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Multivariable tangent planes

  1. Nov 18, 2007 #1
    Ok, I'm pretty much at whit's end trying to figure this review question out. Apparently my teacher forgot to mention that our book couldn't teach us everything we need to know for our test... Anyhow, the question is as follows, and I'm utterly at a loss as to what the answer is:

    Find the points on the surface [ 2(x^2) + (y^2) + 4(z^2) = (1) ] at which the tangent plane is parallel to the plane [ (-3x) + (y) - (2z) = (-9) ]

    I've tried all sorts of guesses at what the correct method would be, but the book I'm learning from offers no examples or other such answers for this type of problem. At this point I can't offer any answers I've arrived at other than the possibilty that a particular line on the given plane might intersect the surface at 2 points, but I have no idea how to find such a line, nor am I sure that this is even the proper approach. Can someone please help me out here?
  2. jcsd
  3. Nov 18, 2007 #2


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    Write out a "generic" (parametric) equation that describes all planes that are parallel to the given plane, z = f(x, y, known constants, unknown parameters).

    Write out the surface as z = g(x, y, known constants). Write out the formula for its tangential plane at any point (x, y, g(x, y, known constants)). Say that formula is z = h(x, y, known constants). Solve f(x, y, k.c., u.p.) = h(x, y, k.c.) for (x, y) for any value of u.p.
    Last edited: Nov 18, 2007
  4. Nov 19, 2007 #3

    Ben Niehoff

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    It sounds like he's trying to get you to apply Lagrange Multipliers. I don't know what book you're using, but this appears in Chapter 12 of Thomas & Finney (which is a common calculus book for college courses).

    First, you want a function whose level surfaces are planes parallel to the given plane. The equation of a plane is

    [tex]ax+by+cz = N[/tex]

    and for a given a, b, c, this generates a family of parallel planes. Then, define

    [tex]f(x,y,z) = ax+by+cz[/tex]

    Now, you want to find maxima and minima of f subject to the constraint

    [tex]g(x,y,z) = 2x^2+y^2+4z^2 - 1[/tex]

    That is, take your equation for the ellipsoid, and collect all the terms on one side so you can express it in the form g(x,y,z) = 0.

    Now, f is maximized on g when a level surface of f is tangent to a level surface of g. Therefore, the normal to the level surface of f will be proportional to the normal to the level surface of g. But we already know how to find the normal to the level surfaces: it's given by the gradient. Therefore,

    [tex]\nabla f = \lambda \nabla g[/tex]

    This is a vector equation, and so it actually yields three separate equations (all with the same [itex]\lambda[/itex]). Combined with the equation

    [tex]g(x,y,z) = 0,[/tex]

    these give you four equations in four unknowns:, x, y, z, and [itex]\lambda[/itex]. Now you just solve this system of equations.
  5. Nov 19, 2007 #4


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    You hardly need Lagrange multipliers! That looks like the kind of problem EVERY calculus book deals with! Two planes are parallel if and only if their normal vectors are parallel. What is a normal vector for the plane (-3x) + (y) - (2z) = (-9) ? (You can copy it right off that equation!) What is the normal vector for any tangent plane to 2(x^2) + (y^2) + 4(z^2) = (1) (hint: the gradient of the function f(x,y,z)= constant points in the direction of the normal to the surface.). It will depend on x,y,z. Choose x,y,z so that the two vectors are parallel (and so that they satisfy the equation of the surface).
  6. Nov 19, 2007 #5
    WHAT? I followed you up to setting the ellipsoid equal to 0, then I have no idea what you're talking about. We haven't done lagrange multipliers. We're about half way through Ch. 11 of James Stewart's Calculus: Concepts and Contexts 3. I don't understand how you're establishing a system of equations, nor how you're combining them with g(x,y,z)=0....
  7. Nov 19, 2007 #6
    Ok now this I can follow, but I'm pretty sure its what I've been trying. Going by your explanation I would set 4x=-3, 2y=1, and 8z=-2, then solve equation for x y and z to get the first point then multiply every value by some constant to get another point, correct? If so, then that is what I tried.
  8. Nov 19, 2007 #7


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    No, that's not what I said. Though the two vectors are parallel they are not necessarily of the same length. All you can say is 4x= -3c, 2y= c, 8z= -2c. That gives x= -3c/4, y= c/2, z= -c/4.
    NOW use the fact that [itex]2x^2+ y^2+ 4z^2= 1[/itex] to find the possible values of c.
  9. Apr 16, 2009 #8
    after we find value of C, what we do from there?
  10. Apr 17, 2009 #9
    hello, anyone there who can help me understand what to do after finding C?
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