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Tangent planes to surfaces

  1. Oct 5, 2008 #1
    Suppose that F(x,y) = x[tex]^{2}[/tex]+y[tex]^{2}[/tex]. By using vector geometry, find the Cartesian equation of the tangent plan to the surface z = F(x,y) at the point where (x,y,z) = (1,2,5). Find also a vector n that is normal to the surface at this point.

    Now the step by step working is given for this question, I however get confused at one part.
    First they intersect the surface z = x[tex]^{2}[/tex]+y[tex]^{2}[/tex] with the plane x = 1

    Therefore z = 1 + y[tex]^{2}[/tex]

    The gradient F[tex]_{y}[/tex](1,2) = 4

    By using the point gradient formula for a straight line, the Cartesian eq for the tangent is:

    z -5 = 4(y - 2), x =1

    If [tex]\lambda[/tex] = y - 2, then the equation of the tangent line in parametric vector form is

    (x y z) = (1 2 5) + [tex]\lambda[/tex](0 1 4)

    Now my question is, where did the (0 1 4) come from?
  2. jcsd
  3. Oct 6, 2008 #2


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    Hi JFonseka! :smile:

    This is the tangent line in the x = 1 plane (hmm … why didn't they choose the easier z = 5 plane? :confused:), so the x term has to be 0.

    And from z -5 = 4(y - 2), the z must increase 4 times as fast as the y, so altogether it's a multiple of (0 1 4). :smile:
  4. Oct 6, 2008 #3


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    I'm not sure I can answer you question because that is not quite the way I would do the problem! In fact, I am pretty sure (x,y,z)= (1, 2, 5)+ [itex]\lambda[/itex](0, 1, 4) is wrong. z= x2+ y2 has zx= 2x= 2 at (1, 2, 5) so that the line z= 5+ 2(x-1) Taking x= 2 in the first equation, z= 5+ 2= 7. But the equation of the plane you give has x= 1 for all y and z.

    Given z= x2+ y2, think of this as a "level surface" of the function F(x,y,z)= x2+ y2- z= 0. The gradient of F is the vector function <2x, 2y, -1> and, at (x,y,z)= (1, 2, 5) that is <2, 4, -1>. The gradient vector is always perpendicular to a "level surface" so this vector, <2, 4, -1> is perpendicular to the surface and so to the tangent plane. A vector equation of a plane with normal vector <2, 4, -1> and containing point (1, 2, 5) is (x, y, z)= (1, 2, 5)+ t(2, 4, -1). That is the equation of the tangent plane and is NOT what you give.
  5. Oct 6, 2008 #4


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    uh-uh … that's the equation of the normal to the tangent plane. :wink:

    (and (0 1 4) is perpendicular to it)
  6. Oct 6, 2008 #5
    I'm going to look at both your answers, thanks!

    I forgot to mention, there is more, I just didn't put it because I was confused at that point.
  7. Oct 6, 2008 #6


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    And just to prove what an imbecile I am, I didn't even notice it was the equation of a line, not a plane! Darn- makes me wish I could just erase that post and pretend I never wrote it. I think I got confused when JFonseca asked about the tangent plane but then started talking about a tangent line.

    The equation of the normal plane to z= x2+ y2 is, of course,
    [tex]2\vec{i}+ 4\vec{j}-\vec{k} \cdot (x-1)\vec{i}+ (y-2)\vec{j}+ (z- 5)\vec{k}[/tex]
    or 2(x-1)+ 4(y-2)- (z- 5)= 0.

    Yes, if we take the line tangent to the curve on z= x2+ y2 with x= 1, then z= 1+ y2 so dz/dy= 2y and dz/dy= 4 at y= 2. dy/dy= 1, of course, and, since x is the constant, 0, here, dx/dy= 0. The tangent vector is <0, 1, 4> and, as tiny-tim pointed out, that is pependicular to my <2, 4, -1> which is perpendicular to the surface itself. I'm not sure why you would want to calculate a tangent to the curve- unless you are also going to calculate the tangent vector to the curve on z= x2+ y2 with y= 2 and then take the cross product of the two tangent vectors to get the normal vector to the surface. Seems a complicated method to me.
  8. Oct 6, 2008 #7
    Ah...so that's how they worked it out, now I get it!

    That's actually what they do in the next few steps, I just never posted that bit because it was similar and also I didn't get how they got that tangent vector <0,1,4>, but now I know. Thanks a lot tiny and hallsofivy!
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