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Math Help - normal line to the curve

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    normal line to the curve

    A curve y=f(x) with the gradient function \frac{dy}{dx}=\frac{x^2}{3x-1}.

    The straight line ky=2x-7 is normal to the curve y=f(x) at (2,m).Find the value of k and m.


    any help will appreciate,.
    Last edited by mastermin346; May 25th 2010 at 08:21 PM.
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    Quote Originally Posted by mastermin346 View Post
    The straight line ky=2x-7 is normal to the curve y=f(x) at (2,m).Find the value of k and m.


    any help will appreciate,.
    There is not enough information given. Go back and check the question.
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    Quote Originally Posted by mr fantastic View Post
    There is not enough information given. Go back and check the question.

    hi sir..

    sorry,my mistake

    the question is. A curve y=f(x) with gradient function \frac{dy}{dx}=\frac{x^2}{3x-1}.
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    Well start by substituting x=2 into \frac{dy}{dx} what you you get? What does that mean?
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    Quote Originally Posted by pickslides View Post
    Well start by substituting x=2 into \frac{dy}{dx} what you you get? What does that mean?
    hi,i get \frac{dy}{dx}=\frac{4}{5} then?it mean the gradient right?
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    Quote Originally Posted by mastermin346 View Post
    hi,i get \frac{dy}{dx}=\frac{4}{5} then?it mean the gradient right?
    Correct, it is the gradient of f(x) at x=2. Now what is the relationship between the gradient and the normal?

    Hint: m_N\times m_T = -1

    How can we use this value?
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    Quote Originally Posted by pickslides View Post
    Correct, it is the gradient of f(x) at x=2. Now what is the relationship between the gradient and the normal?

    Hint: m_N\times m_T = -1

    How can we use this value?
    i get the gradient of normal is -\frac{5}{4}.
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    Quote Originally Posted by mastermin346 View Post

    The straight line ky=2x-7 is normal to the curve y=f(x) at (2,m).
    so maybe its' time to find k

    Making the form y=mx+c

    ky=2x-7\implies y=\frac{2}{k}x-\frac{7}{k}

    \frac{-5}{4}= \frac{2}{k}
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