Post a New Question

Calculus 1

posted by .

I'm asked to find max and min of: L(t)=12+2.8sin((2pi/365)(t-80).

I find the derivative as:

but I get lost afterwards.

  • Calculus 1 -

    You 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, ...

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question