# Directional Derivative Interpretation

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 $$(x_0,y_0)$$

$$\implies \nabla f= $$

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 $$\overrightarrow u = $$ 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

• Think of hiking on a mountainside. The directional derivative in a direction at a point is the rate of change of altitude when you move in that direction from that point. The algebraic formula follows from that understanding. If you have a contour map you can read that off as the rate at which you cross contour lines. Commented Nov 20, 2022 at 18:31

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.

• So directional derivative is the rate of change of a function along a certain direction given by the unit vector So $\frac{d f}{d s}$ can be interpreted as the directional derivative except Commented Nov 21, 2022 at 1:33