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I'm reading up about the more general definition of directional derivatives, but still in the context of $\mathbb{R}^n$. It goes like this:

Let $f:\mathbb{R}^n\to\mathbb{R}$ be a real-valued function on the manifold $\mathbb{R}^n$ and let $v_p$ be a vector tangent to manifold $\mathbb{R}^n$ at point $p$, that is, $v_p\in T_p(\mathbb{R}^n)$. The number $$v_p[f]\equiv\frac{d}{dt}\big(f(p+tv_p\big)\bigg|_{t=0}$$ is called the directional derivative of $f$ w.r.t. $v_p$, if it exists.

Immediately after that there's an exercise:

Let $v=[2,-1,3]^T$ and $p=(2,0,-1)$. Find $v_p[f]$ where:

a) $f(x)=x$

b) $f(x)=x^2-x$

c) $f(x)=\cos(x)$

I'm a bit confused. Isn't the $v_p$ operator supposed to act on a $\mathbb{R}^3\to\mathbb{R}$ function in this case? I was under the impression that if we have an $n$-dimensional manifold and a "well-behaved" point $p$ in it such that the tangent space $T_p(M)$ is also $n$-dimensional, then the operator $v_p$ acts on $\mathbb{R}^n\to\mathbb{R}$ functions. The above seem like $\mathbb{R}\to\mathbb{R}$. Am I missing something?

Community
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Shirish Kulhari
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  • Could you give the name of the book or source you got these questions from? – Andre of Astora Oct 22 '19 at 19:18
  • No, you're entirely correct. At least they should have written $f(x,y,z)= \dots$, but I think these are stupid exercises. – Ted Shifrin Oct 22 '19 at 19:57
  • @TedShifrin: Hm, I guess. The book seems pretty awesome at explaining concepts but there are typos and bizarre mistakes like this from time to time. Thanks so much for clarifying! – Shirish Kulhari Oct 22 '19 at 20:17
  • @VishnuM It is from the book "A Visual Introduction to Differential Forms and Calculus on Manifolds" from Jon Pierre Fortney. It happens to be a good "good night reading". – Cornman May 11 '20 at 03:53
  • @ShirishKulhari I just ordered the book myself. At first I was a little disappointed, but it seems to be a good reading. Then I came to this point, and was about to ask the exact same question... – Cornman May 11 '20 at 03:56
  • @Cornman: From the perspective of a beginner like me, I like the book! I guess your disappointment might be because it's not the really rigorous, graduate text kind of book? I found it good for self-study or at least developing initial concepts. – Shirish Kulhari May 11 '20 at 04:01
  • @ShirishKulhari Yes, exactly. When I bought the book, I wanted something more mathematical. But I think I can benefit from this book too, as it gives more emphasis on simple calculation, which you normally do not do. But I feel like that this helps in understanding the concepts. – Cornman May 11 '20 at 04:03

1 Answers1

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What you have for the function $f(p+tv_p)$ is a composition $$\mathbb R\stackrel C\to\mathbb R^3\stackrel f\to\mathbb R,$$ where $C(t)=p+tv_p$ is the parametrization of a line which passes $p$ and goes the direction $v_p$. Then $f(p+tv_p)$ is the map $$\mathbb R\stackrel{f\circ\ C}\longrightarrow\mathbb R.$$ So by the Chain's Rule $$\frac{d(f\circ C)}{dt}\bigg|_t={\rm grad}f|_{C(t)}\cdot C'|_t,$$ which at $t=0$ gives you $$\frac{d(f\circ C)}{dt}\bigg|_{t=0}={\rm grad}f|_p\cdot v_p.$$

janmarqz
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  • Now by employing a) $f(x,y,z)=x$, b) $f(x,y,z)=x^2-x$,...etc, as Ted's commented, the calculation is simple – janmarqz Oct 23 '19 at 00:48