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Homework Help: Partial Derivatives - Finding tangent in a volume?

  1. May 28, 2010 #1
    Not sure I understand exactly what this question is asking. This is obviously a volume in R3 and so how do you get a tangent inside a volume? Or is it just along the plane y = 2 intersecting the volume? Also, what is a parametric equation...? Thanks for the help:

    The ellipsoid 4x^2 + 2y^2 + z^2 = 16 intersects the plane y = 2 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point (1,2,2).

    I substituted in y = 2 and got the equation 4x^2 + z^2 = 8. Do I just find what the partial derivatives are at the point (1,2)? And would the tangents be combinations of these two derivatives...?
  2. jcsd
  3. May 28, 2010 #2
    You want to find the tangent line to an ellipse. The ellipse lies in the plane y=2, so you don't have to worry about tangent lines in a volume. Solve for the ellipse in the XZ-plane at y=2, and find its tangent at (x,z)=(1,2). You should be able to convince yourself that the tangent to the ellipse lies in the y=2 plane. And (1,2) better be a point on this ellipse (it is) or there is something wrong with the problem statement.
  4. May 28, 2010 #3
    Thanks but I'm not so good with ellipse equations and I dont really understand, if y=2 then the ellipse would be 4x^2 + z^2 = 8 right... but this isnt an equation of two variables is it? Because its an implicit function of z (is this correct?). So solving for z would give a one variable function but then it is not partial derivatives because its only a single variable function right?
  5. May 28, 2010 #4
    I'm not a homework helper. I'm just filling in for a short moment, but, yes, 4x^2 + z^2 = 8 is an equation of two variables.

    For the elliptical curve, all the partial derivatives with respect to y disappear. The curve is stuck in the y=2 plane. So obviously it doesn't change with y.
  6. May 28, 2010 #5


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    Yes, the ellipse, in the plane y= 2 is [itex]4x^2+ z^2= 8[/itex]. To find the equation of the tangent line at (1, 2, 2), you can either think of z as a function of x or x as a function of z.

    Since "z as a function of x" seems more natural, by "implicit differentiation", [itex]4x + 2z z'= 0[/itex] gives, at (1, 2, 2), [itex]4+ 4z'= 0[/itex] so that z'= -1.

    Find z= ax+ b for that slope and point and include y= 2. If you want parametric equations, let x= t, z= at+ b, y= 2.
  7. May 28, 2010 #6


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    Moderator's note: thread moved from "Differential Equations".

    Homework assignments or any textbook style exercises are to be posted in the appropriate forum in our Homework & Coursework Questions forums. This should be done whether the problem is part of one's assigned coursework or just independent study.

    Thank you.
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