orthogonal to the level curve

Question

This is from my textbook, I don't quite understand the context in red

why a zero directional derivative at a point indicates that u is tangent to a level curve? It didn't provide a proof. And how "the gradient ∇ f (a, b) is orthogonal to the level curve f (x, y) = c at the point(a, b)" come from?

Answer 1

The following are not mathematically rigorous, but are intended simply to supply some intuition for the claims:

1. If the function has a zero derivative along ${\mathbf u}$, then it doesn't change along ${\mathbf u}$ and therefore ${\mathbf u}$ must be along (more precisely, tangent to) the level curve at that point.

2. Say that $f$ increases at a rate of $v_x$ in the $x$-direction, and at a rate of $v_y$ in the $y$-direction. In a sense, $v_x$ and $v_y$ tell us how much of the change in $f$ can be attributed to a change in $x$ or a change in $y$, respectively. For instance, if $v_x = 1$, but $v_y = 4$, then changes in $y$ are responsible for $4$ times the change in $f$ that changes in $x$ are.

   With that in mind, it stands to reason that the vector $(v_x, v_y)$ would give us the direction in which $f$ changes most "steeply", and it would therefore be perpendicular to the level curve at that point, in the same way that the steepest direction up a hill at any point is generally perpendicular to the level path around it.

   Note that $\nabla f(x, y)$ is simply this $(v_x, v_y)$.

Answer 2

In response to your first question : From the equation given above, when $D_{\textbf{u}}f = 0$, it means that $\cos(\theta) = 0$ i.e. the angle in between $\textbf{u}$ and $\nabla f(x,y)$ is a right angle ($\theta = \pi/2$). Since we know that $\nabla f(x,y)$ is orthogonal to the level curve at $(x,y)$ (meaning it is orthogonal to all vectors tangent to the curve at this point), this means that $\textbf{u}$ must be tangent to the level curve.

To convince yourself that the gradient is orthogonal to the level curve, check out this simple example :

Let $f(x,y) = \sqrt{4 - x^2 - y^2}$, and consider the level curve $f(x,y) = \sqrt{3}$. Then, since $$\sqrt{3} = \sqrt{4 - x^2 - y^2} \implies x^2 + y^2 = 1,$$ the level curve is the unit circle.

Now observe $\nabla f(x,y) = \Big\langle \frac{-x}{\sqrt{4-x^2-y^2}}, \frac{-y}{\sqrt{4-x^2-y^2}} \Big\rangle$. Take a point on the level curve, say $(0,1)$, and evaluate the gradient at this point giving you $\nabla f(0,1) = \frac{1}{\sqrt{3}}\langle 0,-1\rangle$. Graph this vector so its initial point is the point $(0,1)$ and you will see that the gradient is orthogonal to the curve. Of course, this isn't a proof, but more of demonstration of this theorem.

Answer 3

Along the level curve, we can parametrize it at(a, b) locally as $(x(s), y(s))$, where $s$ is the arc length. So, $$f(x(s), y(s)) = c$$ Or $$ \frac{df}{ds}(x(s), y(s)) = \nabla f \cdot (\frac{dx}{ds}, \frac{dy}{ds}) = 0$$

We know $(x', y')$ is the tangent direction along the level curve. So $\nabla f$ is orthogonal to the tangent line. Any vector $\vec u$, such that $\nabla f \cdot \vec u = 0$, is orthoginal to $\nabla f$. In two dimension, since there is any one direction that is orthonal to $\nabla f$, so $\vec u$ must be parallel to $(x', y')$, the tangent direction of the level curve at (a, b).