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Math Help - 2003 AP BC Calculus MC Questions

  1. #1
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    2003 AP BC Calculus MC Questions

    My teacher assigned us the multiple choice questions from the 2003 BC Calc Ap test and there are four questions that I don't understand how to do.

    10. What is the value of ∑ 2^(n+1)  / 3^n ?
    I simplified it to ∑ 2 (2/3)^n
    When I did the sum of the geometric series from there I got 6 but the correct answer is 4.


    22. What are all values of p for which the infinite series ∑ n/(n^p+1) converges?
    (A) p>0
    (B) p>=1
    (C) p>1
    (D) p>=2
    (E) p>2
    Correct answer is E but I'm not sure where to even start on this problem.


    84. A particle moves in the xy-plane so that its position at any time t is given by x(t) = t^2 and y(t) = sin(4t). What is the speed of the particle when t = 3?
    I figured that you needed to take the derivative of x(t) and y(t) to get dy/dx and then plug in t = 3 but apparently that isn't right. (The correct answer is 6.884)


    Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by Naples View Post
    My teacher assigned us the multiple choice questions from the 2003 BC Calc Ap test and there are four questions that I don't understand how to do.

    10. What is the value of ∑ 2^(n+1)  / 3^n ?
    I simplified it to ∑ 2 (2/3)^n
    When I did the sum of the geometric series from there I got 6 but the correct answer is 4.


    22. What are all values of p for which the infinite series ∑ n/(n^p+1) converges?
    (A) p>0
    (B) p>=1
    (C) p>1
    (D) p>=2
    (E) p>2
    Correct answer is E but I'm not sure where to even start on this problem.


    84. A particle moves in the xy-plane so that its position at any time t is given by x(t) = t^2 and y(t) = sin(4t). What is the speed of the particle when t = 3?
    I figured that you needed to take the derivative of x(t) and y(t) to get dy/dx and then plug in t = 3 but apparently that isn't right. (The correct answer is 6.884)


    Any help is greatly appreciated.
    10) Where does n start n = 0 or n = 1 (it makes a difference)

    22) Compare with \sum \frac{n}{n^p} or \sum \frac{1}{n^{p-1}}

    84) You want to use  \sqrt{x'^2 + y'^2}
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    What is the indexing starting at?
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  4. #4
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    Your problem with number 10 is that the indexing started at 1 not 0 so when you change your indexing you will obtain 4 as the answer.
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    On number 10 is starts at n=1, but I see what you mean, so thank you.
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    Quote Originally Posted by Naples View Post
    On number 10 is starts at n=1, but I see what you mean, so thank you.
    \sum\limits_{n = 1}^\infty  {\frac{{2^{n + 1} }}<br />
{{3^n }}}  = \frac{{\frac{4}<br />
{3}}}<br />
{{1 - \frac{2}<br />
{3}}}<br />
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    How does a = 4/3? This is what I'm getting:

     \sum\limits_{n = 1}^\infty { \frac{2^{n+1}}{3^n} }

     = 2 \frac{2^n}{3^n} = 2 (\frac{2}{3})^n

    a = 2 right?
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    Quote Originally Posted by lilaziz1 View Post
    How does a = 4/3? This is what I'm getting:

     \sum\limits_{n = 1}^\infty { \frac{2^{n+1}}{3^n} }

     = 2 \frac{2^n}{3^n} = 2 (\frac{2}{3})^n

    a = 2 right?
    Write out the first few terms

     <br />
\frac{2^2}{3} + \frac{2^3}{3^2} + \frac{2^4}{3^3} + \cdots,<br />
a is the first term.
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  9. #9
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    Please allow me to be brutally honest.
    In any geometric series this is the fact: \sum\limits_{k = j}^\infty  {ax^k }  = \frac{{ax^j }}{{1 - x}}.
    That is in words: the sum equals the first term divided by one minus the common ratio.

    Can you apply that very simple rule?
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  10. #10
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    ah alright. gotcha danny. thx
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