Problem on using the chain rule to show gradients are perpendicular to level curves - Leading Lesson

Problem on using the chain rule to show gradients are perpendicular to level curves

Let $\mathbf{r}(t) = x(t) \ \mathbf{i} + y(t) \ \mathbf{j}$ be a path along a level curve of $f(x,y)$.

Let $F(t) = f(x(t), y(t))$. Argue that $dF/dt = 0$.

Use the chain rule to express $dF/dt$ as a dot product of two vectors.

Show that the gradient of $f$ is perpendicular to the level curves of $f$.

Solution
Part (a)
Recall that
Level curves and surfaces

The level curves of $f(x,y)$ are curves in the $xy$-plane along which $f$ has a constant value.
Because $\mathbf{r}(t)$ is a level curve of $F$, $F(t)$ is constant. Hence, $dF/dt= 0$.
Part (b)
Recall that
Chain rule with functions of several variables

The two-variable chain rule gives \begin{align}
\frac{d}{dt}f(x(t), y(t)) = &\partial_x f \Bigl (x(t), y(t) \Bigr) \frac{dx}{dt} \\&+ \ \partial_y f \Bigl (x(t), y(t) \Bigr) \frac{dy}{dt}
\end{align}
Definition of the gradient

Direct computation of dot product

We express this value as a dot product of two vectors: \begin{align}
\frac{d}{dt} f(x(t), y(t)) &= \left \langle \partial_x f , \partial_y f \right \rangle \cdot \left \langle \frac{dx}{dt} , \frac{dy}{dt}\right \rangle \\
&= \nabla f \cdot \frac{d \mathbf{r}}{dt}
\end{align}
Part (c)
Combining (a) and (b), we have shown that for all paths along a level curve $$0 = \nabla f \cdot \frac{d \mathbf{r}} {dt}$$
Recall that
Dot product of perpendicular vectors is zero

If the dot product of two vectors is zero, then the vectors are perpendicular.
Thus, $\nabla f$ is perpendicular to $\frac{d \mathbf{r}}{dt}$.
Because $\mathbf{r}(t)$ traces out the level curve, $\frac{d \mathbf{r}}{dt}$ is a tangent vector to the level curve. Hence we have shown that $\nabla f$ is perpendicular to the tangent vector of the level curve. Thus, it is perpendicular to the level curve.

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