1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Calculuate rate of change
  1. Rate of change (Replies: 3)

  2. Rate of change (Replies: 6)

  3. Rate of change (Replies: 4)

  4. Rate of Change (Replies: 2)