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Simple formula for an exponential function. Please help i'm stuck

  1. Feb 21, 2013 #1
    Q
    A cold yam is plaveedd in a hot oven. Newton's Law of Heating tells us that the difference between the oven's temperature and the yam's temperature decays exponentially with time. The yam's temperature is initially 0 deg F, the oven's temp is 300 deg F, and the temp difference decreases by 3% per minute. Find a formula for Y(t), the yam's temperature at time t.

    My A
    Well my thought is that since the Y-axis is the yam's temperature then the function must be increasing. So then the A variable in f(x)=A(B^x) would be 0, which is impossible. I keep trying to use 300 for A but then it is not an increasing function. Am I just reading the question wrong or over thinking this. Please help. Thank you very much!
     
  2. jcsd
  3. Feb 21, 2013 #2

    jbunniii

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    [edit] ignore - I misread the problem. See my next post.
    [strike]The ##y##-axis should represent the difference between the yam and the oven. Therefore the ##y## value at time 0 should be 300 and it should decrease over time.[/strike]
     
    Last edited: Feb 22, 2013
  4. Feb 21, 2013 #3
    But in the question it states "Y(t), YAMS TEMPERATURE at time t" so the Y-axis should be the yams temperature not the difference, because 300 deg F is not the temp of the yam at time 0 Right?
     
  5. Feb 21, 2013 #4
    So you have a differential equation in Y'(t). Solving it and using the given initial condition will give you your function Y(t).
     
  6. Feb 21, 2013 #5

    jbunniii

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    Oops, you're right, I misread. Sorry about that.

    If we assume that the oven's temperature remains constant, then for the difference between the two temperatures, we can write
    $$D(t) = 300 - Y(t)$$
    or equivalently,
    $$Y(t) = 300 - D(t)$$
    and you are given information about the rate of change of ##D(t)##. You can easily translate that into information about the rate of change of ##Y(t)##, because
    $$Y'(t) = -D'(t)$$
     
  7. Feb 21, 2013 #6
    So do you have any suggestions I'm stumped. It canot be 0 for A which is the yams initial temp and I cannot think of what it could be. I'm pretty sure B is 1.03^t.
     
  8. Feb 21, 2013 #7
    Wait sorry didn't see the bottome half. Thank you.
     
  9. Feb 21, 2013 #8

    jbunniii

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    If the difference ##D(t)## decreases by 3% per minute, and ##t## represents minutes, then is of the form ##D'(t) = \lambda D(t)##. What should ##\lambda## be?
     
  10. Feb 22, 2013 #9
    Sorry but what is D' or Y' mean. I have not seen ' used like that before before.
     
  11. Feb 22, 2013 #10

    jbunniii

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    Standard notation for the derivative:
    $$D'(t) = \frac{dD}{dt}$$
     
  12. Feb 22, 2013 #11
    Oh that makes sense. Yeah I'm only in Precalc so Im not quite there yet. Can you explan it in simpler terms. More along the lines of just f(x)=a(b^x) thats all the form we are using so far so basicslly what I have is Y(t)=a(1.03^t) is that incorrect because it is that valure of a that i cannot figur out I know it is in the form f(x)=a(b^x) because its the only for they gave us throughout the whold section and the 4 ahead of it.
     
  13. Feb 22, 2013 #12

    jbunniii

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    I see. If you don't know about derivatives yet, then you can't really derive the answer from first principles. If you'll bear with me, I'll walk through the steps, and hopefully that will be helpful for future problems like this.

    The derivative of a function simply means "the rate of change" of that function. In this case, we know something about the rate of change of the temperature difference. So we assign a function to represent the temperature difference, call it ##D(t)##. We are given that this function decreases by 3% every minute. If ##t## represents time in minutes, that means
    $$D'(t) = -0.03D(t)$$
    where the left hand side means "the rate of change of ##D(t)##", in other words how much does ##D(t)## change in a minute. It decreases by 3% per minute, which means it decreases by an amount equal to 3% of its current value. That is why the right hand side is ##-0.03D(t)##.

    OK, now comes the magic step since you aren't familiar with derivatives. You probably won't understand it all right now, but think of it as a preview of coming attractions, in case you take calculus in the future. We divide both sides by ##D(t)##:
    $$\frac{D'(t)}{D(t)} = -0.03$$
    and now we integrate both sides:
    $$\log(D(t)) = -0.03t + C$$
    where ##\log## is the natural log (base ##e##), and ##C## is some constant that we have to solve for. First, let's undo the log. We undo the log by exponentiating both sides of the equation:
    $$e^{\log(D(t))} = e^{-0.03t + C}$$
    OK, we're done with the calculus part now, so you can take over. The next step is to simplify ##e^{\log(D(t))}## and then solve for ##C##. To solve for ##C##, evaluate the equation at ##t = 0##. You should have enough information to know what ##D(0)## is.
     
  14. Feb 22, 2013 #13
    Well I know D(t)=300-Y(t), but I'm still puzzled I dont know if you need it for this because I think Y(t) is what you are getting at above but how do I get Y(t) for
    D(t)= 300-Y(t). Well actually if you plug 0 in the t for Y(t) it would be 0 correct so then
    D(0)=300. Ok i way over thought that but anyways. so e^log(300)=e^-0.03t+C. Sorry but that is as far with logs as I can go we havnt gone any further.

    On another note is the only way to solve this problem with logs because if so I way over thought this and its probably something alot simpiler like possibly by temp of the yam they really did mean the difference, because for the way the answer of this should look it should only contain a and b constants not C too.
     
  15. Feb 22, 2013 #14

    jbunniii

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    No, don't worry about ##C##, we're going to solve for that. The only thing you need to know about ##\log## is that it cancels out with exponentiation by ##e##, in other words ##e^x## and ##\log(x)## are inverse functions. So ##e^{\log(300)}## is simply ##300##. Also, in order to get ##300## you must have plugged in ##t = 0## to get ##D(0) = 300##, so you have to plug in ##t = 0## on the right hand side too. Now the equation becomes
    $$300 = e^{-0.03(0) + C}$$
    or simply
    $$300 = e^C$$
    So going back to our previous equation,
    $$D(t) = e^{-0.03t + C}$$
    we can use the property that ##e^{x+y} = e^x e^y## to rewrite the right hand side:
    $$D(t) = e^{-0.03t}e^C$$
    We just worked out that ##e^C = 300##, so the equation simplifies to
    $$D(t) = 300 e^{-0.03t}$$
    So that gives us a formula for ##D(t)##. How can you use that to get a formula for ##Y(t)##?
     
    Last edited: Feb 22, 2013
  16. Feb 22, 2013 #15
    Well from above Y(t)=300-D(t) so Y(t)=300-300e^-0.031. I think
     
  17. Feb 22, 2013 #16

    jbunniii

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    Right,
    $$Y(t) = 300 - 300e^{-0.03t}$$
    which can also be written as
    $$Y(t) = 300(1 - e^{-0.03t})$$
    Does this look at all like any of the problems you've done already, or examples in your book? I'm not sure why you thought the answer would have to be of the form ##A(B^t)##.

    By the way, it's always a good idea as a sanity check to make sure the formula works for the initial condition you are given. ##Y(t)## represents the temperature of the yam versus time, and we're given the temperature at time ##0##. So try evaluating ##Y(0)## to make sure it matches.
     
  18. Feb 22, 2013 #17
    Well it just kind of makes me think that I over thought the problem and then led you to too. Because they would not expect us to us logs or e to solve this problem because we have not gone over them as of yet. And the only form that they have had me use for all the sections about exponential functions have only used that form so I didnt think it would be different for just a single problem.
     
  19. Feb 22, 2013 #18
    Well its just that they would not expect us to use logs or e to get the answer. And the ab^ is the only for we have used so I dont think they would change all of that on one problem. or go so far ahead. Also thank you so much for helping me solve this so late at night. I think it checks out Y(0)=0.
     
  20. Feb 22, 2013 #19

    jbunniii

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    Hm, well there's definitely no way to pick values of ##A## and ##B## that will be able to represent ##Y(t)##. This is because we know ##Y(0)## has to be ##0##. So if ##Y(t)## was of the form ##Y(t) = A(B^t)##, we can plug in ##t = 0## and this would give us ##0 = A(B^0) = A##, in other words ##A = 0##. But then ##A(B^t) = 0(B^t) = 0## which would give us ##Y(t) = 0## for all ##t##. That clearly can't be right.
     
  21. Feb 22, 2013 #20
    Yeah that what I was really stuck at. Because the problem says that the formula's value of the Y-axis is the yam's temp, but that is impossible because it makes a=0, so Im not sure if it meant a formula for the difference in temp and just worded it funky or I dont know what else it could have meant.
     
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