Let $f: [a,b] \rightarrow \mathbb{R^n} $ and $g: [c,d] \rightarrow \mathbb{R^n} $ be 2 equivalent paths. Prove that $$\int_{a}^{b} ||f '(t)|| dt = \int_{c}^{d} ||g '(t)||dt. $$

The definition of equivalent paths is as follows :

Two paths $f: [a,b] \rightarrow \mathbb{R^n} $ and $g: [c,d] \rightarrow \mathbb{R^n} $ are equivalent if there exist a $C^{1}$ bijection $\phi: [a,b]\rightarrow [c,d]$ such that $\phi'(t) > 0 $ for all $t \in [a,b]$ and $f = g \circ \phi.$

My thoughts:

I can see that we want to prove that the their lengths are equal, but how can I do that, may be I will take this equation $$f = g \circ \phi,$$ and then differentiate both sides, which leads to,

$$f' = g'(\phi) (\phi)',$$

And then take the norm of both sides then take. Then I will integrate both sides, but I do not know what are the limits of my integration ? because the required includes differentiation along different intervals. Also I did not use the given that $\phi ' > 0$ (which is given in the definition of equivalent paths) could anyone help me please in removing all this discrepancies?


Almost, instead we make a substitution of $ t= \phi(u)$ into the integral with $g'$. Since we have that $t = c \Rightarrow u = a$ and $t = d \Rightarrow u = b$,

$$\int_c^d ||g'(t)|| \ \mathrm{d}t $$

$$\Rightarrow \int_a^b || g'(\phi(u))|| \ \mathrm{d}(\phi(u)) = \int_a^b || g'(\phi(u))|| \ \phi'(u) \ \mathrm{d}u = \int_a^b || g'(\phi(u)) \phi'(u)|| \ \mathrm{d}u = \int_a^b || f'(u)|| \ \mathrm{d}u $$

Where the last equality comes from the identity you proved.

  • $\begingroup$ but we should reach to $f'$ as a function of t ..... or I am incorrect? $\endgroup$ – Smart Apr 18 at 13:11

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