page 139 example 5


(* How fast does the shadow lengthen *)

The sun is at (0,s).

R is the radius of the earth.

h is the height of the building.

a is the angle of the radius through the building.

The top of the building is at

           (R+h) (Cos[a],Sin[a])

The line from the sun to the top of the building is


                     s-(R+h)Sin[a] 
         (y-s) =  -----------------  x
                      -(R+h)Cos[a]) 


        Solving for the point on the earth at the end of the shadow.


        x^2+y^2 = R^2  and


                     s-(R+h)Sin[a] 
         (y-s) =  -----------------  x
                      -(R+h)Cos[a]) 


      2             2                   2                 2
     y   - 2 y s + s  = (s-(R+h) Sin[a])  /( (R+h) Cos[a])  (R^2-y^2)

      
 y^2  (1+(s-(R+h) Sin[a])^2/( (R+h) Cos[a])^2 -2 y s + s^2

                        -(s-(R+h) Sin[a])^2 / (R+h) Cos[a])^2 R^2 = 0


The length of the shadow is

  R(a-ArcSin[y/R])

R = 4000;
s = 93000000;
h = 120/5280;
m = (s - (R+h)Sin[a] )/(-(R+h) Cos[a] );
ans = Solve[(y-s)^2 - m^2 (R^2 - y^2) == 0, y];
y1 = y /. ans[[1]];
y2 = y /. ans[[2]];

f[a_] = R(a - ArcSin[y1/R]); 
g[a_] = R(a - ArcSin[y2/R]); 

d =  2 Pi/86400;

P1 = Plot[ R ( g[a] - g[a+d] ),{a,0.1,Pi/2-0.001},PlotStyle->{RGBColor[1,0,0]}];
P2 = Plot[Pi/360 Csc[a]^2,{a,0.1,Pi/2}];

Show[P1,P2,PlotRange->All]

gp[a_] = D[g[a],a];

P3 = Plot[-gp[a],{a,0.1, 1.5},PlotRange->All,PlotStyle->{RGBColor[1,0,0]}];
P4 = Plot[ Pi/360 Csc[a]^2,{a,0.1,1.5},PlotRange->All];
Show[P3,P4,PlotRange->All]



---
Irvin Roy Hentzel
hentzel@iastate.edu
Professor of Mathematics
515-294-8141
515-294-5454 (FAX)