Gradient is orthogonal to level sets

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}$, and let $c \in \mathbb{R}$. Let $I \subseteq \mathbb{R}$ be an open interval and let $\gamma \colon I \to U$ be a curve that lies in the level set $L_c(f)$, so that $f(\gamma(t)) = c$ for each $t \in I$. Let $t_0 \in I$, and suppose that $\gamma$ is differentiable at $t_0$ and $f$ is differentiable at $\gamma(t_0)$. Then $\nabla f(\gamma(t_0)) \cdot \gamma'(t_0) = 0$.

Remarks

The proof is one step: $f \circ \gamma$ is constant, so its derivative is $0$, and by the chain rule for a curve that derivative is $\nabla f(\gamma(t_0)) \cdot \gamma'(t_0)$. So the gradient is orthogonal to each tangent vector of the level set at the point. Let $\nabla f \ne 0$ and let $f$ be of class $C^1$. Then the level set near the point is a smooth hypersurface, and $\nabla f$ is a normal vector to it. For $f(x, y, z) = x^2 + y^2 + z^2$, the level set at $c = 1$ is the unit sphere, and $\nabla f(p) = 2p$ is normal to the sphere at $p$.

Explanations

L1 human

Along a contour line the height does not change, so the slope along the line is $0$. The gradient gives the slope in each direction by a dot product, so it must be at right angles to the contour line.