(*
CALCULUS 265 Honors
Fall Semester 2000
8:00 to 8:50 MTTF
Carver 0002

Instructor: Hentzel
Office: 432 Carver
Office Hours: 9:00-10:00 MTTF

E-mail  hentzel@iastate.edu
phone  294-8141
Website

http://www.math.iastate.edu/hentzel/honors.265 

Tuesday, October 10:    Section 14.3 

previous
p1016:1,9,11     
p1018:43,48,49   

Main idea: 

Key words: 

Goal:  

Assume that the force of friction is proportional to the

weight of the skier.  Show that the work done on a slope

is independent of path.  The coefficient of friction is u.


     P = ( x(t),y(t) )  

     G = (0,-mg).

     T = ( x', y')/Sqrt[ (x')^2 + (y')^2)

     N = ( y',-x')/Sqrt[ (x')^2 + (y')^2)

    
     F = (G.T)T - u |G.N| T        (||  = absolute value )
     
   Work done = Int F.T ds 

             =  Int  {(G.T)T - u |G.N|}.T ds       


             =  INT  (G.T - u |G.N| ) ds


                      (0,-mg).(x',y') - u |(0,-mg).(y',-x')|
              = INT  ----------------------------------------  Sqrt[(x')^2+(y')^2] dt
                            Sqrt[(x')^2+(y')^2]


              = INT  (0,-mg).(x',y') - u |(0,-mg).(y',-x')| dt


              = mg INT -y' - u |x'| dt          (x' is positive. )

                           stop
              = mg (-y-ux)
                           start

 Notice that: 
    
       The y-value at the start is larger than the y value at the stop.  

       The reverse is true for the x-values.


       So the work done on the skier by gravity increases 

          if the skier descends further. 

       The work done on the skier decreases 

          if he travels further horizontally.   


      The work done = 1/2( (Vstart)^2 - (Vstop)^2).   Vstart is the

velocity at the top and vstop is the velocity at the bottom.  


       One can measure the snow friction if we can measure the velocity of the

skier at the top and bottom and know the horizontal and vertical dispacements of 

the start and stop.