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Math Help - tangent hyperplanes and tanget planes, a bit confused about dimensions

  1. #1
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    tangent hyperplanes and tanget planes, a bit confused about dimensions

    I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).


    Question (Calculus, Binmore and Davies, 3.9(#5*))

    Let f(x,y,z)=ln(x^2+y^2+z^2)

    i. Find the gradient of f at the point (x,y,z)^T

    ii. Find the tangent hyperplane to the hypersurface u=ln(x^2+y^2+z^2)
    where (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T.

    iii. Find the normal and the tangent plane to the contour

    ln(x^2+y^2+z^2)=0 at (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T.



    Answer.

    i. \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T

    ii. To find tangent hyperplane, I want to use the formula

    u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)

    u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}

    u=\frac{2}{\sqrt{3}}(x+y+z)-2

    iii. To me, the equation for the countour ln(x^2+y^2+z^2)=0 looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.

    I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional.

    ???

    I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie

    z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)

    ???
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  2. #2
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    I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii)
    Actually, had a better idea: the contour is f(x,y,z)=0 so the tangent plane to the countour is

    (x-X)f_x+(y-Y)f_y+(z-Z)f_z=0

    so that I get \frac{2}{\sqrt{3}}(x+y+z)-2=0 and x+y+z=\sqrt{3} as per the answer in the book.

    Now I need to figure out the way to find the normal vector. BUT it's probably the coefficients of x, y and z in the above equation, right? So, the normal vector is (1,1,1)^T.

    Any comments are still welcome!
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  3. #3
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    Quote Originally Posted by Volga View Post
    I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).


    Question (Calculus, Binmore and Davies, 3.9(#5*))

    Let f(x,y,z)=ln(x^2+y^2+z^2)

    i. Find the gradient of f at the point (x,y,z)^T

    ii. Find the tangent hyperplane to the hypersurface u=ln(x^2+y^2+z^2)
    where (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T.

    iii. Find the normal and the tangent plane to the contour

    ln(x^2+y^2+z^2)=0 at (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T.



    Answer.

    i. \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T

    ii. To find tangent hyperplane, I want to use the formula

    u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)

    u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}

    u=\frac{2}{\sqrt{3}}(x+y+z)-2

    iii. To me, the equation for the countour ln(x^2+y^2+z^2)=0 looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.
    No, they are not the same thing though the difference is subtle. Consider the lower dimension case of z= x^2+ y^2- 1. That describes a two dimensional surface in a three dimensional space. It has a tangent plane and a normal vector at any given point and, in particular at (1, 0, 0). There, the normal vector is given by 2\vec{i}- \vec{k} and the tangent plane is given by \frac{x- 1}{2}- z= 0 or z= \frac{1}{2}x- \frac{1}{2}. We can get that normal vector by thinking of the surface z= x^2+ y^2- 1 as a "level surface" of the function \phi(x, y, z)= x^2+ y^2- z= 1 and taking the gradient: \nabla \phi= 2x\vec{i}+ 2y\vec{j}- \vec{k}, which, at (1, 0, 1), is 2\vec{i}- \vec{k}.

    But the graph of x^2+ y^2- 1= 0 is a circle in the xy-plane. it is, in fact, the intersection of the xy-plane with the paraboloid z= x^2+ y^2- 1. But because it is in the xy-plane, any normal to it at a given point, say (1, 0) must be in the xy-plane which was not true of the normal vector to the parabloid at (1, 0, 0). Now, instead of thinking of the surface z= x^2+ y^2- 1[/tex] as a level surface of the function [tex]\phi(x, y, z)= x^2+ y^2- z[tex], we think of it as a as a "level curve" of z(x,y)= x^2+ y^2- 1. The (two-dimensional) gradient of that is \nabla z= 2x\vec{i}+ 2y\vec{j} which, at (1, 0), is 2\vec{i}. A tangent line would be x= 1.

    I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional.

    ???

    I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie

    z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)

    ???
    Last edited by HallsofIvy; March 17th 2011 at 04:12 AM.
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  4. #4
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    Now, instead of thinking of the surface z= x^2+ y^2- 1[/tex] as a level surface of the function \phi(x, y, z)= x^2+ y^2- z, we think of it as a as a "level curve" of z=x^2+y^2-1.
    So, in other words, we take a 'slice' of it, to go one dimension lower, by fixing z as a constant, and hence we find ourselvers in xy-plane? So that the grad vector 'loses' z-derivative because z is no longer variable?

    Thanks for your explanation! By the way, as a practice, if I want to calculate normal vector in my example of hypersurface in (ii), I think it will be 4x1 vector as follows:

    u=(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{  \sqrt{3}},-1)(x-\frac{1}{\sqrt{3}},y-\frac{1}{\sqrt{3}},z-\frac{1}{\sqrt{3}},u)^T thus giving the normal vector of

    (\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\s  qrt{3}},-1)^T?

    Question: does the size (length) of the 'normal' vector matter? if what is important is the direction only (eg such normal vector can define a plane), then I can say the vector is k*(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{  \sqrt{3}},-1)^T where k>0? Or the size does matter?
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  5. #5
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    No. The point of a "normal vector" is that it is perpendicular to the surface. Only its direction is relevant. Any multiple of a normal vector is still a normal vector. Sometimes, you might be instructed to find the unit normal vector specifically to avoid any depence upon length. Of course, there are still two unit normal vectors to a surface, in opposite directions.

    There are special cases in which the length of a normal vector might be important. For example, if you are integrating a vector field over a surface, you would want the length of the normal vector to reflect the "differential of surface area". You can get that by using the "fundamental vector product" for the surface.
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