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Max, min, and avg temp

  1. Sep 20, 2004 #1
    Alright, I am given a function that approximates the outdoor temperature on a particular day where t is the time in hours after 9pm.

    I am to find the max and min temps and average temps on tha day between 9am - 9pm. I know i am supposed to derive the given function to find the max and min temps but once i have the derived function, what am i supposed to do with it? The instructor didn't go in depth on how we are supposed to do it. Also, i know that to find the average temp you are supposed to integrate it and do


    Is that all there is to it for finding the average temp?

  2. jcsd
  3. Sep 20, 2004 #2
    9 pm-9am ? t is the time in hours after 9 pm ....
  4. Sep 21, 2004 #3
    thats the way it says on the sheet. anyone else have any input?
  5. Sep 21, 2004 #4
    To answer the first part of your question, think about what you are doing when you derive a function: you are finding a general equation in terms of a variable for the slope of the original function. What can you do with this equation to find the min/max values? (Huge hint: what is the slope equal to at those points?). Don't forget to verify to see which result is max and which is min.
  6. Sep 21, 2004 #5
    I guess the slope would be zero at those points. so i just compute the derivative of T(t)=50+14sin([pi*t]/12) and plug in 0 for t to find the max and mins?
  7. Sep 21, 2004 #6
    No. You do not want to know the temp. at time=0. Remember, the derived equation represents the slope, and as you said yourself, the slope is zero at max/min points.
  8. Sep 21, 2004 #7
    Why don't you draw the graph to find the max and min?
    Average has to do the integration.

    Attached Files:

  9. Sep 21, 2004 #8
    alright.. im guessing i'm supposed to derive the T(t) equation, set that equal to 0 and find the high and low times and then plug those times back in the original equation? Thakns for all your help.

    PS this is prolly wrong. as for graphing it, my teacher would prefer us not to.
  10. Sep 21, 2004 #9
    sometimes it's not necessary to use calculus.
    |sin(alpha)|< or = 1 => the maxium value for this function is sin(alpha)=1
    and the minimum => sin(alpha)=-1from here you can use these value to obtain: for "T"max, T(t)= 50+14=64 and for "T"min, T(t)=50-14 =36.
  11. Sep 22, 2004 #10
    wisky40's way is the fastest to obtain the min. but derivative won't help you to find the max. from the graph i drew,the max happens when t is 6 hours after 9pm, means it happens at 3 am which is out of the range of the question which states from 9 am-9pm. i think for this period, the max is just 50 which happens at t = 12 and 24 corresponds to 9 am and 9 pm respectively.
  12. Sep 23, 2004 #11
    real quick. how would i evaluate the integral of sin[(pi*x)/12]?

    is it just -cos[(pi*t)/12] ?
    Last edited: Sep 23, 2004
  13. Sep 24, 2004 #12
    No. It is
    \frac{-cos\frac{\pi t}{12}}{\frac{\pi}{12}}[/tex]
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