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Homework Help: Calculuate rate of change

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data

    The temparature T of a liquid at a time t minutes is given by the equation
    T = 30 + 20e-0.05t, for t > 0

    (1)Write down the initial tempatature o the liquid, and find the intial rate of change of temparature.

    (2)Find the time at which the temparature is 40

    2. Relevant equations

    3. The attempt at a solution

    When t=0 the initial temparture is 50. I know that to have to find dT/dt, and i and looked at the answer, which says dT/dt = -0.05 x 20e-0.05t, but i cant see how they got that. what happened to the 30? and how did they get to that point?
  2. jcsd
  3. Dec 2, 2007 #2
    They took the derivative of your T equation. Thus derivative of 30 = 0.
  4. Dec 2, 2007 #3
    ok, cheers.

    I still dont see how we get from T = 30 + 20e-0.05t to dT/dt = -0.05 x 20e-0.05t. i thought that the power had to decrease by 1 when you differenciated it and then multiply by the old power, so 20e-0.05 should become -0.05 x 20e-1.05. but thats not what the answer says, which is why i'm confused.
    Last edited: Dec 2, 2007
  5. Dec 2, 2007 #4
    If I have a function [tex] f(x) = e^u [/tex], where u can be any kind of x term, say .05x for instance, then I know that [tex] d/dx = e^u u' [/tex] where u' is the derivative of u, or in this case, the derivative of .05x, which is just .05. Does that make sense?
  6. Dec 2, 2007 #5
    Well there are two because thats how to find derivative of e.

    If you had e^2x, than the derivative would be 2e^2x. (2 comes from derivative of 2x).

    (if you have e^x, than it is 1e^x = e^x because derivative of x is 1)

    So in your problem, what is the derivative of -0.05t? It is -0.05 and thus that is why it comes in the front.

    If you are confused ill try to explain better. Let me know!

    (Sorry i dont know special writing method yet)
    Last edited: Dec 2, 2007
  7. Dec 2, 2007 #6
    You are trying to apply the power rule, but you have [tex]e^x[/tex] not [tex]x^n[/tex] the base is e and the exponent is a function,

    ...chain rule. [tex]\frac{d}{dx}[e^u]=e^u*\frac{du}{dx}[/tex] if I remember correctly.


    Edit: I see that like nine others were typing at the same time as me.when it rains it pours! Take your pick :)
  8. Dec 2, 2007 #7
    Sorry, i eddited my question after you answered it. However i think i get it now, I just forgot that e is its own deriviative.

    so when i said; "i thought that the power had to decrease by 1 when you differenciated it and then multiply by the old power, so 20e-0.05 should become -0.05 x 20e-1.05.", thats not correct because you differenciate e differently

    so it should become -0.05 x 20e-0.05t
  9. Dec 2, 2007 #8
    thanx, thats what i needed. I understand now. :smile:
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