I have just learned about a proof for why the gradient "shows the direction of steepest ascent" involving:

$\partial_{v_0} f(x_0) = (f \circ \gamma)^{\prime}(t_0) = \langle\nabla f(x_0), v_0\rangle$

with: $\gamma (t_0) = x_o, \gamma^{\prime}(t_0)=v_0$

which I now understand. A little later in the script, it is presented that:

$f(x_1) - f(x_0) = \int^b_a \langle \nabla f (\gamma(t)), \gamma^{\prime}(t)\rangle dt$.

for every $C^1$-curve $\gamma:[a,b] \rightarrow U \subset \mathbb{R}^n$ with $\gamma(a) = x_0, \gamma(b) = x_1$, which is without proof or definition. Does anybody have an intuition or explanation as to why this holds true or maybe just a link to where I can read more?

  • $\begingroup$ The first equation tells you what you want to know, if you recall that the directional derivative in direction $v_0$ will therefore be greatest when $v_0$ is in the direction of $\nabla f(x_0)$. (Recall that $u\cdot v\le |u||v|$, with equality holding when $u$ and $v$ point in the same direction.) $\endgroup$ Sep 22 '19 at 1:13
  • $\begingroup$ Oh, I remember, thanks. $\endgroup$
    – psyph
    Sep 22 '19 at 2:59
  • $\begingroup$ @TedShifrin But whats not so obvious to me, is why $f(x_1)-f(x_0) = \int^1_0 df(x_0 + t(x_1-x_0)dt \cdot (x_1-x_0)$ when $[x_0,x_1] = \{ x_0 + t(x_1-x_0): 0 \leq t \leq 1\} \in U \subset \mathbb{R}^n$ would hold true. Specifically, I don't understand why we multiply with $(x_1-x_0)$ in the end. Could you help me with that? $\endgroup$
    – psyph
    Sep 22 '19 at 3:04
  • $\begingroup$ Well, as the answer below says, you're just integrating $(f\circ\gamma)'(t)$ from $0$ to $1$ and using the chain rule to break that derivative down (and then the FTC, as I said below). You can also think of this as a line integral of $\nabla f$ along the path from $x_0$ to $x_1$ (which might be a line segment or might be a more general path). Then it's the FTC for line integrals. (By the way, if you're interested in proofs, you might find my YouTube videos for multivariable calculus/analysis and linear algebra interesting or helpful. They're linked in my profile.) $\endgroup$ Sep 22 '19 at 3:41

By the chain rule, the derivative of $t \mapsto f(\gamma(t))$ is $$f^\prime(\gamma(t)) \circ \gamma^\prime(t)=\langle \nabla f (\gamma(t)), \gamma^{\prime}(t)\rangle$$

Which leads to the required result applying the Funtamental theorem of calculus.

  • $\begingroup$ With an application of the Fundamental Theorem of Calculus at the end. $\endgroup$ Sep 22 '19 at 1:11
  • $\begingroup$ @TedShifrin Indeed. I’ll add this precision in the answer. $\endgroup$ Sep 22 '19 at 3:44

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