Del or Nabla operator on a scalar field

A scalar function takes a point in space as input and assigns a real number to it. It is the perfect tool to quantify properties across 3d space. For example, let $T(x,y,z)$ be a temperature function that takes a position in 3d space $(x,y,z)$ and returns a scalar temperature. Nothing constrains us to temperature of course. Scalar functions describe temperature, height, density, and much more. When we apply $\nabla$ to a scalar function $f$, we obtain $\nabla f$, which we define as the gradient of $f$. We often read that the gradient is the direction of steepest change, and the motivation usually given links to the directional derivative as the direction which maximises change. I propose an alternative approach that builds a more visual intuition. If

$$\vec{v} = \begin{pmatrix} 0 \\ 2 \end{pmatrix} \quad \text{and} \quad \vec{w} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

note that $\vec{v}$ points in the $y$ direction while $\vec{w}$ points in the $x$ direction. If you know how to visually add vectors, you can convince yourself that a linear combination of those two vectors can reach any position in this space. We can reach every possible direction via

$$c_1\vec{v} + c_2\vec{w}$$

For the purpose of understanding the gradient, let us constrain the coefficients $c_1$ and $c_2$ to unit length. That is

$$\sqrt{c_1^2 + c_2^2} = 1$$

This is equivalent to choosing $c_1 = \cos(\theta)$ and $c_2 = \sin(\theta)$. Which $\theta$ will maximise the length of the resulting vector? Since $\theta$ encodes a direction in 2d space, we are looking for the direction that gives us the longest vector. Can you start spotting the link to the gradient as the direction of steepest change?

For a given $\theta$, the gray arrow shows one candidate direction and the green arrow shows the gradient. Notice the gradient is longer. Because of our imposed constraint, the coordinates of the resulting vector are bounded between $0$ and $a$ in $x$ and between $0$ and $b$ in $y$. All possible linear combinations stay within a rectangle when we consider angles less than or equal to $90°$, for the sake of simplicity. In other words, the hypotenuse is the longest possible side given our unit length constraint. That hypotenuse is the gradient direction, and you can obtain it using $\theta = \arctan(2/1)$.

The gradient arrow reaches the far corner of the rectangle while every other candidate direction falls strictly inside it.

Now let us connect this to the gradient proper.

Imagine that $h(x,y)$ tells us the height of a surface at position $(x,y)$. Moving purely in the $x$ direction, with all other variables held constant, the rate of change is the partial derivative

$$\frac{\partial h}{\partial x}$$

Similarly, moving purely in the $y$ direction,

$$\frac{\partial h}{\partial y}$$

A change in any other direction can be expressed as a linear combination of these two rates. If we represent a direction by a unit vector $\hat{u} = (u_x, u_y)$, the directional derivative is

$$D_{\hat{u}}\, h = \frac{\partial h}{\partial x}\, u_x + \frac{\partial h}{\partial y}\, u_y$$

Since $\hat{u}$ has unit magnitude, we require

$$\sqrt{u_x^2 + u_y^2} = 1$$

This is exactly the situation from our geometric warm-up. The partial derivatives $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial y}$ play the role of the side lengths of the bounding rectangle. The directional derivative for each candidate $\hat{u}$ corresponds to one of the arrows drawn inside that rectangle. By the same geometric argument, the direction that maximises the directional derivative is the one pointing to the far corner, whose components are precisely the partial derivatives. That direction is the gradient

$$\nabla h = \begin{pmatrix} \dfrac{\partial h}{\partial x} \\[6pt] \dfrac{\partial h}{\partial y} \end{pmatrix}$$

The gradient is the direction of steepest ascent not only because of an algebraic manipulation, but because it is the hypotenuse of the rectangle whose sides are the individual rates of change. Every other unit direction corresponds to a shorter vector strictly inside that rectangle.