# tangent plane of implicit function

• Oct 14th 2009, 11:49 AM
superdude
tangent plane of implicit function
question: surface Q: $6x^2+3y^2+3z^2-6yz-2y-2z=0$ has exactly one horizontal tangent plane. Find it's equation and the point of tangency. Hint: there is no point on the surface below this plane

work: I took the partial derivatives and set them = 0. So I get x=0, and I used this to deduce that z=y (is this right?).

problem: the equation of a tangent plane requires me to compute f(a,b) but I can't do this since Q is an implicit function.

I haven't learned about vectors yet
• Oct 14th 2009, 12:08 PM
HallsofIvy
Quote:

Originally Posted by superdude
question: surface Q: $6x^2+3y^2+3z^2-6yz-2y-2z=0$ has exactly one horizontal tangent plane. Find it's equation and the point of tangency. Hint: there is no point on the surface below this plane

work: I took the partial derivatives and set them = 0. So I get x=0, and I used this to deduce that z=y (is this right?).

Is that correct for what? It is not the tangent plane but you don't seem to mean that since you ask about the tangent plane below. What did you intend the plane z= y to represent.
(It is not the tangent plane because a horizontal plane is of the form z= constant.)

Quote:

problem: the equation of a tangent plane requires me to compute f(a,b) but I can't do this since Q is an implicit function.

I haven't learned about vectors yet
• Oct 14th 2009, 02:13 PM
shawsend
Quote:

Originally Posted by superdude
question: surface Q: $6x^2+3y^2+3z^2-6yz-2y-2z=0$ has exactly one horizontal tangent plane. Find it's equation and the point of tangency. Hint: there is no point on the surface below this plane

work: I took the partial derivatives and set them = 0. So I get x=0, and I used this to deduce that z=y (is this right?).

problem: the equation of a tangent plane requires me to compute f(a,b) but I can't do this since Q is an implicit function.

I haven't learned about vectors yet

You don't. If $f(x,y,z)=0$ defines an implicit, differentiable function $z(x,y)$, then:

$\frac{\partial z}{\partial x}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$

$\frac{\partial z}{\partial y}=-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}}$

There you go. Set those to 0 to find the extremum. Suppose that point is $(x_0,y_0,z_0)$, then the normal plane to $f(x,y,z)$ at that point is defined to be:

$f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$.

Something like the plot below.

Edit: I'm a little unsure about the normal plane corresponding in this particular case to a horizontal tangent plane at the minimum of the function.
• Oct 14th 2009, 02:36 PM
superdude
This is what I've done:
the equation of a tangent plane is
$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$
So I use implicit differentation to get the partial derivative w.r.t. x

$\frac{dz}{dx}=\frac{-12x^2}{6z-6y-2}$ and I set this equal to 0. Then I use cross multiplication to determine that x=0. Then I find the partial derivative w.r.t y.

$\frac{dz}{dy}=\frac{-6y-6z-2}{6z-6y-2}=0$
I can multiply both sides by ${6z-6y-2}$ but then I'd still have y and z left in the numerator. In my previous post I was thinking that since x=0 I could plug y in for z here, but I realize that's wrong.

So how do I find the value of y?
And once I've done that, refereing back to the tangent plane equation, how do I find f(a,b) since I wasn't given f explicitly?
• Oct 14th 2009, 02:44 PM
shawsend
I got:

$\frac{\partial z}{\partial y}=\frac{6y-6z-2}{6z-6y-2}$

Set that to zero, and get z=y-1/3. Already have x=0 so plug that and z=y-1/3 into f(x,y,z)=0 to get:

$3y^2+3(y-1/3)^2-6y^2-2y-2(y-1/3)=0$

to get y, then you have z too.

Also, if we get the point $(x_0,y_0,z_0)$ and we want a horizontal plane to that point, then we'd need the normal. but an outward normal to the x-y plane is just $(0,0,-1)$ so perhaps the horizontal tangent plane is not the normal plane but rather the equation:

$0(x-x_0)+0(y-y_0)-(z-z_0)=0$ for the horizontal tangent plane at the point $(x_0,y_0,z_0)$

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Using the quadratic formula I get y = +/- $\frac{\sqrt{11}}{2}-\frac{3}{2}$ so how do I know which value to plug into $z=y-\frac{1}{3}$?