Finding the equation of a tangent plane given an implicity defined function and obtaining $8$ roots

Question

I want to find the equation of the tangent plane at the point $P(-1,-1,-1)$ given the equation

$$x^2 + 10 xyz + y^2 + 8z^2 = 0$$

However, I end up getting eight equations, so I have two questions

1. Because $z$ is the dependent variable and $x,y$ are the independent variables, is the equation above an implicitly defined function? If so, I went ahead and solved for $z$, which leads me to my second question

2. Is my work correct, and what is the geometric interpretation of the eight equations if it is? Thanks in advance.

I went ahead and solved for $z$, so I'll spare all the details, but I end up with two possible equations for $z$

\begin{align*} x^2 + 10xyz + y^2 + 8z^2 &= 0\\ \, \, 8(z^2 + \tfrac{5}{4}xyz + \tfrac{25}{64}x^2y^2 - \tfrac{25}{64} x^2y^2) &= -(x^2+y^2)\\ 8(z+\tfrac{5}{8}xy)^2 &= \tfrac{25}{8}x^2y^2 - (x^2 + y^2) \\ z &= -\tfrac{5}{8}xy \pm \tfrac{1}{8}\sqrt{25x^2y^2-8(x^2+y^2)} \end{align*}

The partial derivatives with respect to $x$ and $y$

$$f_x = \tfrac{5}{8}y \pm \frac{x(25y^2-8)}{8\sqrt{25x^2y^2-8(x^2+y^2)}}$$ $$f_y = \tfrac{5}{8}x \pm \frac{y(25x^2-8)}{8\sqrt{25x^2y^2-8(x^2+y^2)}}$$

evaluated at $P(-1,-1,-1)$ are

$$\frac{\partial z}{\partial x}\ = -\frac{1}{2},\frac{4}{3}$$ $$\frac{\partial x}{\partial y} = -\frac{1}{2},\frac{4}{3}$$

So using the equation of a tangent plan, and considering all possible arrangements, which are eight, I get

\begin{align*} z &= -\tfrac{3}{4}-\tfrac{1}{12}(x+1) + \tfrac{4}{3}(y+y1) \\ &= -\tfrac{3}{4}-\tfrac{1}{12}(x+1) - \tfrac{1}{12}(y+1) \\ &= -\tfrac{3}{4}+\tfrac{4}{3}(x+1) - \tfrac{1}{12}(y+1) \\ &= -\tfrac{3}{4}+\tfrac{4}{3}(x+1) + \tfrac{4}{3}(y+1) \\ &= -\tfrac{1}{2}-\tfrac{1}{12}(x+1) + \tfrac{4}{3}(y+1) \\ &= -\tfrac{1}{2}-\tfrac{1}{12}(x+1) - \tfrac{1}{12}(y+1) \\ &= -\tfrac{1}{2}+\tfrac{4}{3}(x+1) + \tfrac{4}{3}(y+1) \\ &= -\tfrac{1}{2}+\tfrac{4}{3}(x+1) -\tfrac{1}{12}(y+1) \\ \end{align*}

Answer 1

Yes, your equation defines $z$ implicitly as a function of $x$ and $y$, around particular points. But it is silly to try to solve for $z$; in general this is impossible. Instead, just use the fact that gradients are perpendicular to level sets.

Your graph is the zero level set of the function $$f(x,y,z)=x^2+10xyz+y^2+8z^2$$ The gradient is $$\langle 2x+10yz, 2y+10xz, 16z+10xy\rangle$$which at your point evaluates to $\langle 8, 8, -6\rangle$. The tangent plane at your point is the plane perpendicular to this vector, which I assume you know how to write down.

If you insist on solving for $z$ you must choose the correct branch (in this case, the correct choice of $+$ or $-$) that contains your point of interest.

Answer 2

The method you are using doesn't make sense. It is pretty apparent from the form in which the equation given and how ugly solving for one of the variables gets that this problem was formulated to test implicit methods.

My approach:

First implicitly differentiate $x^2+10xyz+y^2+8z^2=0$ to get $2xdx+10yzdx+10xzdy+10xydz+2ydy+16zdz =0.$

For a given point, this equation in $dx, dy,$ and $dz$ is a plane: the plane of all vectors tangent to the curve at $\langle x,y,z\rangle$. Thus, we can easily find the normal to this plane and therefore the plane tangent to the original curve at $\langle x, y, z\rangle$.

The equation for the plane at $P(-1, -1, -1)$ simplifies to the following: $$-2dx+10dx+10dy+10dz-2dy-16dz = 0\\\to8dx+8dy-6dz=0\\\to4dx+4dy-3dz=0.$$

Thus the plane in question's normal is $\langle4, 4, -3\rangle$, making the equation of the tangent plane at $P(-1, -1, -1)$ $$4(x+1)+4(y+1)-3(z+1)=0.$$

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To explain what is going on in your attempted solution:

You correctly solve for $z$ in terms of $x$ and $y$, however this does not yield a function (this is evident in $z$ having two possible values at many $(x, y)$ due to the $\pm$).

The reason you get multiple possible values for the partial derivatives of $z$ in the next step is because $z$ acts differently with changes in $(x, y)$ at these different values.

The final step seems to be an effort to congeal these different points into a single one somehow, but this doesn't make any sense due to the actual meaning of the different possible $z$ values.

A correct way to handle the final step here would have been to figure out whether the $\pm$ was a $+$ or $-$ for the point $(-1, -1, -1)$, then take the partial derivatives of that branch of the curve.

Answer 3

$z$ is not a function of $x$ and $y$ (as you noted, there are two sign), but is locally a function (there is a neighborhood in which z is such a function). But this way deal with complicated computations and I'll take another (but similar) point of view.

Let $$f: \mathbb{R}^3 \to \mathbb{R}: (x, y, z) \mapsto x^2 + y^2 + 8 z^2 + 10xyz \\ S = f^{-1}(0) \subseteq \mathbb{R}^3 \\ P=(-1, -1, -1) \in S$$ Let's check if $f$ is a submersion at least in $P$ (gradient is no zero). We have $$Df(x, y, z) = [2x + 10yz, 2y + 10xz, 16z + 10 xy]\\ \therefore Df(P) = [8, 8, -6]$$ The tangent space is just the vector space perpendicular to the gradient, i.e., the kernel of the differential $$ v \in T_PS \iff Df(P) \cdot v = [8, 8, -6] \cdot [v_1, v_2, v_3] = 0$$ So, the equation of the tangent space at $P$ is $$T_PS: \quad \quad 4x + 4y - 3z =0$$ Ooh, you prefer the tangent plane ? Fine, just translate it to $P$ $$A_PS: \quad \quad 4(x+1) + 4(y+1) - 3(z+1) =0$$

Do you prefer see it as a span (well, I prefer...)? Ok. Now, we solve this linear system $$v_3 = \frac{4}{3} (v_1 + v_2) \Rightarrow v = [v_1, v_2, 4/3 (v_1 + v_2)] = v_1 [1, 0, 4/3] + v_2 [0, 1, 4/3] \\ \therefore T_PS = span([1, 0, 4/3], [0, 1, 4/3])$$.