(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

S is the surface with equation [tex] z = x^2 +2xy+2y[/tex]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

[tex]F(x,y,z): z = x^2 +2xy+2y[/tex]

[tex]F_x = 2x + 2y[/tex]

[tex] F_y = 2x + 2[/tex]

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

[tex]z-9=6(x-1) + 4(y-1)[/tex].

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

[tex]F(x,y,z): 0 = x^2 +2xy+2y - z[/tex]

then

[tex]grad F = <2x + 2y,2x + 2,-1>[/tex] 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!

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# Finding Horizontal Tangent Planes on S

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