So I know that the rate of maximum increase of some function (say, $f(x,y)$) is given by the gradient ($\nabla f$), where the direction is the direction of maximum increase of the function, and the magnitude tells you how fast it's increasing.

However, when I get into optimizing not for a maximal $f$, but rather for some other unrelated function (say $T(x,y,z)$) then I'm a bit lost. The particular problem I'm looking at gives specific functions and a specific point, but the generalized version of problem boils down to this:

Find the direction of maximum increase of $T(x,y,z)$ from a point $(x_0,y_0,f(x_0,y_0))$ while constrained to the tangent plane to the surface $z = f(x,y)$ at $(x_0,y_0)$.

If we call the general solution to this problem some vector-valued function $\vec{M}(f,T)$, then my naive attempt was $\vec{M} = \nabla(\nabla f \cdot \nabla T)$. Obviously, this approach is flawed, because $\vec{M}(f,f)$ should yield $\nabla f$, but $\nabla \|\nabla f\|^2 \ne \nabla f$, even though it serendipitously gave the right answer for part a) of the problem.

I was also thinking of parameterizing $f$ in terms of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$, but I don't see how that would provide insight in the general case.

How do I approach this problem?


The directional derivative depends on the angle with the gradient vector: the smaller the angle, the greater the derivative. You are constrained to the directions that lie in a certain plane, the tangent plane to the surface. The direction in the plane that has the smallest angle with the gradient is given by the orthogonal projection of the gradient onto the plane (why? use the fact that the projection finds the nearest point, and try to relate distance to angle).

In practical terms, you should find the unit normal vector $\vec n$ to the surface (obtained by normalizing $\langle, -f_x,-f_y,1\rangle $), and compute $\nabla T-(\nabla T\cdot \vec n)\,\vec n$.

| cite | improve this answer | |
  • $\begingroup$ I didn't quite understand the bit about the orthogonal projection; why does that have the smallest angle with $\nabla T$? And wouldn't the orthogonal projection be normal to the plane (as opposed to the projection which would lie in the plane)? $\endgroup$ – éclairevoyant Nov 5 '13 at 5:56
  • $\begingroup$ Oops, I thought you meant the orthogonal projection relative to the plane, not $\hat n$. $\endgroup$ – éclairevoyant Nov 5 '13 at 5:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.