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jennychan12

AP Calculus AB 2003 #2 http://www.collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_03.pdf I don't understand how to do part c).

  • one year ago
  • one year ago

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  1. bahrom7893
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    total distance is integral of velocity but u gotta take into consideration which way it's going

    • one year ago
  2. jennychan12
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    I looked at the sample respoinses and they did like absolute value of v(t). But I don't understand why. http://apcentral.collegeboard.com/apc/public/repository/ap03_calculusab_q2_27980.pdf

    • one year ago
  3. bahrom7893
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    so you found where it changes direction right?

    • one year ago
  4. bahrom7893
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    the thing is...|dw:1357322365854:dw|

    • one year ago
  5. bahrom7893
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    if u take just the integral, the area on the bottom gets subtracted

    • one year ago
  6. bahrom7893
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    but that's for displacement. So if u want to get the distance you need to add that area instead of subtracting it.. So you have to integrate from start till when it changed direction, then from when it changed direction till when it changed direction again, etc.. and finally take the absolute value of all integrals and add them up

    • one year ago
  7. bahrom7893
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    |dw:1357322510296:dw| see if u start at pt A, displacement, which is the regular integral, is 0, but distance traveled, is 5

    • one year ago
  8. jennychan12
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    i thought to find the distance u have to take integral of the velocity.

    • one year ago
  9. bahrom7893
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    yes you do.. integral of velocity is indeed the area under the velocity curve, which is the displacement

    • one year ago
  10. bahrom7893
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    so in order to get the DISTANCE.. you want to add all the areas that end up below y- axis, instead of subtracting them

    • one year ago
  11. bahrom7893
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    if you were looking for DISPLACEMENT, you would just take the integral of that function from 0 to 3. But, since you're looking for DISTANCE, take the integral of that function from 0 to a, where a is the time at which particle changed direction.

    • one year ago
  12. bahrom7893
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    okay listen, where did the particle change direction?

    • one year ago
  13. jennychan12
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    rad(2pi) and 0

    • one year ago
  14. bahrom7893
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    0?

    • one year ago
  15. jennychan12
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    yeah those were the solutions but only rad(2pi) works

    • one year ago
  16. bahrom7893
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    no it was just t = 2.507

    • one year ago
  17. jennychan12
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    yeah that's rad(2pi)

    • one year ago
  18. bahrom7893
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    so integrate that function from 0 to 2.507, then from 2.507 to 3

    • one year ago
  19. bahrom7893
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    once you get the results, say Int from 0 to 2.507 = A, and Int from 2.507 to 3 = B; just add up the absolute values: |A| + |B|

    • one year ago
  20. bahrom7893
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    that's all he's doing by integrating the absolute value of V over 0 < t < 3

    • one year ago
  21. jennychan12
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    how'd u integrate the function tho? like algebraically?

    • one year ago
  22. bahrom7893
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    isn't that the calculator part?

    • one year ago
  23. bahrom7893
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    i dont think u can integrate exponents inside trig functions yet.. lol

    • one year ago
  24. jennychan12
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    or do u just plug it into the graphing calculator? ohhh -_- ok.

    • one year ago
  25. bahrom7893
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    i mean i dont think there is a way that anyone has found how to do that

    • one year ago
  26. bahrom7893
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    lol yea

    • one year ago
  27. jennychan12
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    ohhh yeah i got the answer when i plugged it into the calculator

    • one year ago
  28. bahrom7893
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    You mean to tell me.... I WASTED MY TIME???!!!!!!

    • one year ago
  29. jennychan12
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    nooooooo...... like after you told me all that...

    • one year ago
  30. bahrom7893
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    k good lol

    • one year ago
  31. jennychan12
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    sorry..if anyone's still viewing this for part d) why did they add 1 to rad(2pi) ?

    • one year ago
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