Thursday, November 5, 2015

Week 9 Discussion

Turn in #6 is due Monday, Nov. 9, 2015!

This is for discussing assignments from WEEK 9, including homework, turn-in #5, in-class work or lessons, or anything else related to the class from these weeks.  Please be sure to include your name at the end of your post for credit.  When answering a question, DO NOT GIVE SOLUTIONS! Provide hints or explain a method that you used, but do not give the final result.  As always, RESPECT IS A MUST! Anyone abusing this forum will be banned from future use (meaning, no extra credit!!!).
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18 comments:

  1. Ari KNovember 7, 2015 at 1:19 PM

    For number 4 on the Turn-In is x(t) the position function?

    ReplyDelete
    Replies
    1. UnknownNovember 7, 2015 at 8:33 PM

      Yes

      Delete
    2. Coach WilsonNovember 7, 2015 at 9:30 PM

      Note that x(t) is NOT the displacement, while the integral of v(t) IS the displacement. The position is the initial value plus the displacement (net change) via FTC part 2.

      Delete
  2. UnknownNovember 7, 2015 at 8:32 PM

    For 5b, what would we use for dt since t doesn't increase at a constant rate?

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    Replies
    1. Coach WilsonNovember 7, 2015 at 9:28 PM

      Your rectangles will not be equal in width. Use the subeintervals given in the table.

      Delete
  3. Rachel F.November 7, 2015 at 10:02 PM

    For problem 2 can I leave my answer with the x values plugged in but not simplified?

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    Replies
    1. AnonymousNovember 8, 2015 at 5:12 PM

      I believe we are allowed to as that is permitted on the AP in May but, it's fairly simple to simplify the values. -Claire Westerlund

      Delete
    2. Coach WilsonNovember 8, 2015 at 5:55 PM

      Indeed.

      Delete
  4. Rachel F.November 7, 2015 at 10:03 PM

    On problem 4, if we wanted to include the right endpoint would we make it a closed bracket?

    ReplyDelete
  5. AnonymousNovember 8, 2015 at 8:35 PM

    Do we have any more turn ins this trimester? -Claire Westerlund

    ReplyDelete
    Replies
    1. AnonymousNovember 8, 2015 at 8:41 PM

      The calendar has one more turn-in (#7) scheduled.
      -Sarah Mostofizadeh

      Delete
    2. Coach WilsonNovember 8, 2015 at 9:43 PM

      Sarah is correct

      Delete
  6. FaythNovember 8, 2015 at 8:57 PM

    I don't totally understand the second half of the average value of a function equation that deals with area over (b-a), can anyone explain to me how this works? Because I keep getting the wrong answer when I use it.

    ReplyDelete
    Replies
    1. Coach WilsonNovember 8, 2015 at 9:43 PM

      The average value of a function over a closed interval [a,b] is the definite integral from X=a to X=b divided by the length of the interval, b-a.

      Delete
  7. AnonymousNovember 8, 2015 at 9:38 PM

    Should we know how to do Simpson's Rule without a calculator?
    -Sarah Mostofizadeh

    ReplyDelete
  8. AnonymousNovember 12, 2015 at 12:43 PM

    What is the need for a dummy variable?
    -Colin Pocock

    ReplyDelete

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