. Conversion between world indices and Lorentz indices should, I think, only be performed for world indices of tensorial type.

Is there something wrong with dsolve?

ode := diff(y(x), x) = sqrt(2*32.2*y(x)):
ics := y(0) = 0:
dsolve({ics, ode});

maple output:   y(x) = 0

The answer should be

y(x) = 16.1 x^2

Wolfram got it

I tried using restart, with(DEtools), still no luck. Though I don't think its necessary to call with(DEtools) for this simple equation.

To make the problem simpler, use dsolve on  dy/dx = √y , y(0)=0.

ode := diff(y(x), x) = sqrt(y(x)):
ics := y(0) = 0:
dsolve({ics, ode});

maple returns  y(x) = 0, which is incorrect.  Should be y(x) = x^2/4

I uploaded the worksheet just in case 'its just me'.

diffeqseperable.mw

Download test.mw
Udp: document attached.

Download test.mw

Hi!

I got a function (from CurveFitting), that produce a polynomial with the some units inside:

As i can see, it can be easy simplified, but i get:

Is where any trick to do it, without stripping units with convert(unit_free) or something?

Thank you!

1. (k-2)*(k^2+5)*(k^3-k^2+7*k+8)/(6*k*(k^2-3*k+8))
2. (k+2)*(k^2+5)*(k^3-5*k^2+13*k-8)/(6*k*(k^2-3*k+8))
3. k^3+3*k^2+11*k-3
4. k^2+2*k+9

Hello everybody

I am currently having trouble 'manually simplifying' an equation that I differentiated using maple. Basically, the equation that I get involves hessian-matrices. Since the resulting equation is rather long, I would like to replace every such matrix by a sign, say H.
This is easy to do for non-matrix equations, i.e.

subs(sin(x)sin(y) = z, sin(x)sin(y) + xy)

will give me , which looks a bit easier.

However, I couldn't figure out how to do a similar thing with matrices. For example, the following code

subs(Matrix(3, 3, [[x, y, z], [y, z, x], [z, x, y]]) = A, Matrix(3, 3, [[x, y, z], [y, z, x], [z, x, y]])+B)

 will not work.
I already thought about converting the matrices first into lists (because for some reason it works for lists). However, I would also like to do the same as above for for functions of matrices, i.e. set , where is a matrix (this also fails, probably for the same reason).

Also, the same thing seems to fail for vectors and general arrays, so I guess the actual problem might be the array type.

I also tried alternative ways, such as or , but the later even gives an error since it cannot handle matrices at all.

This feels a bit like a noob question, but I spent almost 2 days now searching for an answer or a workaround, so my apologies if I missed something trivial.

All the best
Adrian

Hi, I try to solve the below integral. when I press enter key maple dosen't show answer and show the integral again.

int(r*r[bc]*r[tc], r = r[bc] .. r[tc]);

but when I write intgral this way and use " i " as subscript ,maple solve it.

int(r[i]*r[bc]*r[tc], r[i] = r[bc] .. r[tc]);

I just want to know why?

what is difference between first and second integral?

and also is there any way (or any packages) to solve these integrals?

(I read https://www.maplesoft.com/applications/view.aspx?SID=6846&view=html article befor)

thanks.

1. (2*k^3-6*k^2+7*k+15-k*sqrt(k^6-12*k^5+64*k^4-198*k^3+448*k^2-636*k+369))/(-k^4+2*k^3-2*k^2+10*k+15)
2. (k^3+5*k+(-k^2+k)*sqrt(k^4-10*k^3+37*k^2-60*k+180)+30)/(-k^4+k^3+k^2+5*k+30)

hi

please help me for simplify (factor) this equations.

thanks

vel.mw
 

Download vel.mw

I am trying to input data via a data table. Have several problems here.

I used an array because I want the row and column numbers to start at 0.

1st When the table appers after that the document runs hediously slow as in a second or two to enter a digit or letters appear after typing.  Like something is absorbing the computer resources. But I have a fast machine.

2nd Any data I  enter to the table vanishes but does get stored.

3rd I tried to turn it all into a procedure but cant get that to work.
 

Download DataTable_Experiment.mw

Why is maple showing 1D math when evaluating?

I have a trigonometric equation that outputs with a solution in terms of _B1 which I want to remove.

restart: solve({7*cos(2*t)=7*cos(t)^2-5, t>=0, t<=2*Pi}, t, allsolutions, explicit);

output:

{t = arccos((1/7)*sqrt(14))},

{t = 2*Pi-arccos((1/7)*sqrt(14))},

{t = 2*arccos((1/7)*sqrt(14))*_B1-2*_B1*Pi+2*Pi*_Z1-arccos((1/7)*sqrt(14))+Pi}

Is there anyway to get rid of the _B1, or somehow evaluate it by a substitution?

Even numerically the answer still retains the _B1.

{t = 1.006853685}, {t = 5.276331623}, {t = -4.269477938*_B1+6.283185308*_Z1+2.134738969}

Also it would be nice to remove the _Z1 subscript too, as the domain of the equation is [0, 2pi].

I tried removing the 'AllSolutions' command , but then I am missing two solutions:

solve({7*cos(2*t)=7*cos(t)^2-5., t>=0 and t<=2*Pi}, t, Explicit);

 {t = 1.006853685}, {t = 2.134738969}

There should be 4 solutions in the domain [0, 2pi].