Project 44: The Perfecto Skier
Previous project Next project

This project investigates the 3-D effect of sliding down a slippery slope. The component of the force due to graivty acts opposite the gradient, but motion under this acceleration does not produce gradient trajectories. You don't just ooze to the bottom, but zoom right on by. Problem 18.2 of the core text found the gradient trajectories. Exercise 17.7.6 found the tangential component of a force acting on a plane. We begin with a quick reminder about those problems.

Perfecto Ski Bowl has the shape of the surface


  • % Show that the normal vector to the surface of Perfecto above the point (x,y) is

    A skier falls somewhere on the slope above (x0,y0) with initial velocity zero but is without control of direction; his edges are in the air.

    The skier's weight vector is


    where g is Newton's gravitational constant (g=9.8 in mks units). The mountain counteracts the force along its normal, but the rest of his weight vector acts down the mountain.

  • % Show that the force pulling the skier down the mountain when he is at the point above (x,y) is

    Newton's "F = m a" law says that the mass times the acceleration of the skier, , equals the total force on the skier,


    where FM is the force supplied by the mountain. If we assume that the slope is perfectly slippery, this force will be perpendicular to the mountain and act just to keep the skier on the curves of the mountain.

    44.1 The Mountain's Contribution

    A skier at rest will only have the tangential component of gravity acting on him, but in motion, the mountain must contribute a force. We want you to see a simple case of this force without worrying about gravity.

    The vector function


    moves at constant speed around the circle of radius r, given implicitly by x2+y2=r2 and z=0.

  • Compute |X|, and and show that they are all different constants.

    Suppose that we want the speed of the motion to be s. Show that we need to take , so


    The acceleration of a circular "mountain" on a "skier" at constant speed s is s2/r in the direction .

    It is easiest to represent our mountain implicitly


    For example, for Perfecto, we can take the function h[x,y,z]=x2+2y2-z and the implicit equation, h[x,y,z]=0. The reason we want to use this form is so we have a simple way to find a normal to the mountain. In this case, the gradient

  • Verify that is perpendicular to Perfecto Ski Bowl when h=x2+2y2-z.

  • If F is any vector, derive a formula for the component of F that is tangent to the mountain h[X]=j,

    Test your formula in the case F=(0,0,-mg) and h[x,y,z]=x2+2y2-z.

    The force due to the mountain will be a scalar multiple of the perpendicular vector, , and


    We want to compute , knowing that it somehow depends on the curvature of the mountain and the skier's speed.

  • Let X[t] be any parametric function and suppose it lies on the mountain,

    Differentiate this equation once with respect to t and show that

    What does this equation say about the direction of the velocity of the motion X[t]?

    The equation above can be written in components as


    If we differentiate this equation with respect to t again using the Product Rule and Chain Rule, we obtain

    Notice that the last line of this derivative contains the dot product

    We summarize the computation in matrix notation

    where h(2)[X] is the matrix of second partial derivatives of the constraint function h[X]. Notice that this term involves the velocity and second partials related to the curves on the mountain.

    44.2 Gravity and the Mountain

    We use this fact about any motion on h[X[t]]=j together with the fact that the tangential force is perpendicular to to derive a formula for the scalar :


  • Solve these two equations for :

    Test your formula on the case of constant speed s motion around the circle x2+y2=r2, z=0. Newton's law with no tangential force becomes


    Now compute the specific quantities:



  • Recall that the force on the skier is in the direction . Simplify the expressions in this special case and show that the formula agrees with this, .

    44.3 The Pendulum as Constrained Motion

    We may view the pendulum (described in Project 46) as a motion on a circle


    so h[x,y,z]=x2+y2. The only applied force is still gravity which we take as F=(0,-mg,0).

  • Show that the differential equation describing the motion of a pendulum of length L with gravitational constant g may be written:

    44.4 The Explicit Surface Case

    Suppose that the surface constraining the motion can be expressed as an explicit function of x and y,


    The constraint h[x,y,z]=j can be written h[x,y,z]=k[x,y]-z=0 for the general formulas.

  • Show that

    and compute this quantity for the Perfecto Ski Bowl function.

  • Show that

    and

    and compute this quantity for the Perfecto Ski Bowl function.

  • Show that when F=(0,0,-mg),

    and compute this quantity for the Perfecto Ski Bowl function.

  • Let , , , , and . Show that the equations of motion on z=k[x,y] may be written

    and write these equations for the Perfecto Ski Bowl function.

    Notice that the z variable does not appear in the equations for x, y, u, and v. This means that we could solve the pair of second-order equations for x and y, or , , and


    together with initial conditions and find functions x[t] and y[t]. We could then simply let z[t]=h[x[t],y[t]].

    Now put it all together:

    Problem 44.1 Use the phase variable trick from Chapter 23 of the core text to write the equations of motion for the skier as a differential equation:


    In a program, you can "fill in the blanks" in simple steps using computer derivatives and the formulas above. Solve your equations and plot the path of the skier.

    Use the PerfectoHelp program from our website to solve these equations for several choices of the initial conditions. Does the skier stop at the bottom?

    Previous project Next project Close this window