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David's question at Yahoo! Answers (horizontal tangente plane).

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
David's question at Yahoo! Answers (horizontal tangent plane).

Here is the question:

Find the point(s) on the surface at which the tangent plane is horizontal.? z = 3 − x^2 − y^2 + 8y
(x, y, z) = ( )
Here is a link to the question:

Find the point(s) on the surface at which the tangent plane is horizontal.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.


P.S.
Of course I meant in the title tangent instead of tangente (It is hard to forget our mother tongue). :)
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello David,

The equation of the tangent plane to a surface $\phi :z=f(x,y)$ at the point $P_0(x_0,y_0,z_0)$ of $\phi$ is

$\pi: \phi_x(P_0)(x-x_0)+\phi_y(P_0)(y-y_0)-1(z-z_0)=0$

The plane $\pi$ is horizontal if and only if $\phi_x(P_0)=\phi_y(P_0)=0$. In our case if and only if $-2x_0=0$ and $-2y_0+8=0$. We get $x_0=0,y_0=4$.

As $P_0$ belongs to the surface, $z_0=3-0^2-4^2+8\cdot 4=19$. The solution is $(x_0,y_0,z_0)=(0,4,19)$.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Another way to do this: [tex]z= 3 − x^2 − y^2 + 8y[/tex] can be thought of as "level surface": [tex]f(x, y, z)= z+ x^2+ y^2- 8y= 3[/tex]. The gradient, [tex]\nabla f= 2x\vec{i}+ (2y- 8)\vec{j}+ \vec{k}[/tex], is perpendicular to the surface and so the tangent plane (which is, of course, also perpendicular to the normal curve) is parallel to the xy-plane if and only if that gradient is vertical- that is, that 2x= 0 and 2y- 8= 0.