Items tagged with plot3d

Dear all,

 

Is there any way to fill the upper region in plot3d?

e.g. if I put the filled option in 

plot3d(y^3+x^2, x = 0 .. 1, y = -1 .. 1, filled)

then the region between the x-y plane and my plot will be filled. What if I want to fill the upper region?

The reason is that I want to show the upper region is acceptable for me, but I couldn't find any other way. 

If you have any solutions for me, I appreciate it.

Put x=f1(t1,t2,t3), y=f2(t1,t2,t3), z=f3(t1,t2,t3).  

The question is that: How to plot all points of form [x,y,z] with t1 in [a1,b1], t2 in [a2,b2], t3 in [a3,b3].

Thank you.

Hello

I have two vectors of angle Theta and Phi  given by

Theta=[ 0, Pi/3, 2*Pi/3, Pi];

Phi=[ 0, Pi/3 , 2*Pi/3 , Pi,4*Pi/3, 5*Pi/3, 2*Pi, 7*Pi/3];

I would like a general procedure how can I put on the unit sphere only the point 

x=cos(Phi).*sin(Theta);
 y=sin(Theta).*sin(Phi);
 z=cos(Theta);

where

(Theta, Phi) in the set {(0,Pi/3),  (0,Pi),  , (0,5*Pi/3),  

(Pi/3,0),   (Pi/3,2*Pi/3),   (Pi/3,4*Pi/3),   (Pi/3,2*Pi), 

(2*Pi/3,Pi/3),   (2*Pi/3,Pi),   (2*Pi/3,5*Pi/3),   (2*Pi/3,7*Pi/3), 

(PI,2*Pi/3),   (Pi,4*Pi/3),   (Pi,2*Pi) }

can we extract these set of point fro the definition of the two vectors Phi and Theta and then make the plot

of course we use 
Many thanks

 

plot3d([[1,3,5],[10,30,55],[50,70,25]]);

there is no lines in plot3d

plot can do this, plot3d seems different

plot([[1,1],[10,20],[5,7],[2,3]]);

Hi dears,

How can I draw the solution of the following 3d linear inequalities in Maple?

A:={-x-y+3*z >= 0, -x+2*y >= 0, 3*x-2*y-z >= 0, x > 0, y > 0, z > 0}

I am looking forward to hearing from you

Sincerely yours.

 

When trying to perform the following:

p1 := proc (x, y) if x^2+y^2 <= 1 then x*y-y^2 else 0 end if end proc;
plot3d(p1, x = -1 .. 1, y = -1 .. 1);
Error, (in plot3d) expected ranges but received x = -1 .. 1 and y = -1 .. 1
 

I get this strange error message. To the best of my knowledge x and y ARE provided as ranges. What am I missing/not understanding?

If I omit the ranges in plot3d Maple returns a correct plot, but the default range (-10 .. 10) does not display sufficient details

Hi.

I got 2 surfaces given by z=(r,theta)

I have plotted them by converting to cartesian, it gave me a good representation, but i dont get the surface entirely closed. I then used the task template vizual integration and got the solid i needed.

The problem with the rough representation cartesian conertation gives is that the integrals for volume, flux etc becomes so complex maple struggle to calculate it and since the surface is not completly closed it gives me the wrong value.

How can I plot cylindrical and spherical cordinates? 
 

I have checked the following helping page in maple: Set Coordinate System for 3-D Plots

plot_cylindrical.mw

I have the planes:

Plane 1: x + 3y - 5z = 0
Plane 2: x + 4y - 8z = 0
Plane 3: - 2x - 7y + 13z = 0

and I want to plot them in a 3d space with plot3d.

I tried to assign each plane to its own variable like this:

P1 := x+3y:
P2 := x+4y:
P3 := -2x-7y:

plot3d([PlaneOne, PlaneTwo, PlaneThree], x = -8 .. 8, y = -20 .. 20, plotlist = true, color = [blue, red, green])

The planes plot, but they aren't showing correctly.

Here's how they should look, and here's how they look.

Please help. I don't know what I'm doing wrong

There are many conventions for the Euler angles or other angles used to define o rotation of a 3d plot.
In Maple these angles are in the plot option orientation, but I think that the help page is not correct about them.
The same info appears in a worksheet (see ?rotateplot), so I am even more intrigued. [Note that many authors also switch phi and theta in spherical coordinates].

The help file says:

orientation=[theta, phi, psi]
This orientation specified by these angles is obtained by rotating the plot
1. psi about the x-axis,
2. then phi about the (transformed) z-axis, and
3. then theta about the (transformed) y-axis.
  These angles, given in degrees, are the Euler angles for the transformation matrix, using the axes specified. The angle psi is optional and is assumed to be 0 if not given. If the orientation option is not specified, the default value used is [55, 75, 0].


After some tests it seems that y and z should be switched, i.e. keeping the names (and order) for the angles ==>
1. psi about the x-axis,
2. then phi about the (transformed) y-axis, and
3. then theta about the (transformed) z-axis.

 
Am I right?

 

 

Given 3 surfaces:

x^2+y^2=1,z=0(the xy base plane) and z=1-x^2

To plot these I suggested to use cylindrical coordiates knowing x=r*cos(t) and y=r*sin(t)

Which leads to z=1-r^2*cos^2(t)

However i got problems knowing how to plot this object and dearly ask for help.

plothelp.mw

I'd like ot make a 3d graph that is log scaled on at least one of the axis. So far I haven't found a way of doing this that gives a graph that I genuinely like.

The following worksheet shows two ways of making the graph- the first generates the lines on the surface in a very bunched way, the second typesets the tickmarks in a very ugly way.

How can I get a graph with well placed lines and nicely typeset tickmarks?

How do other people make 3d logplots?

 

 

thing := x*log(y)*y^2*sin(1/y)^2;

x*ln(y)*y^2*sin(1/y)^2

 

 

 

``


 

Download logplot3d.mw

 

 

 

Hi All.

I keep getting a incorrect plot of: plot3d(2*x/(x^2+y^2), x = -10 .. 10, y = -10 .. 10)

plot3d(2*x/(x^2+y^2), x = -10 .. 10, y = -10 .. 10)

The negative range excursion is not appearing.

I have tried changing the domains and range settings but to no avail.

I have also tried placing brackets around the numerator and denominator but again to no avail. I also repeated the plot of earlier functions, on the same sheet, below the above function and had no problems with them. See the function below as an example of a good graph plot.

I noticed that the program flashes a negative value graph on screen and then only displays a positive result as shown above.

Good plot of: plot3d((-2*x^2+2*y^2)/(x^2+y^2)^2, x = -10 .. 10, y = -10 .. 10)

plot3d((-2*x^2+2*y^2)/(x^2+y^2)^2, x = -10 .. 10, y = -10 .. 10)

This example shows both the negative and positive f(x,y) values and surfaces.

Can anyone explain what is going on.

What I may be overlooking.

Is there a flaw in Maple 15?

I can get the correct graph using Microsoft Math which is a much less sophisticated program.

Omicron1

 

I've been tasked with generating "phase plots" which are visualizations of complex functions. 

A 2D phase plot is easy to create: Given a complex function F : C -> C colour points (x, y) in R^2 by [ arg(F(x+I*y)), 1, 1, colortype=HSV].   Something like the following seems to do the trick
    
    p := plot3d( 
        1,
        x=-5..5,
        y=-5..5,
        scaling=constrained,
        color=[ argument(f(x+I*y))/2/Pi, 1, 1, colortype=HSV],
        axes=none,
        style=patchnogrid,
    ):

    
    g := plottools[transform]((x,y,z)->[x,y],p);
    plots[display]( g(p) ):

Now, "colouring each point of R^2" is only possible using some type of bijection onto the Riemann sphere or Pseudosphere.

The pseudosphere is:

    x := (u,v) -> sech(u)*cos(v);
    y := sech(u)*sin(v);
    z := u - tanh(u);
    
    return  plot3d(
        [x(u,v),y(u,v),z(u,v)],
        u=0..3,
        v=0..2*Pi,
        numpoints=2^10,
        lightmodel=none,
        color=ColorFunc(u,v) );

In this case I need  ColorFunc to be:

ColorFunc := proc(u,v)
    x, y := v, exp(u);
    ans  := Re( f(x+I*y) );
    if ans > 2*Pi then
       return [0,0,0,colortype=HSV];
    end if;
    return [ans/2/Pi,1,1,colortype=HSV];
end proc;

But it seems that "ColorFunc" cannot be very sophisticated.  Namely, it cannot contain "frems" or even "if" statements because (as far as I can tell) of the order of evaluations.  

It seems possible that I can generate a psuedosphere then change colours AFTER by swapping out the COLOR information in a more sophisticated way.  How can I do this?  I really just need to know how to identify and swap out points from a MESH.

I want to plot the argument for a complex function. The input (x,y) represented in polar coordinates (r,phi) by default puts the cut at -I*Pi. Likewise the argument function:

argument(f(x)) plots the range -Pi..Pi.

However the function f(x)=x^2 could typically be plotted with 2 riemann surfaces on top of each other. When phi becomes 2Pi f(x) becomes 4Pi and only then I want to identify the 0 with 4Pi again since the points are equivalent in the preimage.

On the other hand the function f(x)=sqrt(x) never surpasses its own domain. The values always stay within the argument range of (0,2Pi) (in fact it only goes till Pi, or -Pi/2..Pi/2 in maple) when the preimage is taken to be (0,2Pi). Thus when plotting a preimage value of (x,y) with argument phi and 2Pi+phi they will have the same value since phi=2Pi+phi and I see a step in the plot. This step is actually there since the function has a cut at this point.

This step in the plotting image is also shown for f(x)=x^2 (e.g. at phi=+-Pi/2) but it is not of importance since it just comes from the argument function being constrained to -Pi..Pi.

So is it possible to change this behaviour?

I want to run a specific color red outside and yellow inside on my equation here using MAPLE 8.00:

plot3d([(0.5+cos(5*u))*sin(2*v),(0.5+cos(5*u))*cos(2*v),0.5*(cos(5*u)-0.3*cos(15*u)+0.02*cos(25*v))],u=0..2*Pi,v=0..2*Pi,axes=FRAMED);

is there some one here can help me? thanks...here is the example of color I want Eg: 

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