This study contains brief description of the Akkas gas field in the west dessert of Iraq, as well as some suggestions to develop it.
We can easily understand the meaning of mechanical engineering and the uses of this type of engineering. We never thought mechanical engineering as a profession is field of sports. This is very pop...
Woodsmith Magazine - 063
ROMANIAN SF
Complete the solution of the brachistochrone problem begun in section 2.2 and show that the desired curve is a cycloid with a cusp at the initial point at which the particle is released.
dx 2 dy 2 divided by velocity
Functional we want to extremize is arclength
t
2
ds
v
1
1 x dy
1 x
2
2
1
2 gy
f ( x, x; y )
2 gy , d
0 dy 2 gy(1 x 2 )
2
2 gy
x
[I.1]
This leads to the integral,
x 2
2 gy(1 x 2 ) 2 ga
Hereon, a k . Use the change of variables y
x( y )
y ky
dy
Half-angle formula-- sin ( 2 ) ( e 2
y a
x 2 ya (1 x 2 ) x
i / 2
k k si n ( )
d (k sin ( )) k 2 2
y k
sin 2 ( 2 )
x( y)
y a y
dy
,
sin( 2 ) cos( 2 ) d k sin 2 ( 2 ) d
2 2
cos ( )
[I.2]
[I.3]
) 14 ( ei e i 2e ) 41 (2 cos 2) , making the integral [I.3] into,
e i / 2 2 2i
1 ay
k sin 2 ( 12 ) 2 sin 1
k sin 2 ( 2 ) 2 2
0
x( y ) 12 k (1 cos )d 12 k ( sin ) C0 Even if we reverse the change of variables (using sin( x) Im(e transcendental equation x( y ) 12 k (2 sin
1
y k
1 2
k ( sin ) C 0
e ix ) cos
i x / 2
/2
2 x 2
[I.4]
sin 2 2x 1 2 sin 2 2x ), you get a
)) x0 which you can’t invert to get y(x). You must settle for a (1 2 k y ))
parametric solution. The parameter parameter is as in the change of variables, and you get,
x( ) 12 k ( sin ) x0
y( ) y0 sin 2 ( 12 )
1 2
y0 (1 sin )
[I.5]
oops…off by something…
Show, also, that if the particle is projected with an initial kinetic energy
1 2
2
mv0 that the brachistochrone is still a cycloid
passing through through the two points with with a cusp at at a height z above the the initial point point given given by v0 2 gz . 2
