Results 1 to 3 of 3

Math Help - DQ and IRC

  1. #1
    Newbie crazydaizy78's Avatar
    Joined
    Nov 2007
    Posts
    7
    Awards
    1

    DQ and IRC

    use the difference quotient, w/h=0.01 and estimate the instantaneous rate of change.

    #10. P(r)=100(1+r)^2 where r=0.12

    Can you show me how to set up to solve? I do not seem to grasp the idea of the letters and the corresponding #'s and where to place.

    I don't have the actual answer for this problem, book only gives odds.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by crazydaizy78 View Post
    use the difference quotient, w/h=0.01 and estimate the instantaneous rate of change.

    #10. P(r)=100(1+r)^2 where r=0.12

    Can you show me how to set up to solve? I do not seem to grasp the idea of the letters and the corresponding #'s and where to place.

    I don't have the actual answer for this problem, book only gives odds.
    you want \frac {P(r + h) - P(r)}h (usually for instantaneous rate of change, limits are involved, however, since they gave us constant values for r and h, i assume they don't want us to use that)

    so you want: \frac {P(0.12 + 0.01) - P(0.12)}{0.01}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by crazydaizy78 View Post
    use the difference quotient, w/h=0.01 and estimate the instantaneous rate of change.

    #10. P(r)=100(1+r)^2 where r=0.12

    Can you show me how to set up to solve? I do not seem to grasp the idea of the letters and the corresponding #'s and where to place.

    I don't have the actual answer for this problem, book only gives odds.
    Since you figured out this one (well done ), I'm betting that by now you've figured out the answer to this latest one too.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum