ElectricPavlov

Pretty sure I will need a diagram of the mainsail.....

Since all of the parallel horizontal lines are the same distance apart      WY = YX = XV

WV = 2.8 + 2.8 + 2.8  = 8.4 units

Point-slope form

(y- -1) = -2 (x-0)

y = -2x-1

seconds in december  divided by seconds in november   will result in remander of 1 day's worth of seconds

24 hr x 3600 sec/hr    / 17 =   ......   with   remainder 6

tan 34   =    opp / adj = height / distance to goal

tan34 = (3.05-1.7) / d

d = 2  meters

So your question is looking for where the tangent line has zero slope in the interval between  -1 and 1  inclusive

  If you properly derived the derivative,  set it equal  to 0   and find where in the inteval requested, it = zero

     The derivative of a funtion IS the slope of the function     ....... given f(x)     f ' (x) is the slope.....(  gradient is the same thing as slope.)

1-2x / (sqrt1-x^2)    = 0

1-2x = 0

-2x = -1

x = 1/2

1490/x  =  596/.4

1490 *  .4/596   = x

2.4 cm / 300 p    * 650 p = .......... cm thick             (see how the p cancels out and you are left with ' cm ' ?)

Redux for posterity....figured a simpler way while sleeping last night

RATES     Rate WITH escaltor   1 escalator / 6 sec     Rate against escalator  1 escalator / 30 seconds

NOW it is like the airplane/windspeed or  boat/ current speed calcualtions

rate on still escalator =     ( 1/6 + 1/30 ) / 2 = (6/30) /2 = 6/60 = 1 /10    = rate on still esc    ( 1 esc / 10 sec)

                               then time on still escalator = 10 sec

 The mid segment is right between the lengths of the end segments.....it equals the 'average'  

( 4n+6    + 2n-2  ) /2        =     n+ 12

(6n +4)/2 = n+ 12

3n + 2 = n+ 12

2n = 10

n = 5                   use this 'n' to calc the line lengths

The angles all sum to   360

6 + 12 x       + 120    +   80     +   19x-1     = 360     now you can easily solve for 'x' to compute the angles in question!