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Answer...

Rouben Rostamian 8631
February 14 2018
1 2

Let t be the angle that the vector makes relative to the positive x axis.  Let L be the vector's length.  Then the vector is given by

v := L*<cos(t), sin(t)>;

 

Solution...

Rouben Rostamian 8631
February 14 2018
1 0

In answering your previous questions, Kitonum has shown you the proper syntax for defining position vectors.  Let's call them r1 and r2 in this case.  You need to do:
r1 := t -> 2*t^2*_i+16*_j+(10*t-12)*_k;
r2 := t -> (20-6*t)*_i+4*t^2*_j+2*t^2*_k;

Then you may verify that r1(2) and r2(2) are equal, that is, the airplanes collide at t=2.

The velocities are the derivatives of the position vectors.  They are computed via:

v1 := D(r1);
v2 := D(r2);

The velocities at the time of collision are given by v1(2) and v2(2).  You
should be able to calculate the angle between those vectors.

Additionally:

  1.  Heed rlopez's advice on the proper use of "evaluate" versus "solve".
  2.  You will receive better help if instead of posting code fragments, you upload a complete worksheet.  In the window where you edit your message before sending, observe the big fat green arrow.  Click on it to upload worksheet.

 

Rvachev's method...

Rouben Rostamian 8631
February 11 2018
0 1

Consider the functions "`w__i`(x,y), i=1,2," defined for all x, y, and let
    "`D__i`={(x,y) : `w__i`(x,y)>0 }".

We say that the function w__i characterizes the domain D__i.  One can then verify that

• 

the function w__1+w__2+sqrt(w__1^2+w__2^2) characterizes the domain `union`(D__1, D__2)
• 

the function w__1+w__2-sqrt(w__1^2+w__2^2)characterizes the domain `intersect`(D__1, `D__2.`)

This idea was introduced/popularized by Rvachev; see, e.g.,

http://appliedmechanicsreviews.asmedigitalcollection.asme.org/article.aspx?articleid=1395415

 

This enables us to build a characterizing function for a complicated domain by splitting the domain
into simpler domains. Here I illustrate the method by creating the 8-shaped domain seen at the end of
this worksheet.

> 

 

> 

restart;

A function which is positive on .25 < abs(r) and abs(r) < 1 and nonpositive otherwise:

> 

R := -(r^2-1^2)*(r^2-0.25^2);

-(r^2-1)*(r^2-0.625e-1)

> 

plot(R, r=-2..2, view=-0.5..0.5);

Function of x, yobtained by rotating the graph of R about the vertical axis:

> 

w1 := subs(r=x^2+y^2, R);

-((x^2+y^2)^2-1)*((x^2+y^2)^2-0.625e-1)

w1 characterizes one of the two loops of the eight-shaped figure:

> 

plot3d(w1, x=-2..2, y=-2..2, view=0..0.2);

Translate w1 in the y direction to obtain the characterizing function of the eight-shape's other loop:

> 

w2 := subs(y=y-1.8, w1);

-((x^2+(y-1.8)^2)^2-1)*((x^2+(y-1.8)^2)^2-0.625e-1)

The function w1 + w2 + sqrt(w1^2 + w2^2) is positive where one or both of w1 and w2 are positive,
and nonpositive otherwise.
By taking the max(0, ...) we replace the negative parts by zero:

> 

f := max(0, w1 + w2 + sqrt(w1^2+w2^2));

max(0, -((x^2+y^2)^2-1)*((x^2+y^2)^2-0.625e-1)-((x^2+(y-1.8)^2)^2-1)*((x^2+(y-1.8)^2)^2-0.625e-1)+(((x^2+y^2)^2-1)^2*((x^2+y^2)^2-0.625e-1)^2+((x^2+(y-1.8)^2)^2-1)^2*((x^2+(y-1.8)^2)^2-0.625e-1)^2)^(1/2))

Here is what f  looks like:

> 

plot3d(f, x=-2..2, y=-2..4);

The function g is 1 where f is positive, and zero otherwise:

> 

g := piecewise(f > 0, 1, 0);

g := piecewise(0 < max(0, -((x^2+y^2)^2-1)*((x^2+y^2)^2-0.625e-1)-((x^2+(y-1.8)^2)^2-1)*((x^2+(y-1.8)^2)^2-0.625e-1)+sqrt(((x^2+y^2)^2-1)^2*((x^2+y^2)^2-0.625e-1)^2+((x^2+(y-1.8)^2)^2-1)^2*((x^2+(y-1.8)^2)^2-0.625e-1)^2)), 1, 0)

Here is what g looks like:

> 

plot3d(g, x=-1.4..1.4, y=-1.4..3.1, grid=[200,200],
  style=surface, color=gold, scaling=constrained,
  orientation=[-135,-60,180], axes=none);

Download mw.mw

Translate...

Rouben Rostamian 8631
February 01 2018
0 0

I have no idea what your code is meant to plot, but if you wish to shift the origin of an existing plot to x=1, y=2, here is how:
newplot := plottools:-translate(oldplot,1,2);

 

Solution...

Rouben Rostamian 8631
February 01 2018
0 0

I don't know how to do this in Maple, but it is very easy to solve by hand.


Integrating the differential equation diff(u(x, z, t), z)+C = 0 we obtain u(x, z, t) = -C*z+f(x, t),

where f(x, t) is an arbitrary function.  In view of this, the condition u(x, h(x, t), t) = 0 takes the form

-C*h(x, t)+f(x, t) = 0, and therefore f(x, t) = C*h(x, t). Plugging this back into the previous

expression for u we arrive at the solution "u(x,z,t)=-C*z+C*h(x,t)."

Extra end if...

Rouben Rostamian 8631
January 31 2018
0 1

In line 13 from the bottom, delete the extra "end if".

A different take...

Rouben Rostamian 8631
January 31 2018
1 0

My interpretation of the problem's statement is different from acer's and kitonum's.  I view the first equation as defining γ as the root xx of the equation on the right-hand side. Similarly, the second equation defines ψ as the root xx of the equation on the right-hand side.  With that in mind, here is how to proceed.

> 

restart;
> 

local gamma;
> 

eq1 := gamma*sinh(gamma)+xi*psi*cosh(gamma) = 0;

gamma*sinh(gamma)+xi*psi*cosh(gamma) = 0

> 

eq2 := -gamma^2+beta+sqrt(psi/w)*coth(sqrt(psi/w))-psi = 0;

-gamma^2+beta+(psi/w)^(1/2)*coth((psi/w)^(1/2))-psi = 0

In eq2 we see that we need psi/w >= 0if we are to have real solutions.

From eq1 we see that xi*psi = -gamma*tanh(gamma),  that is,"psi/(w)= -1/(xi*w)*gamma*tanh(gamma)." 

But gamma*tanh(gamma) >= 0, therefore psi/w >= 0 if and only if  xi and w have opposite signs.

> 

params := xi=-1, w=1, beta=2;

xi = -1, w = 1, beta = 2

> 

fsolve(eval({eq1,eq2}, {params}));

{gamma = 1.449541816, psi = 1.298212894}

Download mw.mw

Answer...

Rouben Rostamian 8631
January 29 2018
0 0

You have a set of equation SysEq which are linear in the variables dq.  The coefficient matrix is obtained through
J := LinearAlgebra:-GenerateMatrix(SysEq, dq)[1];

In your case J is an 8x11 matrix.  To display it, do:
evalm(J);

 

Answer...

Rouben Rostamian 8631
January 28 2018
0 7

@assma Here is the code to do that calculation.  I assume that your 6.28 is actually meant to be 2π.  Change it to anything else as needed.

The 2D version

> 

err2 := abs(wr(x,y) - w3(x,y));
> 

range2 := x=0..2*Pi, y=0..2*Pi;

First, be sure that this one works correctly:

> 

maximize(err2, range2);

Only then measure the usage statistics:

> 

CodeTools:-Usage(maximize(err2, range2));
> 

 

The 3D version

> 

err3 := abs(wr(x,y,z) - w3(x,y,z));
> 

range3 := x=0..2*Pi, y=0..2*Pi, z=0..2*Pi;
> 

maximize(err3, range3);
> 

CodeTools:-Usage(maximize(err3, range3));

 

Download mw.mw

 

Answer...

Rouben Rostamian 8631
January 27 2018
2 1
> 

restart;
> 

with(Statistics):
> 

pts := [[0, 0], [1, 13], [2, 21], [6, 45], [12, 54], [15, 77]];

[[0, 0], [1, 13], [2, 21], [6, 45], [12, 54], [15, 77]]

The equation of the straight line that goes through the point (2,21) is  y = 21+a*(x-2),

that is, y = a*x-2*a+21. The LinearFit function does not accept a model involving an additive
constant such as the 21 above, but that's alright.  We subtract 21 from the data, fit with the model

y = a*x-2*a, and then add 21 to the result.

> 

pts1 := (x->x[1])~(pts);

[0, 1, 2, 6, 12, 15]

> 

pts2 := (x->x[2]-21)~(pts);  # note the subtracted 21

[-21, -8, 0, 24, 33, 56]

> 

L := LinearFit(a*x-2*a, pts1, pts2, x) + 21;  # note the added 21

HFloat(4.151724137931035)*x+HFloat(12.69655172413793)

Let's verify that L passes through (2,21):

> 

eval(L, x=2);

HFloat(21.0)

> 

plot([pts, L], x=0..15);

Download linearfit.mw

Explanation...

Rouben Rostamian 8631
January 26 2018
0 3

An object of the form [u1, u2, ..., un] is called a list in Maple. If that list is named X, then X[i] is the ith entry of that list. Each of the ui can itself be a list. If every ui is a list, then we say X is a list of lists. In that case the notation X[i][j] indicates the jth element of the ith list in X.

The argument J of your proc is expected to be a list of list.

  • The first line of the proc creates a list L from the list J as follows. Let's say the ith entry of L is [a,b,c]. Then the ith entry of L is the two-element list
        [a, (b+c*I)/a + 50*(1+I)/abs(J[1][2])]
    where I is Maple's notation for sqrt(-1), and abs() is the absolute value.
    Note: You would want J[1][2] to be nonzero, otherwise you will be dividing by zero.  In the example in the last line of your post is J[1][2] zero.  That's bad.
  • The second line of the proc defines a vector R whose ith entry equals the ith entry in L.
  • The next line says that:
        T := abs(L[1][3]);
        r := L[1][4];
    But L[1][4]  makes no sense since it attempts to pick the 4th entry of the list L[1] which is a list of two entries; there is no 4th entry.  Check for typos.

The rest of the code should be self-evident.

 

A few more options...

Rouben Rostamian 8631
January 24 2018
1 0

In addition to the ways pointed out by Robert Lopez, here are four other methods to calculate the dot product without complex conjugation:

  1.  LinearAlgebra:-DotProduct(<a,b>, <c,d>, conjugate=false);
  2. LinearAlgebra:-DotProduct(<a,b>, <c,d>) assuming real;
  3. <a,b>^%T . <c,d>;
  4. <a,b>^+ . <c,d>;

My preferred method is #4 since it requires the least amount of typing.

 

Does this help?...

Rouben Rostamian 8631
January 19 2018
1 1

@Earl If all you want is the final graphics that you have shown in your worksheet, then it can be done through a direct evaluation of parameter ranges.  This produces better-looking graphics, without the need for "cheating" with a hi-res grid.

Here are the details: mw2.mw
Edit: removed a couple of lines of unused code from an earlier version

Solution...

Rouben Rostamian 8631
January 17 2018
3 2
> 

restart;
> 

with(LinearAlgebra):

Example 1: We have two equations in three unknowns:

> 

eqns :=
   1*x + 2*y + 3*z = 4,
   5*x + 6*y + 7*z = 8;

x+2*y+3*z = 4, 5*x+6*y+7*z = 8

The augmented matrix of our system of equations:

> 

M := GenerateMatrix({eqns}, {x,y,z}, augmented);

_rtable[18446884503773816462]

Here is the solution, expressed as the parametric equations of a line in 3D:

> 

<x, y, z> =~ LinearSolve(M, free=t);

_rtable[18446884503773811046]

Example 2:  The equations are expressed as Ax=B, where A is an `&x`(m, n) matrix and

and B is an m-vector:

> 

m, n := 3, 5;

3, 5

> 

A := RandomMatrix(m,n);

_rtable[18446884503773806830]

> 

B := RandomVector(m);

_rtable[18446884503773809230]

The augmented matrix:

> 

M := < A | B >;

_rtable[18446884503773810190]

Here is the solution -- an afine manifold in R^n:

> 

<(x[i] $i=1..n)> =~ LinearSolve(M, free=t);

_rtable[18446884503773805494]

Not Maple...

Rouben Rostamian 8631
January 16 2018
0 0

Are you sure that's Maple?  To me it looks like a mupad notebook.

How do I know that?

Rename Bugged.mw to Bugged.mw.zip , unzip it, then examine the contents.

 

 

 

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