Undergrad Gradient Vectors: Perpendicular to Level Curves

[AAMAIK]

TL;DR

Gradient vector definition and its properties how its interpreted and what is it good for?

The gradient transforms a scalar function into a vector function where the vector components are the rates of change of the functions with respect to its independent variables.
Also, the properties of the gradient are:
It lies in the plane.
It is perpendicular to the level curves and points towards higher values of the function.
(1)
I am not sure of my interpretation of the definition, but that is how I understood it
Consider a function of two independent variables x and y. If I want to approximate the function in the neighbourhood of the point (x0,y0), then the tangent plane must pass through the same point (x0,y0) and to narrow down the many candidates of the different tangent planes the tangent plane must have the same slopes as the surface in the i and j directions (this is captured by the gradient vector).

The link describes the proof the grad vector is perpendicular to the level curves
https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-36-proof/MIT18_02SC_notes_19.pdf
I don't understand why we take any arbitrary curve r(t) on the level curve surface and the chain rule to prove the property. If my understanding is correct in (1) then I could shown that the gradient is perpendicular to level curves as follows. The tangent plane approximating the function at (x0,y0) will intersect the level surface f(x0,y0)=c in a line that lies on the level surface and if the value of the function does not changes implies that the gradient vector is perpendicular to the change to the position vector along that line of intersection.

 

[Infrared]

The tangent plane approximating the function at (x0,y0) will intersect the level surface f(x0,y0)=c in a line that lies on the level surface

In general, this is not true. For example, a tangent plane to a sphere only intersects the sphere once.

 

[Orodruin]

Summary:: Gradient vector definition and its properties how its interpreted and what is it good for?

It lies in the plane

What plane? The gradient is a vector with the same number of components as your underlying space. If you work in $\mathbb R^3$, then the gradient will be a vector in 3D and may point in any direction.

 

[AAMAIK]

In general, this is not true. For example, a tangent plane to a sphere only intersects the sphere once.

If we are considering a function of two variables then what I meant was the horizontal plane and not the trace on that horizontal plane.

 

[zinq]

There is such a thing as a gradient vector field. Let's assume we have a function f(x,y) defined on the plane. Then if (x₀, y₀) is a point, consider any unit vector u = (a,b). Then a line leaving the point (x₀, y₀) in the direction u is given by L(t) = (x₀, y₀) + t(a,b).

Now consider the values of f(x,y) along that line, or in other words f(L(t)) = f((x₀ + ta, y₀ + tb). Let's see how fast that function of t is increasing: using the chain rule we get d/dt (f(L(t)) = ∂f/∂x * a + ∂f/∂y * b. Or in other words, the dot product of (∂f/∂x, ∂f/∂y) and u. (Where the partial derivatives are to be evaluated at the point (x₀, y₀).) And of course (∂f/∂x, ∂f/∂y) is by definition the gradient of f at (x₀, y₀). So the directional derivative of f in the direction of u is just the dot product of the gradient with u.

From this it's easy to see that at (x₀, y₀), the function f(x,y) increases fastest in the same direction as (∂f/∂x, ∂f/∂y) (this is a good exercise).