Finding a direction to not change the level

[E7.5]

Homework Statement

You are walking on the graph of $f(x,y) = y cos(\pi x) - x cos(\pi y) + 10$, standing at the point $(2,1,13)$. Find an x, y-direction you should walk into stay at the same level.

Homework Equations

$D_u f = \nabla \cdot \textbf{u}$

The Attempt at a Solution

The directional derivative is the rate of change in the direction of $\textbf{u}$, so we want the rate of change in the direction of $\textbf{u}$ to not change, i.e. $\nabla \cdot \textbf{u} = \textbf{0}$. So calculating the gradient gives $<1,1>$. Then $<1,1> \cdot \textbf{u} = \textbf{0}$. So that means $\textbf{u} = <0,0>$, but this is incorrect. Why?

 

[Ray Vickson]

Homework Statement

You are walking on the graph of $f(x,y) = y cos(\pi x) - x cos(\pi y) + 10$, standing at the point $(2,1,13)$. Find an x, y-direction you should walk into stay at the same level.

Homework Equations

$D_u f = \nabla \cdot \textbf{u}$

The Attempt at a Solution

The directional derivative is the rate of change in the direction of $\textbf{u}$, so we want the rate of change in the direction of $\textbf{u}$ to not change, i.e. $\nabla \cdot \textbf{u} = \textbf{0}$. So calculating the gradient gives $<1,1>$. Then $<1,1> \cdot \textbf{u} = \textbf{0}$. So that means $\textbf{u} = <0,0>$, but this is incorrect. Why?

Because going along $\vec{u} = <0,0>$ gets you nowhere: any multiple of 0 is still 0! What is the general criterion for the condition $<u_x,u_y> \perp <1,1>$, expressed as an equation or equations involving $u_x, \, u_y$?

 

[haruspex]

Then $<1,1> \cdot \textbf{u} = \textbf{0}$. So that means $\textbf{u} = <0,0>$,

By definition, you need a nontrivial solution of <1,1>.u = 0. u = <0,0> is not the only solution.