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Indefinite Integrals & The Net Change Theorem

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data





    2. The attempt at a solution

    I integrated the velocity equation and got this: s(t)= sin(t) + (1/2)x

    1. [sin(pi)+(1/2)(pi)] - [sin(0)+(1/2)(0)]
    = pi/2

    2. [sin(pi)+(1/2)(pi)] + [sin(0)+(1/2)(0)]
    = pi/2

    The first answer is right but the second one isn't
     
    Last edited: Apr 24, 2010
  2. jcsd
  3. Apr 24, 2010 #2

    Dick

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    Number one is right. The problem with two is that at e.g. v(pi)=(-1/2). So the particle is travelling backwards part of the time.
     
  4. Apr 24, 2010 #3
    But even if it's traveling backwards, it does not effect the total distance right? Because that will effect the displacement.
     
  5. Apr 24, 2010 #4

    Dick

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    You've got that backwards. Displacement is the total distance between the starting point and the ending point. It doesn't matter how you got from the start to the end. How you did does affect the total distance travelled.
     
  6. Apr 24, 2010 #5
    ok then, but I still can't think of a way to solve it.
     
  7. Apr 24, 2010 #6

    Dick

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    Find out at what time the particle switches from going forward to backwards by solving v(t)=0. Then add the distance it travels forward to the distance it travels backward.
     
  8. Apr 24, 2010 #7
    v(t)=cos(t)+(1/2)=0
    cos (t)= -1/2
    t=2pi/3

    [sin(0)+(1/2)(0)] + [sin(2pi/3)+(1/2)(2pi/3)]
    0 + squareroot(3)/2 + pi/6.......
     
  9. Apr 24, 2010 #8

    Dick

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    (1/2)(2pi/3)=pi/3. So if you mean sqrt(3)/2+pi/3, then, yes, that's the distance it travels forward. Now how far does it travel backwards?
     
  10. Apr 24, 2010 #9
    that was only the forward one; i thought I was solving for both. ok let's see now:
    [sin(2pi/3)+(1/2)(2pi/3)] + [sin(pi)+(1/2)(pi)]
    sqrt(3)/2+pi/3 + 0
    =sqrt(3)/2+pi/3

    forward + backward=
    (sqrt(3)/2+pi/3) + (sqrt(3)/2+pi/3)
    2(sqrt(3)/2+pi/3)
     
  11. Apr 24, 2010 #10

    Dick

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    To get the backwards distance you are integrating v(t) from 2pi/3 to pi and then reversing the sign since you know the result will be negative, and you want the positive distance, right? So shouldn't [sin(2pi/3)+(1/2)(2pi/3)] + [sin(pi)+(1/2)(pi)] be [sin(2pi/3)+(1/2)(2pi/3)] - [sin(pi)+(1/2)(pi)]??
     
  12. Apr 24, 2010 #11
    oh yeah that's right, but won't you still end up with the same answer?
    sqrt(3)/2+pi/3

    then add the forward distance with the backward distance? But then the answer turns out wierd...
     
  13. Apr 24, 2010 #12

    Dick

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    No, I don't get sqrt(3)/2+pi/3 for the backwards distance. The difference between the forward distance and the backwards distance should be the displacement, right? It's not zero. Try the backwards distance again.
     
  14. Apr 24, 2010 #13
    [sin(2pi/3)+(1/2)(2pi/3)] - [sin(pi)+(1/2)(pi)]
    [squareroot(3)/2 + pi/3] - [0+ pi/2] ---sorry about that I assumed it was zero because of the forward direction one, that was a stupid mistake

    squareroot(3)/2 + pi/3 - pi/2
    squareroot(3)/2 -pi/6

    forward + backwards=
    sqrt(3)/2+pi/3 + squareroot(3)/2 -pi/6
    2(sqrt(3)/2)+ 3pi/18
    = sqrt(3) + 3pi/18
     
  15. Apr 24, 2010 #14

    Dick

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    That's better. You can simplify 3pi/18 to pi/6. But now I think it's right.
     
  16. Apr 24, 2010 #15
    Thank you so much; it is right :)
     
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