Direction of Steepest Decent of an Implicit Function.

Question

If given an implicit surface like $x^6z+x^3y^2+y^2z^3=65$ how would I go about finding a 3-D vector that points in the steepest downhill direction at a point?
I know that the gradient of this function is normal to the surface since the surface is a level curve. While the gradient vector would give me the steepest accent between the level curves it wouldn't help me find the steepest downhill direction of the given implicit surface.
Also if I solve the implicit surface for z and take the gradient of that function it would be a 2_D vector.
I am not sure how to go about doing a question like this.

Answer 1

Let $f(x,y,z)=x^6z+x^3y^2+y^2z^3$. Then the surface is given by the implicit equation $$f(x,y,z)=65$$

The steepest descent vector is a vector that is tangent to the surface (i.e. orthogonal to its normal) and that also has the maximum angle with $(0,0,1)^T$. In other words, it is the tangent vector that points the most downwards. Imagine a drop of water at $(x,y,z)$, as it starts following gravity, where shall it go to next along the surface?

As you said, at a point $(x,y,z)$ of the surface (so we have $f(x,y,z)=65$), the normal vector is given by the gradient of $f$ at $(x,y,z)$, so it's $$\nabla f=\left( \begin{array}{c} 6x^5z+3x^2y^2\\ 2x^3y+2yz^3\\ x^6+3y^2z^2 \end{array} \right)$$ Here we write $\nabla f$ for short, when we should really be using $\nabla f(x,y,z)$. Likewise, in what follows we use $\frac{\partial f}{\partial x}$ when we should be writing $\frac {\partial f}{\partial x}(x,y,z)$ etc.

Thus any vector $(u,v,w)^T$ in the tangent plane at that point verifies $$\begin{equation} \langle (u,v,w)^T,\nabla f\rangle = 0 \end{equation}\tag{1}$$ Because we're only interested in the direction of the vector, let's assume the vector to be of norm $1$, that is $$\begin{equation}u^2+v^2+w^2=1\end{equation}\tag{2}$$

Now, among all vectors $(u,v,w)^T$ that satisfy both $(1)$ and $(2)$, we want the one that maximizes the angle with $(0,0,1)^T$. In other words, $(u,v,w)^T$ is the solution of $$\underset{(u,v,w)} \arg \min w \,\,\,\text { under } \left \{ \begin{array}{cc} u^2+v^2+w^2=\langle (u,v,w)^T, (u,v,w)^T\rangle &=1 \\ \langle (u,v,w)^T,\nabla f\rangle&=0 \end{array} \right.$$ Using Lagrange multipliers, this means that the vector $(u,v,w)^T$ is in the span of $\{\nabla f, (0,0,1)^T\}$. That's because the gradient of the objective function (with respect to optimization variables $u$, $v$, and $w$) is $(0,0,1)^T$, and that of the constraints are $(u,v,w)^T$ and $\nabla f$. In other words, there exist $\alpha$ and $\beta$ such that $$(u,v,w)^T = \alpha (0,0,1)^T +\beta\nabla f$$ Using $(1)$ and $(2)$, we can solve for $\alpha$ and $\beta$: $$\left\{ \begin{split} 1 &= \alpha w\\ 0 &= \alpha \frac{\partial f}{\partial z} +\beta \|\nabla f\|^2 \end{split} \right.$$ Thus $\beta = -\frac 1 w \frac{\frac{\partial f}{\partial z}}{\|\nabla f\|^2}$ and we conclude $$\left\{ \begin{split} u&=\frac{1}{\|\nabla f\|}\frac{\frac{\partial f}{\partial z}\frac{\partial f}{\partial x}}{\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}}\\ v&=\frac{1}{\|\nabla f\|}\frac{\frac{\partial f}{\partial z}\frac{\partial f}{\partial y}}{\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}}\\ w&=-\frac{\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}}{\|\nabla f\|} \end{split} \right.$$ This is the general result for any surface given by the implicit equation $f(x,y,z) = \text{constant}$. You can replace $f$ by its expression in your case to find the result.