Directional Derivative Interpretation

Question

I'm trying to understand directional derivatives. If I have a function $f(x,y)$ then $\frac{\partial f}{\partial x}|_{x_0,y_0}=f_x$is the slope of a tangent line to a curve parallel to the X-Z plane and $\frac{\partial f}{\partial y}|_{x_0,y_0}=f_y$is the slope of a tangent line to a curve parallel to the Y-Z plane at point $(x_0,y_0)$

$\implies \nabla f= <f_x,f_y>$

and now if I want to find the slope of a curve obtained by slicing $f(x,y)$ by a plane parallel to some vector $\overrightarrow u = <a,b>$ then we can get the parametric equation of the line parallel to $\overrightarrow u$ through the point $(x_0,y_0)$

as $$ \begin{matrix} x=x_0+as\\ y=y_0+bs\\ \end{matrix} $$

By using the chain rule,

$$ \begin{matrix} \frac{df}{ds}=\frac{\partial f}{\partial x}\frac{dx}{ds}+\frac{\partial f}{\partial y}\frac{dy}{ds} \\ \frac{df}{ds}=\nabla f \cdot \frac{d\overrightarrow r}{ds}\\ \frac{df}{ds}=\nabla f \cdot \overrightarrow u\\ \end{matrix} $$

But why is the professor considering $\frac{d\overrightarrow r}{ds}$ as to $\overrightarrow u$ And, why does $\overrightarrow u$ have to be a unit vector? What is the significance of this? and how is $\frac{df}{ds}$ same as directional derivative. It all seems muddled and awkward

Answer 1

The directional derivative is meant to generalize the notion of partial derivatives - it's a rate of change in a specific direction, similar to how $f_x$ and $f_y$ are the rates of change in the $x$ and $y$ directions respectively. In order for the directional derivative to reduce to the usual partial derivatives when $\vec{u} = \hat{i}$ or $\hat{j}$, we require that $\vec{u}$ be a unit vector. As for why $df/ds$ is the same as the directional derivative, when you slice the graph of your function along a plane, you get a curve lying in that plane. The slope of that curve is the directional derivative, which coincides with the ordinary derivative of the function along the parametrized curve $\vec{r}(s)=\langle x_0 + as, y_0+bs\rangle$.

If we allowed $\vec{u}$ to have a magnitude, we would be scaling our rate of change by another quantity. A good example is the following: Let's say you're walking over a hot surface with temperature $T(x,y)$. One could ask how quickly the temperature you experience is changing with time (i.e. a derivative with respect to time), or the rate of change of temperature you experience per unit distance you've walked (i.e. a derivative with respect to space). The directional derivative is meant to capture the latter, since if $x$ and $y$ have the same units, the directional derivative will have the same units as $f_x$ and $f_y$.

To determine time rate of change, suppose you're walking with position $\vec{r}(t)$ and velocity $d\vec{r}/dt$. By the chain rule, we arrive at $$\frac{d}{dt}f(\vec{r}(t)) = \left.\nabla f\right|_{\vec{r}(t)} \cdot\frac{d\vec{r}}{dt}$$ Which is almost a directional derivative, except now $d\vec{r}/dt$ may not be a unit vector. In fact, the time rate of change of temperature is precisely the directional derivative multiplied by your speed.