Earl

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18 years, 255 days

MaplePrimes Activity


These are replies submitted by Earl

@Kitonum As do often in the past, you've given me a new insight.

@Kitonum A brilliant answer! Please explain.

@acer Thank you for these displays. It is enlightening to view the one with range m = 0..1 first along the t axis, then along the m axis.

@Carl Love Thanks to you and vv I believe I now understand the graph.

@vv Is this correct?

For each value of m on the graph, the y coordinate is the maximum for all real values of t, which value does not appear.

Could Maple create a plot of a 3D surface where, for each surface point, the x and y coordinates are m and t and the z coordinate is the maximum?

@vv Based on your hints, I will spend some time investigating.

@vv Thank you so much. Is there anywhere in Maple where complicated syntax such as this is explained or exampled?

@vv Thank you very much. Replacing floats with fractions produces correct results.

@Kitonum I have and continue to learn from you. Thank you.

@Kitonum A brilliant answer!

What would the transform be if the sphere were truncated by a plane z=f(x,y) e.g. z=x+y-2?

What would the transform be if the sphere were truncated by a non-planar surface e.g. a sphere radius 5 centred at [-2,0,0]?

@Preben Alsholm Thank you! Adding 'implicit' caused Maple 2016 to produce output similar to that of Maple 9.5 i.e. expressions for y(_T) and for x(_T). Apparently Maple 9.5 does not require 'implicit' to do this.

As you pointed out, the help page does provide information and examples.

@Joe Riel Please see my reply to Dr. Lopez above. Am I correctly interpreting the hint you have given me?

@rlopez Thank you for  your reply above and your email (I'll reply to that as well)

Joe Riel's hint to consider the cone tangent to the unit sphere at the path of transport lit a light for me. I imagined the following:

On a rectangular sheet of paper (i.e. a 2d Euclidean space) draw successive images of a vector parallel transported along a horizontal line. Then draw on the paper the shape which will form the tangent cone. While still flat, translate the vector images without changing their orientation, from the roots on the horizontal line to roots on the arc which will be the line on the cone tangent to the unit sphere.

The translated vector images will appear to rotate horizonally vis-a-vis the cone's tangent line, as they will when the cone is formed from the paper cutout and placed on the unit sphere.

In this sense, the vectors on the unit sphere maintain the parallelism they had before Euclidean space was transformed into the tangent cone.

Does this sound correct?

 

@Carl Love The ode is the first equation in the output of EulerLagrange(L, t, [theta(t)]) where

L = KE - PE and 

KE := (1/2)*BeadMass*((RingRadius*(diff(theta(t), t)))^2+(RingRadius*cos(theta(t)-(1/2)*Pi)*`Dφ0`)^2)

and 

PE = BeadMass*g*RingRadius*(1+sin(theta(t)-(1/2)*Pi)).

BeadMass = 1, RingRadius = 1, the ring's angular velocity `Dφ0` is 1, g is 9.81.

The ring is initially in the xz plane and centred at the origin. Theta is the angle a radius to the bead makes with the negative z axis and phi is the azimuth angle the ring makes with the positive x axis as its diameter rotates in the xy plane.

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