Oliver K

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17 years, 45 days

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These are answers submitted by Oliver K

convert(sin(x)/cos(x),tan);

The bottom surface (p3) may be overkill, leave it out if you want.

restart: with(plots): with(plottools):
p1 := implicitplot3d(x^2+y^2 = 1, x = -2 .. 2, y = -2 .. 2, z = 0 .. 3-sqrt(3), numpoints = 50000, scaling = constrained):
p2 := implicitplot3d(x^2+y^2+(z-3)^2 = 4, x = -2 .. 2, y = -2 .. 2, z = 3-sqrt(3) .. 6, numpoints = 500000, scaling = constrained):
p3 := implicitplot(x^2+y^2 <= 1, x = -2 .. 2, y = -2 .. 2, filledregions = true, coloring = ["OliveDrab", yellow], numpoints = 50000):
f := transform(proc (x, y) options operator, arrow; [x, y, 0] end proc):
display([p1,p2,f(p3)],lightmodel=light4,axes=framed,style=patchnogrid);

Let me add that the first plot is made with a lesser number of points. Since it is already looking like a bug-ridden  Hula Hoop in Standard Worksheet mode, increasing the number of dots doesn't do any good. This is something where the GUI for plots really needs to be improved. Initially, i tried plot3d with style=point, but this was even worse in terms of performance.   

Let me try something. Execute this in classic and standard worksheet to see the difference in plots. in classic, fine dots are shown and rotating is much smoother.

 restart; with(Statistics): with(plots): with(Optimization): with(plottools):

m := 10000: #number of points
 
R := 2: # R is the 'big' radius of the torus, r is the (normal distributed) 'thickness'

 r := Sample(Normal(0.2, .3), m):alpha := Sample(Uniform(0, 2*Pi), m):phi := Sample(Uniform(0, 2*Pi), m):

 f := proc (R, r, alpha, phi) options operator, arrow; [(R+r*cos(alpha))*cos(phi), (R+r*cos(alpha))*sin(phi), r*sin(alpha)] end proc:

# f is the torus procedure (parametric form)

 L := seq(f(2, r[i], alpha[i], phi[i]), i = 1 .. m):
 p1:=pointplot3d([L],style = point, colour = black, axes = normal, scaling = constrained, view = [-3 .. 3, -3 .. 3, -1 .. 1],tickmarks=[5,5,5]):
 seq(L[i]-f(r1, r2, alpha[i], phi[i]), i = 1 .. m):
 ListTools[Flatten]([%]):
 S := LSSolve(%)[2]; #Least Squares Fit
assign(S):
display([p1,torus([0, 0, 0], r2, r1)], scaling = constrained, colour = blue,style=hidden);




 

See also ?InversePlot

In case you want to know the asymptotic behaviour for x->+-infinity,i'd suggest the asympt command:

asympt(sqrt(9*x^2-4*x+2)+3*x-5,x);

The terms up to linear order give the (not necessarily horizontal/vertical) asymptotic line.

For x->-infinity:

f:=eval(sqrt(9*x^2-4*x+2)+3*x-5,x=-x);

asympt(f,x);

Again, drop the nonlinear terms here.

You may also check out the VectorCalculus package. It contains -among other vector operation goodies- a diff command that natively maps diff to a vector.

I doubt there is a way to compute a closed form solution for this antiderivative, but you may use a procedure to evaluate it with unknowns:

f := proc (a, b, c, d) int(sqrt(1+16*a^2*y^6+24*a*y^5*b+16*a*y^4*c+8*a*y^3*d+9*b^2*y^4+12*b*y^3*c+6*b*y^2*d+4*c^2*y^2+4*c*y*d+d^2), y = 0 .. .2, numeric) end proc;

f(1, 1, 1, 1);

0.3204569392

 

This happened to me, too. It helped me to plot it a second time.

There is nothing to simplify. You probably want Maple to show 1/sqrt(2), but it is general convention to have rational denominators, not only in Maple. Anyway, i think there's no (meaningful) way to force Maple to do this.   

Hi,
remove the brackets from m und use selectremove:

sols:=solve(eval(Determinant(eigen), lambda = 1), m);

selectremove(is,[sols], real);

 

 

Try this:

restart:
with(algcurves);
eqs:= x = 3*t/(t^2+1), y = 3/(t^2+1);
implicitize([eqs], [t = 0 .. 1], 2, symbolic = true, useFNV = false);

 

Regards




 

 

 

 

Check the help page for the keyword queen (select Search for text), there is already an example that makes use of Maple's Task Programming Feature.

Regards

Hi Brian,

example: type

plot(a*x^2+b,x=-2..2,y=-5..5):

This gives me a warning, but ignore it. Rightclick on the plot command and select explore

Regards

 

Very well,
someone  will be pleased to see this.

Thanks, Robert

 

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