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A model of basal metabolism.

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A model for the basal metabolism rate, in kcal/h, of a young man is given by the formula below, where t is the time in hours measured from 5:00 AM. What is the total basal metabolism of this man over a 24 hour period?
    R(t) = 85 - 0.18cos (Pi*t/12)
    ∫(0to24) R(t)dt= ? kcal

    2. Relevant equations

    The substitution rule??

    3. The attempt at a solution

    I tried to R(t)=85-0.18cos((πt)/(12)) = ∫(85-0.18cos((πt)/(12)))dt but i want to know from here what is the part that i have to use the u-substitution and from there evaluate. At first i thought that u= (πt)/(12) which i think it equaled= 0.26167t than i'm left with ∫85.0-0.18cos(0.26180t) dt.

    I went and looked around the internet just to confirm that my u substitution was right and i found a similar attempt to it. However, I am a little confused how this person reached from here ∫(85-0.18cos((πt)/(12)))dt to here 85.0t-0.68755sin0.26180t.

    How did he get 0.68755sin????
  2. jcsd
  3. Mar 21, 2012 #2


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    First of all, don't plug in numbers until the end. It just makes things ugly and cluttered. Do the u-substitution. Let u = πt/12. That way you can have an integral of the form cos(u)du, which you know how to evaluate. So, with this substitution:

    - what is du in terms of dt?

    - what do the limits of integration become?
  4. Mar 21, 2012 #3
    ok let me seee.... Let me know if i'm doing this right.


    ∫85-0.18cos(u)du=∫-sin(u)??? =∫85-0.18sin(0.26167t)


    Is that right??
  5. Mar 21, 2012 #4


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    Again, don't compute decimal values for things, and then carry all those ugly digits through all of your steps. It's just unnecessary and clutters things up. What's wrong with writing this?[tex] du = \frac{\pi}{12}dt [/tex]So, wherever you see a dt in your original integral, replace it with [itex] \frac{12}{\pi}du[/itex]. You didn't answer my second question about the limits of integration (i.e. 0 hours and 24 hours). If t = 0 hr, then what is u equal to? This tells you the lower limit of your new integral. If t = 24 hr, what is u equal to? This tells you the upper limit of your new integral.

    So far, after doing the substitution, we have [tex]\frac{12}{\pi} \int_?^
    ? (85 - 0.18\cos u)\,du [/tex]where I want you to fill in the ?'s

    Let's address this part before moving on to the rest.
    Last edited: Mar 21, 2012
  6. Mar 21, 2012 #5
    the ?'s are upper limit 24hours and lower limit 0 hours. I went right to the decimals because I did not know what du without taking the derivative was since is a fraction I was thinking in Quotient rule or product rule. I know i have to replace my du with what it is in my original integral but i'm not sure how did you flipped the fraction. I would love to see how did you do that.
  7. Mar 22, 2012 #6


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    :grumpy: ... Those are the limits of integration of the original integral, which was with respect to t. You need to figure out the limits of integration of the new integral, which, after the substitution, is now with respect to u. To determine these new limits, answer these questions:

    - at t = 0, what is u equal to?
    - at t = 24, what is u equal to?

    :confused: I actually have no idea what you are trying to say here.

    If[tex] u = \frac{\pi}{12} t [/tex]then [tex]\frac{du}{dt} = \frac{\pi}{12}[/tex]Now, I want to emphasize that what I am about to write below is an abuse of notation, because we are NOT supposed to think of du/dt as being a fraction. Nevertheless, it's a pretty standard thing that you see see people doing when integrating by substitution, and it works. We can write the above as[tex]du = \frac{\pi}{12}dt[/tex]Now simply re-arranging the above equation in order to solve it for dt, you get [tex] dt = \frac{12}{\pi}du[/tex]
    I did it by...basic algebra. As I explained above in boldface, all I did was rearrange the equation to solve for dt, so that I would know what to replace dt with.
  8. Mar 22, 2012 #7
    nevermind i got what you were doing. However upper limit of u is t=24 u=pi/12(24)=2pi and lower limit of u is t=0 u=pi/12(0)=0 12/pi ∫upper limit (2pi) , lower limit (0) 85-0.18cos(u)du than what?
  9. Mar 22, 2012 #8


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    Just out of curiosity, are you taking a course in integral calculus right now, or are you just attempting to teach yourself?

    Regarding your question. Step 1 is this: you have a function that is actually a sum of two different terms. In fact you can consider them to be two different functions g(u) = 85 and f(u) = -0.18cos(u) that are being added together.

    What is the integral of the sum of two different functions? [g(u) + f(u)]du = ?
  10. Mar 22, 2012 #9
    I am taking calculus one and just learning integrals. Integral calculus is Cal 2 so no, I am not teaching my self integral calculus but I am trying to learn it. for the integral of g(u)=85=85u and for the integral of f(u)=-0.18cos(u)=-9/50sin(u).
  11. Mar 22, 2012 #10
    ∫ 85u-9/50∫ sin(u)
  12. Mar 22, 2012 #11
    Note: cal 1 is learning how to differentiate. cal 2 is learning how to integrate. I am in baby steps towards cal 2.
  13. Mar 22, 2012 #12


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    The answer I was going for was that the integral of the sum of two functions is equal to the sum of the integrals of those two functions. I.e.

    [g(u) + f(u)]du = g(u)du + f(u)du

    I think that you know this, you just didn't say it outright.

    Umm, yeah, that's sort of correct. There are a couple of things wrong with your notation. You are missing the du's. Also, it's really weird that you still have the integration signs in there after you've finished integrating. You wouldn't include the integration sign anymore, because...you've already computed the integral. For instance, you'd write cos(u)du = -sin(u).

    You would NOT write cos(u)du = - sin(u)du because this is just not true. The integral cos is not equal to the negative integral of sine. The integral of cos equals negative sine.

    In any case, getting back to the problem at hand: now what needs to done? Remember that this is a definite integral that you are evaluating, which means that it has limits. You have to do something with those limits...
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