Suppose the level curve \(C\) is parametrized by a differentiable vector function
\begin{equation}
\mathbf r(t) = \langle x(t),\, y(t)\rangle,\tag{7.6}
\end{equation}
and let \(\mathbf v(t) = \langle x'(t),\, y'(t)\rangle\) denote its velocity vector. Because every point of \(C\) lies on the level curve, its coordinates satisfy
\begin{equation}
f\bigl(x(t),\, y(t)\bigr) = c \qquad \text{for all } t.\tag{7.7}
\end{equation}
The left-hand side is constant in \(t\text{,}\) so its derivative is zero. Differentiating both sides with respect to \(t\) and applying the chain rule gives
\begin{align}
\frac{d}{dt}\, f\bigl(x(t), y(t)\bigr) \amp= \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt}\notag\\
\amp= 0.\tag{7.8}
\end{align}
We recognize the middle expression as the dot product of the gradient vector with the velocity vector:
\begin{align}
\left\langle \frac{\partial f}{\partial x},\, \frac{\partial f}{\partial y} \right\rangle \cdot \left\langle \frac{dx}{dt},\, \frac{dy}{dt} \right\rangle \amp= \nabla f \cdot \mathbf v(t)\notag\\
\amp= 0.\tag{7.9}
\end{align}
Therefore, at each point of
\(C\text{,}\) the gradient
\(\nabla f\) is orthogonal to the velocity vector
\(\mathbf v(t)\text{,}\) as shown in
Figure 7.12. Since the velocity vector points in the direction of motion along the curve, it is tangent to
\(C\text{;}\) hence
\(\nabla f\) being orthogonal to
\(\mathbf v(t)\) means precisely that
\(\nabla f\) is perpendicular to the level curve
\(C\text{.}\) This holds at every point where
\(\nabla f \neq \mathbf 0\text{.}\)
