Kitonum

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11 years, 269 days

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Should be  exp(-5*t)  instead of  e^(-5*t) .
e is just a symbol in Maple not the number 2.71828...

@Carl Love  for this. I just didn’t know about this command and thought that  LinearAlgebra:-Equal  was meant.  By the way, what you wrote can be written easier

with(Student:-MultivariateCalculus):
L1:=<2-4*t, 5+6*t>:  L2:= <-6-12*t, 17+18*t>: 
Equal(Line~([L1,L2])[]);

 

@tomleslie  It is interesting how, using the  Equal  command, you will establish the identity of the lines  L1  and  L2  defined by their  parametric equations:

L1:=<2-4*t, 5+6*t>L2:=<-6-12t, 17+18t> 

@rlopez  Unfortunately, the commands in this package do not distinguish between coincident and parallel straight lines.
The example below shows this (here the lines are given by a point and a direction vector):

with(Student[MultivariateCalculus]):
l1 := Line([0, 2, 6], <10, 15, 20>):
l2 := Line([0, 2, 6], <2, 3, 4>):
AreParallel(l1, l2);

                                          true

Obviously  l2  is the same line as  l1 .

@Jjjones98  In the code of  sys.mw  replace the name of matrix  A  with  M , because  A  is the name of the unknown A :

with(LinearAlgebra):
M,b:=GenerateMatrix(Sys1, [A,B,C,E,L,F]);
Determinant(`<|>`(seq(`if`(k=2,b,M(..,k)),k=1..6)))/Determinant(M);

  

@Jjjones98  For  C=Sol[3]  should be

Determinant(`<|>`(seq(`if`(k=3,b,A(..,k)),k=1..6)))/Determinant(A);

The command  `<|>`(seq(`if`(k=3,b,A(..,k)),k=1..6))
replaces the third column in the matrix  A  with a column of free members  b .

@vv A very simple and clear solution, vote up. The key here, of course, is the application of the  expand  command to  w^(2*sigma)  and  w^(sigma-1) , which then allows factorization to be done:
expand( w^(2*sigma)) = (w^sigma)^2
and
expand( w^(sigma-1)) = (w^sigma)/w

@Jjjones98  Yes you are right. Maple returns the value of  A=Sol[1], but for some reason does not return  Sol[2], and so on. But you can still get the values of the rest of the unknowns if you use explicit formulas, for example, Cramer's rule. So the meaning of  B=Sol[2]  will be
Determinant(`<|>`(seq(`if`(k=2,b,A(..,k)),k=1..6)))/Determinant(A);
and so on. Unfortunately, the expressions are still very bulky, and I don’t know how to simplify them and whether this is possible in principle.


Edit.

@Carl Love  In the general case, the result may depend on the order in which the substitutions occur. Compare:

A:=a/x+b*x+1:
S:={a=0, b=2};
eval(eval(A, S), x=0);
eval(A, S union {x=0});

 

@Yaongyaong  To find the surface area of a paraboloid lying inside a cone, you can use the standard formula through the double integral:

z:=12-x^2-y^2:
int(sqrt(1+diff(z,x)^2+diff(z,y)^2),[y=-sqrt(9-x^2)..sqrt(9-x^2), x=-3..3]);

 Output:                           37*sqrt(37)*Pi*(1/6)-(1/6)*Pi


Another way:

Student:-Calculus1:-SurfaceOfRevolution(12-x^2,x=0..3, axis =vertical, output=value);

Output:                                 (1/6)*(37*sqrt(37)-1)*Pi 

These 2 expressions are not equal:

restart;
A:=Ve*k21*ke*x1+Ve*k21*x1^2+k12*ke^2*x1[t]+2*k12*ke*x1*x1[t]+k12*x1^2*x1[t]+k21*ke^2*x1[t]+2*k21*ke*x1*x1[t]+k21*x1^2*x1[t]+Ve*ke*x1[t]+ke^2*x1[t,t]+2*ke*x1*x1[t,t]+x1^2*x1[t,t]:
B:=x1[t,t]*(x1+k2)^2+(k12+k21)*x1[t]*(x1+ke)^2+Ve*x1[t]*ke+(k21*Ve)*x1*(x1+ke):

expand(A-B);

Output:    -k2^2*x1[t, t]-2*k2*x1*x1[t, t]+ke^2*x1[t, t]+2*ke*x1*x1[t, t]

@acer  We can replace the last 2 lines of your code with one line:

subs(1-isolate(e_nsp2,gamma), isolate(e_nsp1,gamma), set_3);

Maple sequentially performs these 2 substitutions.

@lhasif  Thank you. The surface will look more expressive if you use the following code:

with(Student:-Calculus1):
f := x->(x-1)^2:
g := x->x+1:
A:=SurfaceOfRevolution(f(x), x = 0 .. 3, distancefromaxis = 4, surfaceoptions=[color=khaki, style=patch], output=plot):
B:=SurfaceOfRevolution(g(x), x = 0 .. 3, distancefromaxis = 4, surfaceoptions=[color=khaki, style=patch], output=plot):
plots:-display(A,B, caption=``, lightmodel=light3, scaling=constrained);

                      

For even greater expressiveness, we can make a cutout on this surface in order to clearly see the curve to be rotated and the axis of rotation. But in this case, we must specify the surface and these curves by explicit equations:

f := x->(x-1)^2;
g := x->x+1;
A:=plot3d([[x,(4-f(x))*cos(phi),(4-f(x))*sin(phi)+4],[x,(4-g(x))*cos(phi),(4-g(x))*sin(phi)+4]],x=0..3,phi=-Pi/2..5*Pi/6, style=surface, color=khaki):
B:=plots:-spacecurve([[t,0,f(t)],[t,0,g(t)]], t=0..3, color=red, thickness=3): # The line to be rotated
C:=plots:-spacecurve([t,0,4], t=-1..4, linestyle=3, color=blue, thickness=2): # Axis of rotation
plots:-display(A,B,C, axes=normal, scaling=constrained, view=[-1..4,-4..4,0..8], orientation=[-125,75]);

                 

 

@Teep You can do it similarly:


 

``

restart; with(plots)

X := Vector([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

Y1 := Vector([1, 3, 6, 2, 9, 4, 2, 7, 4, 9])

Y2 := Vector([1, 2, 1, 4, 11, 10, 7, 8, 9, 12])

display(plot([], x = 1 .. 10), seq(`$`(Statistics:-ColumnGraph([Y1[1 .. i], Y2[1 .. i]]), 5), i = 1 .. 10), insequence)

 

``


 

Download MaplePrimes_Example_animChart.mw

 

@Teep    I don’t understand what you call an animated column chart. Write the code for a non-animated chart, I'll try to animate it.

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