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If Integral with Sine Limits What is Second Derivative?

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data

    If f(x) = ∫sin x0 √(1+t2)dt and g(y) = ∫3y f(x)dx, find g''(pi/6)?

    2. Relevant equations

    FTC: F(x) = ∫f(x)dx
    ab f(t)dt = F(b) - F(a)

    Chain Rule:
    f(x) = g(h(x))
    f'(x) = g'(h(x))h'(x)


    3. The attempt at a solution

    I tried u-substition setting u = tan(x) for the first dirivative with the limits of sine, and it got really weird and bad. I also tried trigonometric substition and I got a similar bad and ugly answer.

    g'(y) = f(x) = ∫0sin x √(1+t2)dt

    How do I take the second derivative of g(y)?
     
  2. jcsd
  3. Aug 27, 2014 #2

    haruspex

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    That doesn't really make any sense. In the definition of g, x is a 'dummy' variable. It doesn't exist outside of the integral. Have another go.
     
  4. Aug 27, 2014 #3

    Zondrina

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    Apply the fundamental theorem to ##g(y)##:

    $$g'(y) = \frac{d}{dy} \int_3^y f(x) dx$$
     
  5. Aug 27, 2014 #4
    If g = ∫f(x), why isn't g' = f(x)? Is the second derivative of g not equal to the first derivative of f? If it doesn't, then I guess I have no clue where to start. Do I need to integrate √(1+t^2)dt?
     
  6. Aug 27, 2014 #5
    Maybe I don't understand how to apply FTC..

    g'(y) = f(y) - f(3) ?

    If that's true then I still don't know where to go from here...
     
  7. Aug 27, 2014 #6

    Zondrina

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    You are forgetting the chain rule:

    $$g'(y) = \frac{d}{dy} \int_3^y f(x) dx = f(y) \frac{d}{dy} [y] - f(3) \frac{d}{dy} [3] = f(y)$$

    What is ##f(y)##? Take the second derivative of ##g(y)## now.
     
  8. Aug 27, 2014 #7
    Why do I need to use the chain rule? Are y and 3 functions? How can you tell when you need to use the cahin rule?

    Besides, if y is the variable, then isn't d/dy y just equal to 1 and d/dy 3 = 0? Does that make g'(y) = f(y)? Am I going in completely the wrong direction here?
     
  9. Aug 27, 2014 #8

    Zondrina

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    The fundamental theorem stated in a general form is:

    $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)$$

    Quick proof:

    $$\int_{a(x)}^{b(x)} f(t) dt = F(t) |_{a(x)}^{b(x)} = F(b(x)) - F(a(x))$$

    Where ##F(t)## is an anti derivative of ##f(t)##. Taking the derivative of the expression:

    $$\frac{d}{dx} F(b(x)) - F(a(x)) = f(b(x)) b'(x) - f(a(x)) a'(x)$$

    Where ##f(t)## is the derivative of ##F(t)## and we have applied the chain rule to the composition of the functions.

    We applied this theorem just now to take the first derivative with respect to ##y## of ##g(y)##. You find this is equal to ##f(y)##. Write out what ##f(y)## is and then take the second derivative.
     
  10. Aug 27, 2014 #9
    Okay, I think I follow most of the way now... I understand how FTC works now.

    I guess now I'm just confused about how and what to take the second derivative of:

    Do I need to write ∫sin x0 √(1+t2)dt in terms of y or is there something more I can do with g'(y) by doing the chain rule again somehow?
     
  11. Aug 27, 2014 #10

    Zondrina

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    Yes, you literally want to plug in ##y## for ##x## in ##f(x)##. This will change the limits on the integral, namely you now have a ##\sin(y)##. Then you need to take the derivative again to get ##g''(y)##.
     
  12. Aug 27, 2014 #11

    haruspex

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    Think about what differentiation means. g'(y) means you make a small change in y and observe the consequent change in g(y). Think of the integral as a sum. If you increase the upper limit slightly on ∫yf(x), making it ∫y+dyf(x), how much does the 'sum' increase by?
     
  13. Aug 27, 2014 #12
    Okay, let me try!

    g'(y) = f(y) = ∫sin(y)0 sqrt(1+t2)dt

    g''(y) = f'(y)

    Do I do this one with FTC as well or should I do a u-subs with x = tan(u)?

    If I try it with FTC...

    f'(y) = √(1+sin2y) d/dy sin(y) - √(1 + 02) d/dy (0)

    f'(y) = √(1+sin2y) cos(y) = g''(y)

    g''(pi/6) = √(5/4)((√3)/2)

    Is that right?
     
  14. Aug 27, 2014 #13

    Zondrina

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    Everything looks good, it seems you can apply the theorem now.

    You should spend some time asking yourself what the implications of this theorem are. Try to get a geometric grasp to aid in visualizing.
     
  15. Aug 27, 2014 #14
    Thank you very much for all of your help! I will see what I can do to figure out the implications and to better visualize it.
     
  16. Aug 27, 2014 #15

    Ray Vickson

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    You wrote something that makes no sense: you should have ##g'(y) = f(y)##, not ##f(x)##. You cannot have unrelated ##x## and ##y## on opposite sides of the same equation!
     
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