I just have a question about directional derivatives.

I know that for a unit vector $v$, it is $\nabla f \cdot v$ (where $v$ is the direction we want to find the derivative of).

The proof of this using limits makes sense, but how would you look at this geometrically?

The gradient vector is the direction in which the function increases the most, but then having a dot product of this with a unit vector apparently gives the derivative going in that direction.

I'm not quite sure how to interpret the dot product geometrically.

Thank you!

  • $\begingroup$ Thanks, I'm new and in a bit of a rush so I have no idea how to put in the fancy looking letters haha $\endgroup$
    – 14tim4
    Dec 20, 2016 at 23:08
  • 1
    $\begingroup$ You are looking for this $\endgroup$
    – IAmNoOne
    Dec 20, 2016 at 23:13
  • $\begingroup$ Makes sense, but I'm a little unsatisfied. I guess my question is really about the geometry of a dot product. What is the significance of the number resulting from a.b? $\endgroup$
    – 14tim4
    Dec 20, 2016 at 23:24
  • $\begingroup$ @14tim4: Is your question about the directional derivative or the dot product? They are separate things which potentially deserve different explanations. For the dot product: $\mathbf{a}\cdot\mathbf{b}$ is, in some sense, "the amount of $\mathbf{a}$ going in the direction of $\mathbf{b}$" (think of $\mathbf{a}$ and $\mathbf{b}$ as arrows). It's the length of the projection of $\mathbf{a}$ onto the line spanned by $\mathbf{b}.$ [Since the dot product is commutative, we can actually also swap $\mathbf{a}$ and $\mathbf{b}$ in the above description.] $\endgroup$
    – Will R
    Dec 20, 2016 at 23:27
  • $\begingroup$ If it helps, $\nabla f$ is not really a geometric object; it's more akin to a function, and obtaining the directional derivative $\nabla_v f$ from it is like evaluating a function. The idea of the "gradient vector" that you described is the application of a trick: for any linear functional $\omega$, there exists some vector $w$ such that you can compute the values of the linear functional by $\omega(v) = v \cdot w$. It is traditional to use this trick to convert everything into 'geometry', but it really does obscure what's going on. $\endgroup$
    – user14972
    Dec 21, 2016 at 0:19

2 Answers 2


You appear to be asking two questions, one about the directional derivative, the other about the dot product. Since your question appears to be mostly about the directional derivative, I will give an answer explaining the geometric meaning of the directional derivative. For ease of visualization I restrict attention to functions $\mathbb{R}^{2}\to\mathbb{R},$ but $\mathbb{R}^{2}$ can in principle be replaced by $\mathbb{R}^{n}.$

As always, a picture is worth a thousand words: if you can glean the meaning of the directional derivative from this .gif I found online then you may not need to read what I have to say. I hope the animation helps either way.

A gif I found online.

It's worth noting that the little green segment at the base of the figure in this animation is meant to represent the value of the gradient of the function at the given point.

Let $\mathbf{v}\in\mathbb{R}^{2}$ be a unit vector, considered as a vector in the $(x,y)$-plane in $\mathbb{R}^{3}.$ The vector $\mathbf{v}$ determines a unique plane $\Pi = \Pi(\mathbf{v})$ which contains the origin, the point $(0,0,1)$ and $\mathbf{v}$ itself. For example, if $\mathbf{v}=\begin{pmatrix}1\\0\end{pmatrix}$ then the plane $\Pi$ is the $y=0$ plane, i.e., the $(x,z)$-plane. (Draw a sketch to make sure you are comfortable with this idea; if you think I have explained it poorly, please ask.)

We want to consider this same plane, but "going through $\mathbf{x}$", that is, we want to consider the plane $\mathbf{x}+\Pi = \{\mathbf{x} + \mathbf{p}:\mathbf{p}\in\Pi\}.$ All we've done is taken the plane $\Pi,$ which went through the origin, and physically moved it so that the point on $\Pi$ which was previously at $(0,0,0)$ is now at $\mathbf{x}.$

Consider the directional derivative of $f\colon\mathbb{R}^{2}\to\mathbb{R}$ at $\mathbf{x}\in\mathbb{R}^{2}$ in the direction of $\mathbf{v}.$ Its value is equal to the value of the limit: $$\lim_{h\to0}\frac{f(\mathbf{x}+h\mathbf{v})-f(\mathbf{x})}{h}.$$ This can be taken as the definition of the directional derivative; in particular, the above expression turns out to be equal to $\nabla{f}\cdot\mathbf{v}.$ [Note: This equality holds only because $\mathbf{v}$ is a unit vector. If $\mathbf{v}$ is not a unit vector then we need to introduce a scaling factor to account for this.]

Consider the graph of $f,$ that is, the set $$\Gamma = \Gamma(f) = \{(x,y,z)\in\mathbb{R}^{3}: z=f(x,y)\}.$$ This is (more-or-less) a surface in $3$-space. If we look at the points where the graph of $f$ intersects $\mathbf{x}+\Pi,$ then we get a curve in space, which we'll call $\gamma = \gamma(f,\mathbf{v}) = \Gamma\cap(\mathbf{x}+\Pi).$

We can imagine the $z$-axis in $\mathbb{R}^{3},$ which is contained in $\Pi,$ together with the line $\ell = \{k\mathbf{v}:k\in\mathbb{R}\},$ defining a coordinate system on $\Pi$: the vertical axis is given by the $z$-axis, and the horizontal axis is given by $\ell,$ with $0$ at the origin and the values on $\ell$ increasing in the direction of $\mathbf{v}.$

That is, we can imagine the curve $\gamma$ as being the graph of some function $\mathbb{R}\to\mathbb{R},$ except that instead of drawing this graph on the usual $(x,y)$-plane we are drawing it on $\mathbf{x}+\Pi.$

The expression for the directional derivative given as a limit is telling you that the value of the directional derivative (of $f$ at $\mathbf{x}$ in the direction of $\mathbf{v}$) is the slope, under the coordinate system I have described above, of the line which is contained in $\mathbf{x}+\Pi$ and tangent to the curve $\gamma,$ that is, "the slope of the tangent line" in an appropriate plane and coordinate system.

  • $\begingroup$ I found the .gif via the references of these UCL lecture notes. Unfortunately, it doesn't say who made the .gif, so I can't give any more credit than the author of the notes does. $\endgroup$
    – Will R
    Dec 21, 2016 at 0:14
  • $\begingroup$ I wish I could upvote this multiple times because that animation was gold. $\endgroup$ Aug 13, 2021 at 18:48

Say, we have $f:\Bbb R^n\to\Bbb R$, a $v\in\Bbb R^n$ unit vector and a point $x_0\in\Bbb R^n$.

Then look at the graph of $f$ in a one dimension higher space. $$\Gamma_f:=\ \{({\bf x}, f({\bf x}))\,\mid\,{\bf x}\in\Bbb R^n \}\ \subseteq\ \Bbb R^{n+1}$$ Let $e_{n+1}$ be the last element of the standard basis of $\Bbb R^{n+1}$ (that is, $e_{n+1}=(0,0,\dots,0,1)^T$).

The directional derivative $\partial_vf(x_0)\ =\nabla f(x_0)\cdot v$ is the slope of the tangent line of ($\Gamma_f$ intersected with the 2d plane planned by $v$ and $e_{n+1}$).

In other words, take values of $f$ only over the line of direction $v$, which contains $x_0$.


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