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Math Help - Secant Line Slope and Tangent Line Slope

  1. #1
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    Secant Line Slope and Tangent Line Slope

    I have answered these problems already but I am not sure what the correct answers were.

    Point P=(3,1/9) lies on 1/x^2

    Find slope of secant line PQ for certain values of X:

    X=2.5, mSecant=-.097778 <You may want to check the mSecant Val.
    X=2.9, mSecant=-.077950 for the values.
    X=2.99, mSecant=-.074446
    X=2.999, mSecant=-.074111
    X=3.5, mSecant=-.058957
    X=3.1, mSecant=-.070528
    X=3.01, mSecant=-.073705
    X=3.001,mSecant=-.074037

    b)Estimate the value of the slope of the tagent line at curve P. (use 2.999 and 3.001)

    and

    Using the slope from B find equation for tagent line to the curve at (3,1/9)
    Last edited by mr fantastic; September 7th 2009 at 01:01 PM. Reason: Restored deleted question.
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    Quote Originally Posted by spursstar13 View Post

    b)Estimate the value of the slope of the tagent line at curve P. (use 2.999 and 3.001)
    f(x) = \frac{1}{x^2} \Rightarrow f'(x) = \frac{-2}{x^3}

    Find f'(2.999) and f'(3.001) and take an average?


    Quote Originally Posted by spursstar13 View Post



    Using the slope from B find equation for tagent line to the curve at (3,1/9)
    Tangent line is y-y_1 = m_T(x-x_1)

    In your case y-\frac{1}{9} = m_T(x-3)

    m_T will be the answer to the previous question
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  3. #3
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    Quote Originally Posted by spursstar13 View Post
    I have answered these problems already but I am not sure what the correct answers were.

    Point P=(3,1/9) lies on 1/x^2

    Find slope of secant line PQ for certain values of X:

    X=2.5, mSecant=-.097778 <You may want to check the mSecant Val.
    X=2.9, mSecant=-.077950 for the values.
    X=2.99, mSecant=-.074446
    X=2.999, mSecant=-.074111
    X=3.5, mSecant=-.058957
    X=3.1, mSecant=-.070528
    X=3.01, mSecant=-.073705
    X=3.001,mSecant=-.074037
    Yes, these all look good.

    b)Estimate the value of the slope of the tagent line at curve P. (use 2.999 and 3.001)
    You have -.074111 and -.074037 for those two values. The best way to use those two numbers is to average them: [tex]\frac{-.074111-.074037}{2}= \frac{.148148}{2}= 0.074074.

    and

    Using the slope from B find equation for tagent line to the curve at (3,1/9)
    so y= 0.074074(x- 3)+ 1/9.
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    you guys both are truely awesome. but a special thanks to the second for giving a direct answer. I appreciate it and if you need anything P.M. ME!
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