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Math Help - Equation of Line Tangent to Intersection of Surfaces

  1. #1
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    (PLEASE HELP) Equation of Line Tangent to Intersection of z = arctan(xy) and x = 2

    Can someone help me with this problem? I'm trying to find the equation of the line tangent to the intersection of the surface z = arctan(xy) with the plane x = 2 at the point (2, 1/2, pi/4).

    Since x is held constant we are finding the partial derivative wrt to y, i.e.
    d(arctan(xy))/dy = 1/(1+(xy)^2) * x = x/(1+(xy)^2) where d/dy is a partial derivative. The slope of the line is given by the value of (partials) dz/dy at (2,1/2) which is precisely 1. I know now that the parametric equations will be given by:

    x = 2
    y = 1/2+ ?
    z = pi/4 + ?

    Can somebody explain to me how to get the terms containing t in the parameterizations for y(t) and z(t)? I would really appreciate it.
    Last edited by sdh2106; October 5th 2009 at 04:48 PM. Reason: need help sooner
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by sdh2106 View Post
    Can someone help me with this problem? I'm trying to find the equation of the line tangent to the intersection of the surface z = arctan(xy) with the plane x = 2 at the point (2, 1/2, pi/4).

    Since x is held constant we are finding the partial derivative wrt to y, i.e.
    d(arctan(xy))/dy = 1/(1+(xy)^2) * x = x/(1+(xy)^2) where d/dy is a partial derivative. The slope of the line is given by the value of (partials) dz/dy at (2,1/2) which is precisely 1. I know now that the parametric equations will be given by:

    x = 2
    y = 1/2+ ?
    z = pi/4 + ?

    Can somebody explain to me how to get the terms containing t in the parameterizations for y(t) and z(t)? I would really appreciate it.
    You have this intersection's equation z = \arctan \left(2y\right)

    The general equation of the tangent line (on the yz-plane) has the form {\color{red}\boxed{\color{black}{z - z_0 = {{z'}_0}\left(y - y_0\right)}}}

    In your case y_0 = \frac{1}{2},{\text{ }}{z_0} = \frac{\pi }{4}

    Find {z'}_0

    z' = \frac{dz}{dy}\arctan \left(2y\right) = \frac{2}{1 + 4y^2} \quad \Rightarrow \quad {{z'}_0} = \frac{2}{1 + 4 \left(1/2\right)^2} = \frac{2}{2} = 1.

    Finally you have z = y + \frac{\pi }{4} - \frac{1}{2} = y + \frac{\pi -2}{4}.
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  3. #3
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    I appreciate the help, but I was trying to find out how to parameterize this information. If you see my original question, all I am trying to do is find the terms that contain t in y(t) and z(t).
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