# The integration of the norm of the derivative of 2 equivalent paths are equal.

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?

## 1 Answer

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.

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