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!

Fundamental THM of Calc?

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Heres the question:

    f(t) = Integral from c to t of a(s)ds, and h(t) = ef(t)

    Compute h' (t) using:

    2 fundamental theorems of Calculus

    and

    The Chain Rule


    any help please ?



    2. Relevant equations

    Fundamental Theorems of calc

    g(x) = Integral from a to x of f(t) dt

    and

    Integral from a to b of f(x) dx = F(b) - F(a)

    and chain rule



    3. The attempt at a solution

    using the chain rule i know the answer is f'(t)*ef(t)

    but how do you do it using the 2 fundamental thrms??

    Help Please
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 8, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Nooo. The derivative of h(t) is f'(t)*e^(f(t)) isn't it? Now start working on the derivative of f(x).
     
  4. Sep 8, 2009 #3
    so basically f(t) = Integral from c to t of a(s)

    f(t) = a(t) - a(c)

    then f'(t) = ??
     
  5. Sep 8, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Wrong again. f(t)=A(t)-A(c) where A(t) is an antiderivative of a(t), i.e. A'(t)=a(t). Tell me again what f'(t) is. I'll warn you in advance, I think you'll get it wrong. Prove I'm wrong about that and think about it.
     
  6. Sep 8, 2009 #5
    ya I understand that, part 2 of the Fund thm of calc. but can u please explain your whole process on what your doing here cause your just pulling out these rules which still don't help me understand the process of setting this up.
     
  7. Sep 8, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm not 'pulling out any rules'. I'm just using the chain rule which says (h(f(x)))'=f'(x)*h'(f(x)) with h(x)=e^x (so h'(x)=e^x) and the fundamental theorem of calculus which says the integral from c to x of f(x) is F(x)-F(c) where F is an antiderivative of f. These are the rules you are told to use. Just apply those to your problem.
     
  8. Sep 8, 2009 #7
    alright, this isn't really helping [maybe I'm just not getting it] but thanks for trying, appreciate it.
     
  9. Sep 8, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I appreciate the thanks, but I don't think you are trying very hard. If f(t)=A(t)-A(c) and A'(t)=a(t), what's the derivative of f(t) (remembering that the derivative of a constant is 0)? I really am trying to help.
     
  10. Sep 8, 2009 #9
    well since A'(t) = a(t) then wouldn't the derivative of f(t) just be A'(t) and A'(c)?

    so:
    f(t)=A(t)-A(c)
    f'(t) = a(t) - a(c) ?

    **The main problem is Im in Calc C and all this stuff was stuff I did in Calc B, which was all the way back in Feb.**
     
  11. Sep 8, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Excuses understood. But A(c) is a CONSTANT. I don't think it's derivative will be nonzero. Hint, hint.
     
  12. Sep 8, 2009 #11

    ok well since its a constant then like you said the derivative is zero. [ i wasn't grasping the whole a(c) is a constant part] which leaves the derivative of f(t) = a(t) only?


    and i appreciate you being patient with this, i probably would of stopped helping by now :yuck:
     
  13. Sep 8, 2009 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, absolutely.
     
  14. Sep 8, 2009 #13
    so now this means [ going back to the original problem] that since:
    h(t) = ef(t)
    and
    h'(t) = f'(t)*ef(t)

    h'(t) = a(t)*eA(t) -A(c)
    ?????
     
  15. Sep 8, 2009 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Right. Or h'(t)=a(t)*e^(f(t)). Yes?
     
  16. Sep 8, 2009 #15
    yess, thanks a lot once again appreciate your patience, and 2 thumbs up from me lol
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook