My Math Forum maximum rate of energy consumption

 Calculus Calculus Math Forum

 October 26th, 2012, 10:47 AM #1 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 maximum rate of energy consumption totally lost here. what is the maximum value? I figured out the time (noon) but not the last part. Thanks!
 October 28th, 2012, 06:58 PM #2 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 Re: maximum rate of energy consumption oops sorry was over the upload count for my imageshack, here's the problem I need help with (got everything else, need the last part):
 October 28th, 2012, 07:20 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: maximum rate of energy consumption You need to find P(0) right?
 October 28th, 2012, 08:12 PM #4 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 Re: maximum rate of energy consumption I believe so since the starting point is at noon.
 October 28th, 2012, 08:24 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: maximum rate of energy consumption You use t = 0, not because this is noon, but because this is where the power consumption is at a maximum. $P'(t)=-\frac{125\pi}{6}\sin$$\frac{\pi t}{12}$$$ $P''(t)=-\frac{125\pi^2}{72}\cos$$\frac{\pi t}{12}$$$ Now, equating the first derivative to 0, we find: $\frac{\pi t}{12}=k\pi$ where $k\in\mathbb{Z}$ $t=12k$ Since the period of the sinusoidal energy function is 24 hr, we need only consider $k\in\{0,1\}$ So, we look at: i) $t=0$ we find $P''(0)<0$ so this extremum is a maximum. ii) $t=12$ we find $P''(0)>0$ so this extremum is a minimum. So, we find the maximum power consumption occurs at $t=0$. Now evaluate the power function at this point to find the maximum power consumption.
 October 29th, 2012, 08:27 AM #6 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 Re: maximum rate of energy consumption ok got it, thanks Mark!

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