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    Hey,
    Looking at this integrtion question

    The temperature (in degrees cenigrade) measured at a weather station on 21st Sept is given by

    T(t) = 15 - 10cos^2(pi*t/24) - t/10

    0 < t < 24

    where t is the time measured in hours. Find average temperature over the entire day, and average during the hours of daylight 6<t<18

    It's just integrating the expression I can't do, the worst part is I have my working from when I have done it before but I haven't copied it all down and now I can't see how I integrated that expression last time to get the answer :|

    Thanks for any help
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    (Original post by RHCPfan)
    Hey,
    Looking at this integrtion question

    The temperature (in degrees cenigrade) measured at a weather station on 21st Sept is given by

    T(t) = 15 - 10cos^2(pi*t/24) - t/10

    0 < t < 24

    where t is the time measured in hours. Find average temperature over the entire day, and average during the hours of daylight 6<t<18

    It's just integrating the expression I can't do, the worst part is I have my working from when I have done it before but I haven't copied it all down and now I can't see how I integrated that expression last time to get the answer :|

    Thanks for any help
    Use double angle formulae for cos^2

    cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1 and rearrangle for cos squared.

    and x \mapsto \frac{\pi}{24}\cdot t
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    (Original post by RDKGames)
    Use double angle formulae for cos^2

    cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1 and rearrangle for cos squared.

    and x \mapsto \frac{\pi}{24}\cdot t
    Ah yeah so that gives me

    T(t) = 15 - 5 - 5cos(pi*t/12) - t/10
    = 10 - 5cos(pi*t/12) - t/10

    thanks!

    I'm just a little unsure how to do the, - 5cos(pi*t/12)? Sorry my brain is failing as i'm sitting looking at my previous correct answer for this!
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    (Original post by RHCPfan)
    Ah yeah so that gives me

    T(t) = 15 - 5 + 5cos(pi*t/12) - t/10
    = 10 + + 5cos(pi*t/12) - t/10

    thanks!

    I'm just a little unsure how to do the, + 5cos(pi*t/12)? Sorry my brain is failing as i'm sitting looking at my previous correct answer for this!
    Reverse chain rule. Cos goes to sine and you divide the 5 by the coefficient of t; in this case pi/12
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    (Original post by RDKGames)
    Reverse chain rule. Cos goes to sine and you divide the 5 by the coefficient of t; in this case pi/12
    Ah yeah! Thanks so now I have it

    (-60/pi)sin(pi*t/12)

    Thank you
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    (Original post by RDKGames)
    Reverse chain rule. Cos goes to sine and you divide the 5 by the coefficient of t; in this case pi/12
    The chain rule in general doesn't work in reverse: the method that you have described only works when the derivative of the inner function is a constant, or in other words, the inner function is linear. What you are really doing is making a substitution and skipping out the steps that you can do in your head.
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    (Original post by HapaxOromenon3)
    The chain rule in general doesn't work in reverse: the method that you have described only works when the derivative of the inner function is a constant, or in other words, the inner function is linear. What you are really doing is making a substitution and skipping out the steps that you can do in your head.
    Exactly. No need to fuss around explaining this to him if he had done it before.
 
 
 
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