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jennychan12

  • 2 years ago

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).

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

  2. jennychan12
    • 2 years ago
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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

  3. bahrom7893
    • 2 years ago
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    so you found where it changes direction right?

  4. bahrom7893
    • 2 years ago
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    the thing is...|dw:1357322365854:dw|

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

  6. bahrom7893
    • 2 years ago
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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

  7. bahrom7893
    • 2 years ago
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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

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

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

  10. bahrom7893
    • 2 years ago
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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

  11. bahrom7893
    • 2 years ago
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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.

  12. bahrom7893
    • 2 years ago
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    okay listen, where did the particle change direction?

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

  14. bahrom7893
    • 2 years ago
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    0?

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

  16. bahrom7893
    • 2 years ago
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    no it was just t = 2.507

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

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

  19. bahrom7893
    • 2 years ago
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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|

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

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

  22. bahrom7893
    • 2 years ago
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    isn't that the calculator part?

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

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

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

  26. bahrom7893
    • 2 years ago
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    lol yea

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

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

  29. jennychan12
    • 2 years ago
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    nooooooo...... like after you told me all that...

  30. bahrom7893
    • 2 years ago
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    k good lol

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

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