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April 19, 2014

April 19, 2014

Posted by **Leo** on Monday, July 8, 2013 at 12:40am.

I find the derivative as:

L'(t)=(5.6*pi/365)*cos[(2pi/365)(t-80)]

but I get lost afterwards.

- Calculus 1 -
**Steve**, Monday, July 8, 2013 at 3:39amYou don't really need any derivatives for this. You know sin(z) has a max od 1 and a min of -1. So, when (2pi/365)(t-80) is an odd multiple of pi/2, sin is 1 or -1, so the min/max of L is 12±2.8

If you want to find the values of t where these extrema occur, then recall that max/min occur when L'=0. So, when does

cos[(2pi/365)(t-80)] = 0?

cos(z)=0 when z is an odd multiple of pi/2. So, for integer values of k, we need

(2pi/365)(t-80) = (2k+1)*pi/2

t-80 = (2k+1) pi/2 * 365/2pi

t-80 = 365(2k+1)/4

t = 80 + 365(2k+1)/4

t = 80±365/4, 80±1095/4, ...

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