Directional derivatives (intuition)

Question

Let $U\subset \mathbb R^n$ be an open subset and $f:U\to \mathbb R$ a differentiable function.

We know the directional derivative is defined as

$$\frac{\partial f}{\partial v}(a)=\lim_{t\to 0}\frac{f(a+tv)-f(a)}{t}$$

Looking at this limit, we can easily see that the directional derivative is the derivative at the point $t=0$ of the composition $f\circ\lambda:(-\epsilon,\epsilon)\to\mathbb R$, where $\lambda:(-\epsilon,\epsilon)\to\mathbb R^n$ is the path $\lambda(t)=a+tv$. (here we choose $\epsilon$ small enough so that $(-\epsilon, \epsilon)\subset U$).

I'm trying to get this intuitively drawing a picture of the paths $\lambda$ and $f\circ\lambda$ without success. I need help to understand the paragraph above visually.

Answer 1

Just as the ordinary derivative in single-variable calculus can be seen as the slope of the tangent to the graph of the function, the directional derivative give the slope of the tangent hyperplane in the direction of $v$, i.e., it’s the rate of change of the function when you ”look” in the direction given by $v$.

It’s probably easiest to visualize for a differentiable function $f:\mathbb R^2\to\mathbb R$. The graph of $z=f(x,y)$ is some surface in $\mathbb R^3$. If we take a vertical slice through this surface by intersecting it with the plane $P(a,v)$ given by $v_y(x-x_a)-v_x(y-y_a)=0$ we get a differentiable planar curve. This plane intersects the $x$-$y$ plane in the line $\lambda:t\mapsto a+tv$, so the intersection curve is your $f\circ\lambda$. The derivative $(f\circ\lambda)'(a)={\partial f\over\partial v}(a)$ is then just the slope of the tangent to this planar curve at $(x_a,y_a,f(x_a,y_a))$. Since $f$ is differentiable at $a$, the surface $z=f(x,y)$ has a well-defined tangent plane at this point. If we intersect the tangent plane with $P(a,v)$, we’ll find that the intersection line is exactly the tangent line to $f\circ\lambda$, so another way to describe the directional derivative is as the slope of the tangent hyperplane in the direction of $v$.

Answer 2

The image of $\lambda$ is a short line segment in $\mathbb{R}^n$ that goes from $a-\epsilon v$ to $a + \epsilon v$ (in particular, it passes through $a$ since it is a line segment).

$f \circ \lambda$ on the other hand is a real valued function on the interval $(-\epsilon , \epsilon)$, and it's value at a point $x$ is exactly the value of $f$ over the respective point in $\lambda(x) \in \mathbb{R}^n$, e.g. $f \circ \lambda (0) = f (\lambda(0)) = f(a)$.

Answer 3

$$ \frac{d}{dt} \Bigr|_{t = t_0} (f \circ c)(t) = \nabla f\Bigr|_{c(t_0)} \cdot c'(t_0) = \frac{d}{dt}\Bigr|_{t = 0} f(c(t_0) + c'(t_0)t)$$

Let $\textbf{p} = c(t_0)$ and $v = c'(t_0)$. Then by the above, $(f \circ c)'(t_0)$ is the initial rate of change in $f$ at $f(\textbf{p})$ in the direction $v$.