# Finding Horizontal Tangent Planes on S

1. Dec 2, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
S is the surface with equation $$z = x^2 +2xy+2y$$a) Find an equation for the tangent plane to S at the point (1,2,9).
b) At what points on S, in any, does S have a horizontal tangent plane?

3. The attempt at a solution
$$F(x,y,z): z = x^2 +2xy+2y$$
$$F_x = 2x + 2y$$
$$F_y = 2x + 2$$

Evaluated at (1,2) gives answers 6 and 4, respectively. My equation for a plane is:

$$z-9=6(x-1) + 4(y-1)$$.

I think any horizontal plane should have normal vector <0,0,k>, where k is some scalar. I'm pretty sure that S has no such normal vector. But if
$$F(x,y,z): 0 = x^2 +2xy+2y - z$$
then
$$grad F = <2x + 2y,2x + 2,-1>$$ It seems like I can let (x,y) = (-1,1) to zero the x-, y-components of the gradient. Plugging (-1,1) into the definition of z gives z = 1. This suggests to me that there is a point (-1,1,1), at which there is a horizontal tangent plane. Yet I feel pretty sure that this isn't true!

2. Dec 3, 2011

### ehild

You made a little mistake when writing out the equation of the tangent plane. The y coordinate of the fixed point is 2, you wrote 1.

A surface in 3D is of the form F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal to the surface, and the tangent plane of the surface at a given point. You want a horizontal tangent plane, so a vertical gradient:(0,0,a). That means Fx=2x+2y=0, Fy=2x+2=0 --->x=-1, y=1, so your result for the x,y coordinates are correct. Plugging into the original equation for x and y, you got z=x2+2xy+2y=1, it is correct. Why do you feel it is not?

ehild

3. Dec 3, 2011

### TranscendArcu

When I graphed F(x,y,z) in MatLab (and it's possible I graphed it incorrectly), I observed that the the resulting paraboloid is always "tilted". Below is a picture from my plot:
http://img440.imageshack.us/img440/687/skjermbilde20111203kl85.png [Broken]
How can this surface have a horizontal tangent anywhere when it is tilted like this?

Last edited by a moderator: May 5, 2017
4. Dec 3, 2011

### ehild

Try to plot z out for -2<x<0 and 0<y<2

ehild

5. Dec 3, 2011

### TranscendArcu

http://img7.imageshack.us/img7/139/skjermbilde20111203kl10.png [Broken]
Hmm. I'm not seeing the a point in this picture where the gradient is pointing directly upwards. Everything still looks kind of tilted.

Last edited by a moderator: May 5, 2017
6. Dec 3, 2011

### ehild

The function is equivalent with z=(x+2y-1)(x+1)+1 and z=1 along the lines x=-1 and y=(1-x)/2. I attach a plot of the surface near the point (-1,1)

ehild

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7. Dec 4, 2011

### HallsofIvy

Staff Emeritus
I presume that by "horizontal" you mean perpendicular to the z-axis.

The simplest way to find a tangent planes for a surface is to write it in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\nabla F$. Here, you can write $F(x, y,z)= x^2+ 2xy+ 2y- z= 0$.

What is $\nabla F$? That will be vertical (and so tangent plane horizontal) when its x and y components are 0.

8. Dec 4, 2011

### ehild

@HallsofIvy: The OP has shown the solution in his first post, he only can not believe it, as the surfaces he got with MatLab look tilted. If you could give advice how to plot surfaces with MatLab, that would be real help for him.

ehild

9. Dec 4, 2011

### TranscendArcu

If anyone is familiar with MatLab, this is the code I've been using:

Note that the "%" mark my annotations. I included them so that hopefully you can follow what I'm doing more easily.

10. Dec 4, 2011

### TranscendArcu

Ha! I figured it out. I forgot a "."

I should have written

z = x.^2 + 2*x.*y +2*y;

Everything makes sense now.

11. Dec 4, 2011

### ehild

You see: it is worth typing something out again and again. Is your plot similar to my one? It was made with Origin. I would like to see your final plot... Please...

ehild

12. Dec 4, 2011

### TranscendArcu

http://img259.imageshack.us/img259/2104/skjermbilde20111204kl10.png [Broken]It looks like it could have a horizontal tangent plane right around (-1,1,1)

Last edited by a moderator: May 5, 2017
13. Dec 4, 2011

### ehild

It is really nice!!! And a missing dot made you sceptical about the truth of Maths??!!!:uhh:

ehild