Posted by Anonymous on Sunday, April 5, 2009 at 8:54pm.
Take the derivitive of D with respect to t. Set to zero, solve for the solutions. Those will be max/min day lengths. You can use the second derivitative to find out which.
for 3,4, set dD/dt equal to 2/60 and solve for t.
wait to take the derivative, should i multiple out the stuff inside the ()?
dD/dt = A(2pi/365)*cos [(2pi/365)*(t-80)]
=0 for maximum or minimum
(2pi/365)*(t-80) = n*pi/2
when n=1 gives maximum , then (t-80) = 365/4, for minimum n = 3
and t-80 = 3*365/4
(3) = dD/dt = +2/60
(4) = dD/dt = -2/60
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