Conference rusure::math

    
    how about this way,
    
    since s= SQRT( x^2 + y^2)
    express the above it terms of polar coordinate, and then do
    d s
    --------- = 0  for max, or min, and solve for Angle ?
    d (angle)
    
    
    

    re .-1
    
    I should have mentioned that I don't have an elegant solution to this
    myself.
    
    I didn't think of using polar coordinates. Whether this is beneficial
    or not depends on the form of the equation of the path in p.c. or more
    accurately the form of the path length integral which tends to be yucky
    even for simple curves.
    
    Differentiating the path length integral with respect to theta, setting
    this to 0 and solving for theta is what I had in mind.
    
    Let's see what you come up with!

>    since s= SQRT( x^2 + y^2)
>    express the above it terms of polar coordinate, and then do
>    d s
>    --------- = 0  for max, or min, and solve for Angle ?
>    d (angle)


    The equation for s above is only for a straight line; Or explicitly it
    gives the straight line distance between the origin and the point (x,y).
    You need

    ds = SQRT( (dx)^2 + (dy)^2 )

I can't believe that this problem is too hard so I have to assume that no-one
wants to wade through a few pages of algebra. Nevertheless, for those readers
satisfying the condition "0 < curiosity <= laziness", I present my solution.



Let � be the launch angle relative to the horizontal
Let g be the magnitude of the acceleration due to gravity (assumed > 0)
Let v be the launch speed (assumed > 0)

Suppose 0 <= � < pi/2 since the boundary case (pi/2) needs special treatment.

The following parametric equations describe the motion of the projectile.

(1) x = vt cos�
(2) y = vt sin� - �gt�

Eliminating t by substituting x/(v cos �) for t from (1) in (2) gives

        sin�           g
(3) y = ---- x  -  --------- x�
        cos�       2v� cos��

Also solving (2) for y = 0 in order to find the start and end of the path gives

t = 0 (obviously) or t = (2v sin�)/g

    and substituting for t in (1) gives

                 2v� sin� cos�
x = 0 or x = r = ------------- which are the end points (r for range)
                       g

To simplify (3), make the substitution z = x - �r  which will centre the path
on the y-axis. Replacing x by z + �r in (3) gives

    sin�                  g
y = ---- (z + �r)  -  --------- (z� + rz + �r�)
    cos�              2v� cos��

  = sin�     sin�   v� sin� cos�       g              g       2v� sin� cos�
    ---- z + ---- � ------------ - --------- z� - --------- � ------------- z
    cos�     cos�         g        2v� cos��      2v� cos��         g

                 g       v� v� sin�� cos��
           - --------- � -----------------
             2v� cos��           g�

    v� sin��   v� sin��       g
  = -------- - -------- - --------- z�
       g          2g      2v� cos��

    v� sin��       g
  = -------- - --------- z�
       2g      2v� cos��

with end points z = -�r and z = +�r.

Reverting to use of x instead of z then

        v� sin��       g
(4) y = -------- - --------- x�
           2g      2v� cos��

Differentiating (4) w.r.t. x in order to apply the path length integral formula
we get
          g
y' = - -------- x
       v� cos��

Let � = g/(v� cos��) so y' = -�x and define L(�) as the path length. Then
by symmetry we can integrate from x = 0 to x = �r and multiply the result by 2
so

L(�) =  2 integral(0,�r) sqrt(1 + ��x�) dx

To find the indefinite integral make the subsitution "sinh u = �x" whence
sqrt(1 + ��x�) = cosh u  and  dx = (cosh u)/� du.

         1
integral - cosh�u du
         �

  1
= - integral (� cosh 2u + �) du
  �

  1
= - (� sinh 2u + �u)
  �

  1
= - (� sinh u cosh u + �u)
  �

  1
= - (��x sqrt(1 + ��x�) + � arcsinh �x)
  �

                      1
= �x sqrt(1 + ��x�) + -- arcsinh �x
                      2�

Evaluating at the end points, the above expression is 0 when x = 0 and for
    x = �r

�x = g/(v� cos��) . (v� sin� cos�)/g = tan�  and so sqrt(1 + ��x�) = sec�

           1   v� sin� cos�            1   v� cos��
L(�) = 2 ( - � ------------ � sec�  +  - � -------- � arcsinh tan�)
           2         g                 2      g

       v�
     = -- (sin� + cos�� arcsinh tan�)
       g

Now arcsinh x = ln(x + sqrt(x� + 1)) (*) so

arcsinh tan� = ln(tan� + sec�) = ln((1 + sin�)/cos�) = -ln(cos�/(1 + sin�))
    and cos�/(1 + sin�) = tan(pi/4 - �/2) (*) so

(5) L(�) = v�/g . (sin� - cos�� ln tan(pi/4 - �/2))

This expression for L(�) is not defined at �=pi/2. However it can be shown that
limit(�->pi/2) L(�) exists and is v�/g which is the correct value for L(pi/2).


Differentiating (5) for L w.r.t. � and equating with 0 we get

                                        cos�� sec�(pi/4 - �/2)(-�)
cos� - 2cos�(-sin�)ln tan(pi/4 - �/2) - -------------------------- = 0
                                             tan(pi/4 - �/2)

                               �cos� sec�(pi/4 - �/2)
1 + 2sin� ln tan(pi/4 - �/2) + ---------------------- = 0
                                  tan(pi/4 - �/2)

But �sec�t/tan t = 1/(2 sin t cos t) = 1/sin 2t and with t = pi/4 - �/2, so that
2t = pi/2 - �, in the last term of the above expression we get 

                                    cos�
1 + 2sin� ln tan(pi/4 - �/2) + ------------- = 0
                               sin(pi/2 - �)

but sin(pi/2 - �) = cos� so

2 + 2sin� ln tan(pi/4 - �/2) = 0

(6) sin� ln tan(pi/4 - �/2) = -1

This is as far as I can go with algebra. Solving this numerically gives

� = 56.47� (2 d.p.)

and I state without proof that this is a local maximum and that the value of L
at that point exceeds the value at the end points and thus the above point
gives the maximum over the interval [0,pi/2].


(*) left as an exercise for the reader

Nice work!  I double-checked it all, and also computed
� = 0.985514737862315475 radians = 56.46583512745235�.