tangent hyperplanes and tanget planes, a bit confused about dimensions

**Volga** · March 16th 2011, 08:35 PM

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

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

???

**Volga** · March 17th 2011, 02:08 AM

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!

**HallsofIvy** · March 17th 2011, 03:57 AM

Originally Posted by Volga

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

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

???

**Volga** · March 17th 2011, 05:38 AM

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?

**HallsofIvy** · March 17th 2011, 05:47 AM

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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