Project Help Notebook by Keith Stroyan

Help NoteBook Copyright 1998 by Academic Press, Inc. - All rights reserved.
Modify this Notebook to prepare your project report.

Project 44: The Perfecto Skier

by A. Student and B. Pupil

with technical help from Da LabMonitor and Help Notebook

This help notebook is NOT a substitute for the print project, but may help you with the paper and pencil work as well as give you some technical computing help.

Special Mathematica Functions used in this NoteBook

This horrible stuff sets various defaults and defines some functions needed to draw the geometric side of the lexicon.  Ignore the details here and go on to the project.

Off[General::spell];
Off[General::spell1];
Off[ParametricPlot::ppcom];
Off[ParametricPlot3D::ppcom];
Needs["Graphics`Colors`"];

SetOptions[Graphics,Axes->False,AxesOrigin->{0,0},AspectRatio->Automatic];
SetOptions[Graphics3D,ViewPoint->{5,2,1},Boxed->False,AspectRatio->Automatic];
SetOptions[ParametricPlot3D,ViewPoint->{5,2,1},AspectRatio->Automatic];
SetOptions[Plot3D,ViewPoint->{5,2,1},AspectRatio->Automatic];
SetOptions[ParametricPlot,AspectRatio->Automatic];
SetOptions[Plot,AspectRatio->Automatic];

colorQ[a_]:=(Head[a]===RGBColor);
grayQ[a_]:=(Head[a]===GrayLevel);
thickQ[a_]:=(Head[a]===Thickness);
absthickQ[a_]:=(Head[a]===AbsoluteThickness);
notruleQ[u_] := !(Head[u]===Rule);
ruleQ[u_] := (Head[u]===Rule);

Norm::usage = "Norm[expr] returns the Euclidean length of the
		vector, 'expr'.";

Norm[a_?VectorQ] := Sqrt[a.a];

Angle[a_?VectorQ,b_?VectorQ] := ArcCos[(a.b)/(Norm[a] Norm[b])]

Proj[a_?VectorQ,b_?VectorQ] := ((a.b)/(b.b))b

tol = .5*10^-6;

OrthUp[a_?VectorQ] :=
	Module[{b,c,d,f},
		d = N[a/Norm[a]];
		If[Chop[d[[3]]^2 - 1,tol] == 0, b={1,0,0} ,
		b = {0,0,1} - d[[3]] d; b = b/Norm[b];];
	Return[b]];
	
OrthSet[a_?VectorQ] :=
	Module[{b,c,d},
		d = N[a/Norm[a]];
		If[Chop[d[[1]],tol]==0,b={1,0,0},
			 If[Chop[d[[2]],tol]==0,b={0,1,0},
			  b={-d[[2]],d[[1]],0}]];
			b=b/Norm[b];c = Cross[d,b];
	Return[{b,c/Norm[c],d}]];

				
Rotate[a_?VectorQ,theta_] :=
	Module[{u,v,r},
		u = OrthSet[a];
		v = Transpose[u];
		r = N[{{Cos[theta],-Sin[theta],0},{Sin[theta],Cos[theta],0},{0,0,1}}];
	Return[v.r.u]];

Vector2D[tail_,tip_,opts___]:=
	Module[{ntail,ntip,sopts,Punit,up,arrow},
	ntail = N[tail];
	ntip = N[tip];
	
	sopts = Join[Select[{opts},colorQ],Select[{opts},
				thickQ],Select[{opts},absthickQ],Select[{opts},grayQ]];

If[Chop[ntail.ntail+ntip.ntip,tol] == 0,
			(*too small to plot or not number*)
			Return[Graphics[Point[{0,0}]]];
	,						
If[Chop[(ntip-ntail).(ntip-ntail),tol] == 0,
		Return[Graphics[{Join[sopts,{Line[{ntail,ntip}]}]}]];
	,
	Punit = (.075 Norm[ntip-ntail]/Norm[ntip-ntail]) (ntip-ntail) ;
	up = 0.15 {-Punit[[2]],Punit[[1]]};
	arrow = Line[{ntail,ntip,ntip-Punit+up,ntip,ntip-Punit-up}];
	]];
Return[Graphics[{Join[{AbsoluteThickness[2]},Join[sopts,{arrow}]]}]];
];

Vector3D[tail_,tip_,opts___]:=
	Module[{ntail,ntip,sopts,Punit,up,u2,u3,u4,rotmat,arrow},
	ntail = N[tail];
	ntip = N[tip];
	
	sopts = Join[Select[{opts},colorQ],Select[{opts},
				thickQ],Select[{opts},absthickQ],Select[{opts},grayQ]];

If[Chop[ntail.ntail+ntip.ntip,tol] == 0,
			(*too small to plot or not number*)
			Return[Graphics3D[Point[{0,0,0}]]];					
	,		
If[Chop[(ntip-ntail).(ntip-ntail),tol] == 0,
		Return[Graphics3D[{Join[sopts,{Line[{ntail,ntip}]}]}]];
	,
	Punit = (ntip-ntail);
	rotmat = Rotate[Punit,N[Pi/2]];
	up = OrthUp[Punit];
	Punit = .075 Punit;
	up = .15 Norm[Punit] up;
	u2 = rotmat.up;
	u3 = rotmat.u2;
	u4 = rotmat.u3;
	arrow = Line[{ntail,ntip,
				ntip - Punit + up,ntip,
				ntip - Punit + u2,ntip,
				ntip - Punit + u3,ntip,
				ntip - Punit + u4}];
	]];
Return[Graphics3D[{Join[{AbsoluteThickness[2]},Join[sopts,{arrow}]]}]];
];

aRCtAN[y_,x_] := Module[{v},
	If[ Chop[N[x],tol] == 0 ,If[ N[y] < 0, v = -N[Pi/2], v = N[Pi/2]],
		v = N[ArcTan[y/x]];
		If[ x < 0 && y < 0, v = N[Pi -  v],v]];
	Return[v]];

Arc2D[X_,Y_,opts___] := Graphics[{opts,
		Disk[{0,0},1,{Min[
							aRCtAN[
						N[X[[2]]],N[X[[1]]]
								]
							,
							aRCtAN[
							N[Y[[2]]],N[Y[[1]]]
								]
						]
						,
					Max[
							aRCtAN[
						N[X[[2]]],N[X[[1]]]
								]
							,
							aRCtAN[
							N[Y[[2]]],N[Y[[1]]]
								]
						]}
				]}
				];
Arc3D[X_,Y_] := 
	Module[{t,U,V},
		U[V_]:= N[(1/Sqrt[V.V]) V];
		Return[Graphics3D[
		Polygon[Append[Table[U[t X + (1-t) Y],{t,0,1,0.1}],{0,0,0}]]
				]
		]
	];

(*Zero Arrow causes trouble. Put in Vector?*)

Vector[tail_,tip_,opts___]:=
	If[Length[N[tail]]==2,Return[Vector2D[tail,tip,opts]],
		Return[Vector3D[tail,tip,opts]]];
		
Arrow[vec_,opts___]:= 
	Module[{a,b},
				If[Length[N[vec]]==2,
				a=ToExpression["Tail"]/.Select[{opts},ruleQ]/.
					{ToExpression["Tail"]->{0,0}};
				Return[Vector2D[a,a+vec,opts]],
				a=ToExpression["Tail"]/.Select[{opts},ruleQ]/.
					{ToExpression["Tail"]->{0,0,0}};
				Return[Vector3D[a,a+vec,opts]]]
				];

Frame2D[delta_] := 
      Graphics[{
            {Green,Line[{{-delta/4,0},{delta,0}}]},
                    Text["x",{1.01 delta,0}],
            {Green,Line[{{0,-delta/4},{0,delta}}]},
                    Text["y",{0, 1.01 delta}]}];

Frame3D[delta_] := 
      Graphics3D[{
            {Green,Line[{{-delta/4,0,0},{delta,0,0}}]},
                    Text["x",{1.01 delta,0,0}],
            {Green,Line[{{0,-delta/4,0},{0,delta,0}}]},
                    Text["y",{0, 1.01 delta,0}],
            {Green,Line[{{0,0,-delta/4},{0,0,delta}}]},
                    Text["z",{0,0,1.01 delta}]}];

View2D[stuff_] := Show[stuff,Frame2D[1]];
View3D[stuff_] := Show[stuff,Frame3D[1]];

My Project Introduction

Write this last.  Begin with the basic question and state your main finding clearly, but with few technical details.  You might even modify the title of this section to make it more interesting.

Beginning Geometric Explorations

Your Derivation of the Equation of a Perpendicular to an Explicit Surface

When a surface is given by  z = f[x,y], a vector perpendicular to this surface at a particular point {x1,y1,z1} on the surface is given by the vector

v = (({{∂(f/∂)x[x1, y1]}, {∂(f/∂)y[x1, y1]}, {-1}}))

This is because ...  HINTs: What vector is perpendicular to the tangent plane?  What is the equation of the tangent plane in local coordinates?

A Computer Illustration of our Formula

The next program plots an interesting  z = f[x,y]  and puts a normal vector at a chosen {x1,y1,f[x1,y1]}

Clear[x, y, dx, dy, f, fX, fY];
f[x_, y_] := 3 ((x^2) + (y^2)) Exp[-((x^2)/3) - ((y^2)/5)];
      

fX[x_, y_]  :=  Evaluate[D[f[x, y], x]];
fY[x_, y_]  :=  Evaluate[D[f[x, y], y]];

x1 = 0.5;
y1 = -2.5;
z1 = f[x1, y1];

scale = 3;
surface = 
        Plot3D[f[x, y], {x, x1 - scale, x1 + scale}, 
          {y, y1 - scale, y1 + scale}, 
          PlotRange -> {z1 - scale, z1 + scale}, 
          BoxRatios -> Automatic, 
          AxesLabel -> {"x", "y", "z"}, 
          Ticks -> 
            {{x1 - 1, x1 + 1}, {y1 - 1, y1 + 1}, {z1 - 1, z1 + 1}}, 
          DisplayFunction -> Identity];
tangent = 
        Plot3D[z1 + fX[x1, y1] (x - x1) + fY[x1, y1] (y - y1), 
          {x, x1 - (scale/3), x1 + (scale/3)}, 
          {y, y1 - (scale/3), y1 + (scale/3)}, 
          PlotRange -> {z1 - scale, z1 + scale}, 
          BoxRatios -> Automatic, PlotPoints -> 2, 
          DisplayFunction -> Identity];
combinedPlot = 
        Show[{surface, tangent, 
            Arrow[{-fX[x1, y1], -fY[x1, y1], 1}, 
              Tail -> {x1, y1, z1}, Red]}, 
          DisplayFunction -> $DisplayFunction];
Print["At {", x1, " , ", y1, " , ", z1, 
        "}, the normal vector V ="];
MatrixForm[{-fX[x1, y1], -fY[x1, y1], 1}]
Print["The general normal V ="];
MatrixForm[{-fX[x, y], -fY[x, y], 1}]

Perpendicular and Tangential Components of a Vector

HINT 44.2 asks you to find the tangential component of a weight vector.  Following are some reminders from the Algebraic-Geometric Lexicon NoteBook.  Your goal is to give a paper and pencil derivation of the formula in the print project HINT 44.2.

Perpendicuar Projection hint

Clear[X, Y, Z];
X = {3, 4, 5};
Y = 4 {1, 2, 1};
Z = (((X . Y )/Y) . Y) Y;
View3D[
        {Arrow[Z, Yellow], Arrow[X, Blue], Arrow[Y, Red], 
          Graphics3D[Line[{Z, X}]]}];
Print["Projection of ", MatrixForm[X], 
        " in the direction of ", MatrixForm[Y], " is ", 
        MatrixForm[Z]];

Assuming that you entered the surface perpendicular illustration above, this can also be illustrated as follows:

W  =  {0, 0, 2};
perp = {-fX[x1, y1], -fY[x1, y1], 1};
pW =  (W . (perp/perp) . perp) perp;
combinedPlot  =  
        Show[combinedPlot, 
          Arrow[pW, Tail -> {x1, y1, z1}, Yellow, AbsoluteThickness[4]], 
          Arrow[W, Tail -> {x1, y1, z1}, Blue]];

Displacement Vectors hint

Clear[X, Y, Z];
X = {3, 4, 5};
Y = {1, -2, 1};
Z = Y - X;
Print["Z = ", Z];
View3D[
        {Arrow[X, Blue], Arrow[Y, Blue], 
          Arrow[Y - X, Tail -> X, VenetianRed], Arrow[Z, Blue], 
          Graphics3D[Text["Z", 1.1 Z]]}];

Assuming that you entered the surface perpendicular illustration above, this can also be illustrated as follows:

Z  =  W - pW;
combinedPlot  =  
        Show[combinedPlot, 
          Arrow[Z, Tail -> {x1, y1, z1}, VenetianRed, 
            AbsoluteThickness[3]]];

Uniform Circular Motion as a Special Constrained Motion:

HINT 44.3  Do this with paper and pencil and type your work here.

The important part of the computation is to have a clear geometric interpretation of your computation as requested in the print hint.

Following is a computer computation.

Clear[X, t, ω];
X[t_] =  {r Cos[ω t], r Sin[ω t], 0};
MatrixForm[X[t]]
Norm[X[t]]
Simplify[Norm[X[t]]]

Now the first derivative

MatrixForm[X'[t]]
Norm[X'[t]]
Simplify[Norm[X'[t]]]

Now the second derivative

MatrixForm[X''[t]]
Norm[X''[t]]
Simplify[Norm[X''[t]]]

A Computer Illustration

Clear[X, t, ω];
X[t_] =  {r Cos[ω t], r Sin[ω t]};
ω = 2/3;
r = 2;
Do[
arc = 
        ParametricPlot[X[t], {t, 0, s}, AxesLabel -> {"x", "y"}, 
          PlotRange -> {{-3, 3}, {-3, 3 }}, 
          PlotStyle -> {Red}, 
          DisplayFunction -> Identity]; 
Show[arc, 
        Arrow[X[s], VenetianRed], Arrow[X'[s], Tail -> X[s], Blue], 
        Arrow[X''[s], Tail -> X[s], DarkGreen], 
        DisplayFunction -> $DisplayFunction];
, 
      {s,  .2, 3 π,  .7}]

Implicit Equations for a Surface and Its Perpendicular Vector

Derivation of the equations of constrained motion are easiest using the implicit form of a constraint surface.

h[X] = j (a constant), where  X = {x,y,z}.

For example, when  h[x, y, z] = (x^2) + 2 (y^2) - z

h[X] = 0

becomes

(x^2) + 2 (y^2) - z = 0

which is equivalent to  z = (x^2) + 2 (y^2) the Perfecto Ski Bowl equation.

The equation of the tangent plane to such an implicit surface  h[x,y,z] = j is the implicit plane in {dx,dy,dz}-coordinates given by

∂(h/∂)x dx  +  
          ∂(h/∂)y dy  +  
          ∂(h/∂)z dz  = 0	or	∇h • dX = 0

This simply says  ∇h (at a fixed {x,y,z}) is perpendicular to the unknown  dX = {dx,dy,dz}.  Since {dx,dy,dz}-coordinates are parallel to {x,y,z}-coordinates, and since the tangent plane approximates the surface, this means  ∇h[x,y,z]  is perpendicular to the surface at {x,y,z}.

HINT 44.4  do the computation for Perfecto in the implicit form and type your work here (or do it with Mathematica.)

HINT 44.5  use perpendicular projection of any vector  F  onto the perpendicular direction  ∇h  to give a perpendicular component of that vector (Subscript[F, N]).  Then use dispalcement vectors to find the tangential component (Subscript[F, T]).  Note that  F  =  (Subscript[F, T]) + (Subscript[F, N]).

An Arbitrary Motion on an Implicit Surface

HINT 44.6  work the print hint and put your work here.  Following is a Mathematica approach (that is a little awkward.)

Clear[h, x, y, z, t, j];
D[h[x[t], y[t], z[t]] == j, t]

The general second derivative is even messier,

D[t^6, {t, 2}]

Clear[h, x, y, z, t, j];
D[h[x[t], y[t], z[t]] == j, {t, 2}]

The important observation is that ANY motion  X[t]  that satisfies the implicit equation  h[X[t]] = j, satisfies:

∇h • (((d^2) X)/(dt^2)) = -(({{(dx/dt), (dy/dt)  (dz/dt)}}))(({{(∂^2) (h/∂)(x^2), (∂^2) (h/(∂x ∂y)), (∂^2) (h/(∂x ∂z))}, {(∂^2) (h/(∂x ∂y)), (∂^2) (h/∂)(y^2), (∂^2) (h/(∂y ∂z))}, {(∂^2) (h/(∂x ∂z)), (∂^2) (h/(∂y ∂z)), (∂^2) (h/∂)(z^2)}}))(({{(dx/dt)}, {(dy/dt)}, {(dz/dt)}}))

The Machine Computed General Case of Gradient and Hessian Matrix of Second Paartials

Clear[x, y, z, h]; 
    gradh[x_, y_, z_]  =  
      Evaluate[{D[h[x, y, z], x], D[h[x, y, z], y], D[h[x, y, z], z]}];
grad2h[x_, y_, z_] = gradh[x, y, z] . gradh[x, y, z];
MatrixForm[gradh[x, y, z]]
grad2h[x, y, z]

d2h[x_, y_, z_]  =  
        Evaluate[
          {D[gradh[x, y, z], x], D[gradh[x, y, z], y], D[gradh[x, y, z], z]}];
      
MatrixForm[d2h[x, y, z]]

A Special Case Gradient and Hessian

h[x_, y_, z_] = x^2 + 2  y^2 - z
MatrixForm[gradh[x, y, z]]
grad2h[x, y, z]
MatrixForm[d2h[x, y, z]]

Newton's Law Constrained to the Mountain

HINT 44.7

Work the hint here.

The General Computer Computation for λ

We use the general "phase" variables  u, v, w for velocity, that is, (dx/dt) = u, (dy/dt) = v, (dz/dt) = w and express  λ  in terms of a general implicit constraint function  h and variables  x, y, z, u, v, w:

Clear[x, y, z, h, u, v, w]; 
    gradh[x_, y_, z_]  =  
      Evaluate[{D[h[x, y, z], x], D[h[x, y, z], y], D[h[x, y, z], z]}];
grad2h[x_, y_, z_] = gradh[x, y, z] . gradh[x, y, z];
d2h[x_, y_, z_]  =  
        Evaluate[
          {D[gradh[x, y, z], x], D[gradh[x, y, z], y], D[gradh[x, y, z], z]}];
      
vel = {u, v, w};
q  = Simplify[vel . d2h[x, y, z] . vel];
λ = Simplify[-((m q)/grad2h)[x, y, z]]

A Special Case of λ

The Perfecto equations make  λ  the following:

h[x_, y_, z_]  :=  x^2  +  2 y^2  -  z;
λ
MatrixForm[λ gradh[x, y, z]]

HINT 44.8

Work the hint here.

The Circular Skier

Clear[h];
h[x_, y_, z_] :=  x^2  +  y^2;
MatrixForm[Simplify[λ gradh[x, y, z]]]

The Pendulum "Skier"

HINT 44.9  Use the general formulas above to show by hand that the equations of motion "F= ma" for the pendulum are

(dx/dt) = u

(dx/dt) = v

(du/dt) =  (g/(L^2)) x y  -  (((u^2) + (v^2))/(L^2)) x

(dv/dt) =  -(g/(L^2)) (x^2)  -  (((u^2) + (v^2))/(L^2)) y

m A = (Subscript[F, T]) +  λ ∇h

Mathematica can now write the whole equation of motion  "F = m A" as

Clear[x, y, z, u, v, w, h, F, fX, fY, fZ, q, λ, vel]; 
    gradh[x_, y_, z_]  =  
      Evaluate[{D[h[x, y, z], x], D[h[x, y, z], y], D[h[x, y, z], z]}];
grad2h[x_, y_, z_] = gradh[x, y, z] . gradh[x, y, z];
d2h[x_, y_, z_]  =  
        Evaluate[
          {D[gradh[x, y, z], x], D[gradh[x, y, z], y], D[gradh[x, y, z], z]}];
      
vel[t_] = {u[t], v[t], w[t]};
q[t_]  = Simplify[vel[t] . d2h[x[t], y[t], z[t]] . vel[t]];
λ[t_] = 
        Simplify[-((m q[t])/grad2h)[x[t], y[t], z[t]]];
F[x_, y_, z_] = {fX[x, y, z], fY[x, y, z], fZ[x, y, z]};
fT[x_, y_, z_] := 
        Simplify[
          F[x, y, z] - 
            ((F[x, y, z] . gradh[x, y, z] )/grad2h)[x, y, z] 
              gradh[x, y, z]];
\(eqs\  = \ 
      {
\(x'\)[t]\  == \ u[t], 
\(y'\)[t]\  == \ v[t], 

        \(z'\)[t]\  == \ w[t]\ , 

        {\(u'\)[t], \(v'\)[t], \(w'\)[t]} == \ 
          fT[x[t], y[t], z[t]]/m + \ λ[t]\ gradh[x[t], y[t], z[t]]/m}
        \)

The Perfecto Case

h[x_, y_, z_]  :=  x^2  +  2  y^2  - z;
fX[x_, y_, z_] = 0;
fY[x_, y_, z_] = 0;
fZ[x_, y_, z_] = -m g;
eqs

The Pendulum Case

h[x_, y_, z_]  :=  x^2  +  y^2;
fX[x_, y_, z_] = 0;
fY[x_, y_, z_] = -m g;
fZ[x_, y_, z_] = 0;
Simplify[eqs]

Re-Writing the Explicit Form

HINT 44.10

If the implicit constraint can be written in the form  h[x,y,z] = k[x,y] - z,

h[X] = j	⟺	k[x,y] - z = 0	(⟺	z = k[x,y])

Then

∂(h/∂)x =  ∂(k/∂)x	and	∂(h/∂)y =  ∂(k/∂)y	and	∂(h/∂)z =  -1

so that the gradient  ∇h=(({{∂(k/∂)x}, {∂(k/∂)y}, {-1}}))	and  |∇h| = ???

The Computation with Mathematica

Clear[x, y, z, u, v, w, h, k];
h[x_, y_, z_]  =  k[x, y]  -  z;
gradh[x_, y_, z_] = 
        Evaluate[{D[h[x, y, z], x], D[h[x, y, z], y], D[h[x, y, z], z]}];
      
% // MatrixForm

HINT 44.11

(∂^2) (h/∂)(x^2) =  
        (∂^2) (k/∂)(x^2)	and	(∂^2) (h/(∂x ∂y)) =  
        (∂^2) (k/(∂x ∂y))	and	(∂^2) (h/(∂x ∂z)) =  ??

(∂^2) (h/(∂y ∂x)) =  ?	and	(∂^2) (h/∂)(y^2) =  ?	and	(∂^2) (h/(∂y ∂z)) =  ??

(∂^2) (h/(∂z ∂x)) =  ?	and	(∂^2) (h/(∂z ∂y)) =  ?	and	(∂^2) (h/∂)(z^2) =  ??

Finish the hint here.

The Computation with Mathematica

d2h[x_, y_, z_]  =  
        Evaluate[
          {D[gradh[x, y, z], x], D[gradh[x, y, z], y], D[gradh[x, y, z], z]}];
      
% // MatrixForm
Expand[{u, v, w} . d2h[x, y, z] . {u, v, w}]

HINT 44.12

Work this by hand.

The Computation with Mathematica

Clear[x, y, z, u, v, w, h, k, F, fX, fY, fZ, q, λ, vel, m, g];
      
h[x_, y_, z_]  =  k[x, y]  -  z; 
    gradh[x_, y_, z_]  =  
      Evaluate[{D[h[x, y, z], x], D[h[x, y, z], y], D[h[x, y, z], z]}];
grad2h[x_, y_, z_] = gradh[x, y, z] . gradh[x, y, z];
d2h[x_, y_, z_]  =  
        Evaluate[
          {D[gradh[x, y, z], x], D[gradh[x, y, z], y], D[gradh[x, y, z], z]}];
      
vel[t_] = {u[t], v[t], w[t]};
q[t_]  = Simplify[vel[t] . d2h[x[t], y[t], z[t]] . vel[t]];
λ[t_] = 
        Simplify[-((m q[t])/grad2h)[x[t], y[t], z[t]]];
fX[x_, y_, z_] = 0;
fY[x_, y_, z_] = 0;
fZ[x_, y_, z_] = -m g;
F[x_, y_, z_] = {fX[x, y, z], fY[x, y, z], fZ[x, y, z]};
fT[x_, y_, z_] := 
        Simplify[
          F[x, y, z] - 
            ((F[x, y, z] . gradh[x, y, z] )/grad2h)[x, y, z] 
              gradh[x, y, z]];
fT[x, y, z]

HINT 44.13

Work this by hand here.

Computation with Mathematica

Clear[x, y, z, u, v, w, h, k, F, fX, fY, fZ, q, λ, vel, m, g];
      
h[x_, y_, z_]  =  k[x, y]  -  z; 
    gradh[x_, y_, z_]  =  
      Evaluate[{D[h[x, y, z], x], D[h[x, y, z], y], D[h[x, y, z], z]}];
grad2h[x_, y_, z_] = gradh[x, y, z] . gradh[x, y, z];
d2h[x_, y_, z_]  =  
        Evaluate[
          {D[gradh[x, y, z], x], D[gradh[x, y, z], y], D[gradh[x, y, z], z]}];
      
vel[t_] = {u[t], v[t], w[t]};
q[t_]  = Simplify[vel[t] . d2h[x[t], y[t], z[t]] . vel[t]];
λ[t_] = 
        Simplify[-((m q[t])/grad2h)[x[t], y[t], z[t]]];
fX[x_, y_, z_] = 0;
fY[x_, y_, z_] = 0;
fZ[x_, y_, z_] = -m g;
F[x_, y_, z_] = {fX[x, y, z], fY[x, y, z], fZ[x, y, z]};
fT[x_, y_, z_] := 
        Simplify[
          F[x, y, z] - 
            ((F[x, y, z] . gradh[x, y, z] )/grad2h)[x, y, z] 
              gradh[x, y, z]];
\(eqs\  = \ 
      {
\(x'\)[t]\  == \ u[t], 
\(y'\)[t]\  == \ v[t], 

        \(z'\)[t]\  == \ w[t]\ , 

        {\(u'\)[t], \(v'\)[t], \(w'\)[t]} == \ 
          fT[x[t], y[t], z[t]]/m + \ λ[t]\ gradh[x[t], y[t], z[t]]/m}
        \)

The Whole Enchalada 44.14

Clear[x, y, z, u, v, k, g, t];
pkpx[x_, y_] :=  Evaluate[D[k[x, y], x]];
pkpy[x_, y_] :=  Evaluate[D[k[x, y], y]];
P[x_, y_] :=  
        (Evaluate[D[k[x, y], x]])^2 + (Evaluate[D[k[x, y], y]])^2 + 1;
      
Q[x_, y_, u_, v_] :=  
        u^2 Evaluate[D[k[x, y], {x, 2}]] +  
          2 u v Evaluate[D[k[x, y], {x, 1}, {y, 1}]]  +  
          v^2 Evaluate[D[k[x, y], {y, 2}]];
equns={x'[t]==u[t],y'[t]==v[t],u'[t]==-(((g + Q[x[t], y[t], u[t], v[t]]))/P)[x[t], y[t]]
                       pkpx[x[t], y[t]],v'[t]==-(((g + Q[x[t], y[t], u[t], v[t]]) 
                    )/P)[x[t], y[t]] pkpy[x[t], y[t]]}

k[x_, y_]  =  x^2  +  2 y^2;
equns

Start the Perfecto Skier above x=1, y=1 at rest and let 'er slide:

g = 9.8;
ivprob={x[0] == 1,y[0] == 1,u[0] == 0,v[0] == 0,x'[t]==u[t],y'[t]==v[t],u'[t]==-(((g + Q[x[t], y[t], u[t], v[t]]))/P)[x[t], y[t]]
                       pkpx[x[t], y[t]],v'[t]==-(((g + Q[x[t], y[t], u[t], v[t]]) 
                    )/P)[x[t], y[t]] pkpy[x[t], y[t]]}
tFinal = 5;
soln = NDSolve[ivprob, {x[t], y[t], u[t], v[t]}, {t, 0, tFinal}];
xx[t_] = x[t] /. soln[[1, 1]];
yy[t_] = y[t] /. soln[[1, 2]];
path = ParametricPlot3D[{xx[t], yy[t], k[xx[t], yy[t]]}, {t, 0, tFinal}];
      
surface = 
        Plot3D[k[s, t], {s, -3/2, 3/2}, {t, -3/2, 3/2}, 
          Mesh -> False];
Show[path, surface];

Plot[xx[t], {t, 0, tFinal}];
Plot[yy[t], {t, 0, tFinal}];