calculus 2 - exercices of true and false Can someone explain me why the first two questions are true and the last two  are wrong?
1

Consider a vector function $r: [0, 3] \to \mathbb R ^ 2$ regular. Plot by $C$ the curve defined by $r$. Consider the vector function $\sigma: [0, 3] \to \mathbb R ^ 2$ defined by $\sigma (\tau) = r (3 - \tau)$. The curve defined by $\sigma$ is the inverse curve of curve $C$.
  (True)

2

Let $C$ be a regular parametric curve and $r (s)$, $s \in [0, L]$, $L> 0$, the parameterization by arc length. Then $\|r'(s)\| = 1$, $\forall s \in ] 0, L [$
  (True) 

In my resolution of this exercise, I only arrived to: $\| r '(s) \| = s'(s)$
3

The whole level curve of a real function of two variables can be written as the graph of a real function of a real variable
  (False)

4

$(x, y) = (t, t), t \in [0, 1]$ is the parametrization by arc length of the straight line segment $(0, 0)$ and end point $(1, 1)$.
  (False)

 A: *

*The definition of "inverse curve" in Wikipedia and in Wolfram MathWorld is to invert the image of the curve in a circle. Under that definition, your first statement is false. Perhaps the intended statement was about a "reverse curve," where the image is the same as the first curve but the curve is traversed in reverse order. Under the conditions given, $r$ and $\sigma$ are reverse curves, since for $\tau\in [0,3]$ we have $3-\tau\in [0,3]$ and $\sigma(\tau)=r(3-\tau)$ so the curves have the same image. And since $r(0)=\sigma(3)$ and $r(3)=\sigma(0)$, the initial and terminal points of the two curves are swapped.

*The definition of $s$, the arclength parameter, is, for a specified $t_0$,
$$s(t)=\int_{t_0}^t \|\mathbf r'(\tau)\|\,d\tau$$
Thus
$$\|\mathbf r'(s)\|
 =\left\|\frac{d\mathbf r}{ds}\right\|
 =\left\|\frac{d\mathbf r/dt}{ds/dt}\right\|
 =\left\|\frac{\mathbf r'(t)}{\|\mathbf r'(t)\|}\right\|
 =\frac{\|\mathbf r'(t)\|}{\|\mathbf r'(t)\|}
 =1
$$


*A counterexample to this statement is $f(x,y)=x^2+y^2$. Then the level curve for $1$ is $x^2+y^2=1$ which is a circle. This fails the vertical-line test and thus cannot be the graph of a function $y$ of $x$.


An even more drastic example is $f(x,y)=1$. Then the level curve for the value $1$ is the entire plane and all other level curves are empty. None of these can be the graph of a function $y$ of $x$.


*The derivative of the given curve $(t,t)$ is $(1,1)$, which has the magnitude $\sqrt 2$. By statement 2 above, for arc length parameterization this would have been the value $1$.

