Thomas Madden

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Robert, Thanks, and my apologies to ichyfat and Acer for the misleading post above. It occurs to me that graph is the same in either case I stated above. This is simply more evidence that the we just can't get by without mathematicians:) Thomas
Here are the others For y=0, set x = t and then from the equation 3x^2-z=0 we get z = 3*t^2 spc1:=spacecurve([t,0,3*t^2], t=-2..2, thickness=2, color=green): For x = 0, set y = t and then from the equation 5y^2+z=0 we get z = -5*t^2 spc2:=spacecurve([0,t,-5*t^2], t=-2..2, thickness=2, color=blue): Here is the whole thing: restart: with(plots): S:=[solve(3*t^2-5*y^2=0,y)]; p1:=plot3d(3*x^2-5*y^2, x=-2..2, y=-2..2): spc:=spacecurve({[t, S[1], 0],[t,S[2],0]}, t=-2..2, thickness=2, color=red): spc1:=spacecurve([t,0,3*t^2], t=-2..2, thickness=2, color=green): spc2:=spacecurve([0,t,-5*t^2], t=-2..2, thickness=2, color=blue): display([p1,spc,spc1,spc2]);
Here are the others For y=0, set x = t and then from the equation 3x^2-z=0 we get z = 3*t^2 spc1:=spacecurve([t,0,3*t^2], t=-2..2, thickness=2, color=green): For x = 0, set y = t and then from the equation 5y^2+z=0 we get z = -5*t^2 spc2:=spacecurve([0,t,-5*t^2], t=-2..2, thickness=2, color=blue): Here is the whole thing: restart: with(plots): S:=[solve(3*t^2-5*y^2=0,y)]; p1:=plot3d(3*x^2-5*y^2, x=-2..2, y=-2..2): spc:=spacecurve({[t, S[1], 0],[t,S[2],0]}, t=-2..2, thickness=2, color=red): spc1:=spacecurve([t,0,3*t^2], t=-2..2, thickness=2, color=green): spc2:=spacecurve([0,t,-5*t^2], t=-2..2, thickness=2, color=blue): display([p1,spc,spc1,spc2]);
I am not sure why the plots are not showing above. I tried to edit it without success.
I am not sure why the plots are not showing above. I tried to edit it without success.
Thanks DJ, I had looked at this, but it seems to work only for expressions and Jerry had asked how to convert the coordinates of a point. restart: changecoords([1,1,0],[x,y,z],spherical,[r,theta,phi]); [1, 1, 0] Also, I think it only goes from Cartesian to other. Additionally, in regard to what J. Tarr had referred to, the convention here is to give spherical coordinates ordered [r, theta, phi] as opposed to [r, phi, theta] in VectorCalculus. It appears that phi is still measured from the positive z-axis in this case. changecoords(x,[x,y,z],spherical,[r,theta,phi]); r sin(phi) cos(theta) changecoords(y,[x,y,z],spherical,[r,theta,phi]); r sin(phi) sin(theta) changecoords(z,[x,y,z],spherical,[r,theta,phi]); r cos(phi)
Thanks DJ, I had looked at this, but it seems to work only for expressions and Jerry had asked how to convert the coordinates of a point. restart: changecoords([1,1,0],[x,y,z],spherical,[r,theta,phi]); [1, 1, 0] Also, I think it only goes from Cartesian to other. Additionally, in regard to what J. Tarr had referred to, the convention here is to give spherical coordinates ordered [r, theta, phi] as opposed to [r, phi, theta] in VectorCalculus. It appears that phi is still measured from the positive z-axis in this case. changecoords(x,[x,y,z],spherical,[r,theta,phi]); r sin(phi) cos(theta) changecoords(y,[x,y,z],spherical,[r,theta,phi]); r sin(phi) sin(theta) changecoords(z,[x,y,z],spherical,[r,theta,phi]); r cos(phi)
Thanks, Georgios. It does seem odd, especially since "convert" works in the limited cases above. I'm not complaining though, since I love the way VectorCalculus displays vectors in terms of basis vectors.
Thanks, Georgios. It does seem odd, especially since "convert" works in the limited cases above. I'm not complaining though, since I love the way VectorCalculus displays vectors in terms of basis vectors.
Yes, thanks Georgios, that is what I was using as an example of a simple method. But, if you look at the OP, Jerry wanted to convert between cartesian ,spherical and cylindrical and I don't think "convert" can do this. If it can I am not sure how to enter the information.
Yes, thanks Georgios, that is what I was using as an example of a simple method. But, if you look at the OP, Jerry wanted to convert between cartesian ,spherical and cylindrical and I don't think "convert" can do this. If it can I am not sure how to enter the information.
I think it's just a convenience, and sort of a nice one, since is saves you having to write the parameterization explicitly. Up until today I never looked at the coords option in plot3d. I just defined my own parameterization explicitly as you suggest. However, plot3d(1,theta=0..2*Pi,phi=0..Pi/2,coords=spherical,axes=boxed,scaling=constrained); is kind of a nice way to get a quick plot of a hemisphere.
I think it's just a convenience, and sort of a nice one, since is saves you having to write the parameterization explicitly. Up until today I never looked at the coords option in plot3d. I just defined my own parameterization explicitly as you suggest. However, plot3d(1,theta=0..2*Pi,phi=0..Pi/2,coords=spherical,axes=boxed,scaling=constrained); is kind of a nice way to get a quick plot of a hemisphere.
William, from your post it is not clear what you are trying to do. The graph of abs(x) + abs(y) + abs(x - y) is a surface in R^3. If you would like to plot it you can use plot3d(abs(x) + abs(y) + abs(x-y), x=-2..2, y=-2..2,axes=boxed); Perhaps you can tell us more specifically what you are trying to do. Thomas
William, from your post it is not clear what you are trying to do. The graph of abs(x) + abs(y) + abs(x - y) is a surface in R^3. If you would like to plot it you can use plot3d(abs(x) + abs(y) + abs(x-y), x=-2..2, y=-2..2,axes=boxed); Perhaps you can tell us more specifically what you are trying to do. Thomas
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