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Math Help - Equation of tangent line to curve

  1. #1
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    Equation of tangent line to curve

    The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

    x^2y + y^2x + xyz + z^2 = 4
    x + y + z = 3

    Find the equation of the tangent line to the curve at the given point.


    Attempt at a solution

    GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

    At (1,1,1):

    GRAD f(1,1,1) = 3i + 3j + 3k

    Tangent line has normal of (3,3,3).

    Therefore:

    3(x - 1) + 3(y - 1) + 3(z - 1) = 0
    3x + 3y + 3z = 9
    x + y + z = 3

    I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?
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  2. #2
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    Accidental double post.
    Last edited by HallsofIvy; October 26th 2009 at 01:20 PM.
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  3. #3
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    Quote Originally Posted by My Little Pony View Post
    The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

    x^2y + y^2x + xyz + z^2 = 4
    x + y + z = 3

    Find the equation of the tangent line to the curve at the given point.


    Attempt at a solution

    GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

    At (1,1,1):

    GRAD f(1,1,1) = 3i + 3j + 3k

    Tangent line has normal of (3,3,3).

    Therefore:

    3(x - 1) + 3(y - 1) + 3(z - 1) = 0
    3x + 3y + 3z = 9
    x + y + z = 3

    I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?
    Since the curve is formed by both surfaces, don't you think that would be a good idea? A single curve does not define a surface, much less its tangent line.

    But you started correctly: (3,3,3) is normal to the first surface and so is normal to any line in the surface. But your answer x+ y+ z= 3 is the equation of the plane tangent to the surface there.

    The difficulty with using just one surface is that there exist an infinite number of vectors perpendicular to a give vector. There exist and infinite number of tangent lines to that surface.

    Instead, find the normal to the second surface (which is actually a plane) at (1, 1, 1). The vector you want must be perpendicular to both of those- in other words it is their cross product.

    Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.
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  4. #4
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    Quote Originally Posted by My Little Pony View Post
    The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

    x^2y + y^2x + xyz + z^2 = 4
    x + y + z = 3

    Find the equation of the tangent line to the curve at the given point.


    Attempt at a solution

    GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

    At (1,1,1):

    GRAD f(1,1,1) = 3i + 3j + 3k

    Tangent line has normal of (3,3,3).

    Therefore:

    3(x - 1) + 3(y - 1) + 3(z - 1) = 0
    3x + 3y + 3z = 9
    x + y + z = 3

    I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?
    Quote Originally Posted by HallsofIvy View Post
    . . .
    Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.
    I think you do. If you eliminate z using the plane, then you have

     <br />
x^2y + y^2x + xy(3-x-y) + (3-x-y)^2 = 4<br />

    or

     <br />
3xy + (3-x-y)^2 = 4\;\;\;(1)<br />

    which is the curve projected into the xy plane.

    @ little pony. As for the tangent line. It you let x = t, (so \frac{dx}{dt} = 1) then from (1) you can find \frac{dy}{dt}. For the last \frac{dz}{dt} = - \frac{dx}{dt} - \frac{dy}{dt} so now you have the three conponents of the tangent vector. You use the point (1,1,1) to get the tangent line.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.
    Ah sorry, I believe I computed the gradient incorrectly.

    The normal of the first equation should actually be (4,4,3).

    Thanks a bunch for your help!
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