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Math Help - equation of the normal curve

  1. #1
    rcs
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    equation of the normal curve

    What is the equation of the normal line to the curve y = x^2 + 2x -1 at point (1,2)

    attempt: dy/dx = 2x + 0
    dy/dx = 2x,

    im stuck here
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  2. #2
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    Re: equation of the normal curve

    Quote Originally Posted by rcs View Post
    What is the equation of the normal line to the curve y = x^2 + 2x -1 at point (1,2)

    attempt: dy/dx = 2x + 0
    dy/dx = 2x,

    im stuck here
    Hint: The normal line is perpendicular to the tangent line. The gradients of perpendicular lines multiply to give -1.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: equation of the normal curve

    Your derivative is not quite right:

    y=x^2+2x-1

    \frac{dy}{dx}=2x+2

    The slope m of the normal line, however, is:

    m=-\frac{dx}{dy}=-\frac{1}{\frac{dy}{dx}}

    So, for x=1, what is the slope of the normal line?
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  4. #4
    rcs
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    Re: equation of the normal curve

    Quote Originally Posted by MarkFL2 View Post
    Your derivative is not quite right:

    y=x^2+2x-1

    \frac{dy}{dx}=2x+2

    The slope m of the normal line, however, is:

    m=-\frac{dx}{dy}=-\frac{1}{\frac{dy}{dx}}

    So, for x=1, what is the slope of the normal line?
    m = 2(1) + 2 = 4 is the slope.

    how is equation of the normal line ? how is that done?

    thanks
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: equation of the normal curve

    You have found the slope of the tangent line. The normal line is perpendicular to the tangent line, so what must the slope of the normal line be?
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  6. #6
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    Re: equation of the normal curve

    hmmmm cant get it
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  7. #7
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    Re: equation of the normal curve

    Quote Originally Posted by rcs View Post
    hmmmm cant get it
    You can't get it or you haven't tried? I suggest you read my previous post.
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  8. #8
    rcs
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    Re: equation of the normal curve

    really cant get it ... i tried but i really cant./. sorry
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: equation of the normal curve

    You found the slope of the tangent line m_{\text{T}}, but you need the slope of the normal line m_{\text{N}}. As Prove It pointed out, you require:

    m_{\text{N}}m_{\text{T}}=-1
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  10. #10
    rcs
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    Re: equation of the normal curve

    i dont have idea how it looks like... i cant visualize really
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  11. #11
    MHF Contributor MarkFL's Avatar
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    Re: equation of the normal curve

    A normal line is perpendicular to the curve at the point of intersection, thus it is perpendicular to the tangent line at that point.

    As far as the product of perpendicular slopes being -1, consider that the difference in their angles of inclination must be \frac{\pi}{2}:

    \tan^{-1}(m_1)-\tan^{-1}(m_2)=\frac{\pi}{2}

    Using the identity \tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2} we may write:

    \tan^{-1}(m_1)-\tan^{-1}(m_2)=\tan^{-1}(m_1)+\cot^{-1}(m_1)

    \tan^{-1}(-m_2)=\cot^{-1}(m_1)

    \tan^{-1}(-m_2)=\tan^{-1}\left(\frac{1}{m_1} \right)

    m_1m_2=-1
    Last edited by MarkFL; September 16th 2012 at 06:18 AM.
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