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Thread: Problem finding normal line equation using partial derivative and gradient vector

  1. #1
    Senior Member x3bnm's Avatar
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    Problem finding normal line equation using partial derivative and gradient vector

    I'm trying to solve a problem but the answer is different from the book.
    So wondering what am i doing wrong. The problem is like this:

    Find equations for the lines that are normal to function:
    $\displaystyle
    y - \sin{(x)} = 1$ at $\displaystyle P_0\left(\pi, 1)
    $


    The way i did was first find the gradient vector:
    $\displaystyle
    \nabla f = \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j
    $

    Then get the derivative $\displaystyle D_uf\left|_{(\pi,1)}$ which is(because we know that slope
    of a function is it's derivative at that point):
    $\displaystyle
    D_uf\left|_{(\pi,1)} = \sqrt{\left(-\cos{(x)}\right)^2 + (1^2)} = \sqrt{1^2 + 1^2} = \sqrt{2}
    $

    Finally the equation of normal line is:

    $\displaystyle
    y - 1 = \sqrt{2}\left(x - \pi\right)
    $

    Is this right or am i wrong? The answer back of my book is $\displaystyle y = x - \pi + 1 $
    Why book's answer is different? Any ideas?
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  2. #2
    A Plied Mathematician
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    Well, one problem is that you can't use the slope of the tangent line as the slope of the normal line! They are related: once you find the slope of the tangent line, you can use a simple formula to find the slope of the normal line. Second, I'm not sure I agree with your method of computing the slope of the tangent line. Can't you just use the normal Calc I method? Your method got the wrong answer. (I agree with the book's answer completely, by the way).
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  3. #3
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    Quote Originally Posted by x3bnm View Post
    I'm trying to solve a problem but the answer is different from the book. Find equations for the lines that are normal to function:
    $\displaystyle y - \sin{(x)} = 1$ at $\displaystyle P_0\left(\pi, 1)$
    The way i did was first find the gradient vector:
    $\displaystyle \nabla f = \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j$.
    The answer back of my book is $\displaystyle y = x - \pi + 1 $
    Why book's answer is different? Any ideas?
    Here is another way to look at it.
    The vector $\displaystyle \nabla f(\pi ,1)$ is the direction vector of the normal line to the curve at the point $\displaystyle (\pi,1)$.
    That is $\displaystyle \nabla f(\pi ,1)=i+j$ and the normal is $\displaystyle <\pi,1> +t<1,1>.$
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  4. #4
    Senior Member x3bnm's Avatar
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    >Well, one problem is that you can't use the slope of the tangent line as the slope of the normal line!

    Okay.

    >Can't you just use the normal Calc I method?

    Can you kindly explain this method?
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  5. #5
    A Plied Mathematician
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    Just solve your original equation for y = sin(x) + 1. Take its derivative: y' = cos(x). Plug in the value of x in question to get y'(pi) = -1. That's your slope of the tangent line. Then go from there.
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  6. #6
    Senior Member x3bnm's Avatar
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    Oops! looks like we posted at the same time. Sorry.
    I thank you all for your help. Again those who are interested:

    The slope of the normal line is the negative reciprocal of slope of the tangent line.
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  7. #7
    A Plied Mathematician
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    No, we didn't post at the same time. I'm just fast that way, sometimes.
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