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Math Help - 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:
    <br />
y - \sin{(x)} = 1 at P_0\left(\pi, 1)<br />


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

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

    Finally the equation of normal line is:

    <br />
y - 1 = \sqrt{2}\left(x - \pi\right)<br />

    Is this right or am i wrong? The answer back of my book is 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:
    y - \sin{(x)} = 1 at P_0\left(\pi, 1)
    The way i did was first find the gradient vector:
    \nabla f =  \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j.
    The answer back of my book is y = x - \pi + 1
    Why book's answer is different? Any ideas?
    Here is another way to look at it.
    The vector \nabla f(\pi ,1) is the direction vector of the normal line to the curve at the point (\pi,1).
    That is \nabla f(\pi ,1)=i+j and the normal is  <\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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