Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Tangent planes to surfaces
  1. Tangent Plane (Replies: 6)