Let $\Omega$ be a subset of $\mathbb{R}^2$, a convex region whose boundary $\partial\Omega$ is the image of a 1-periodic function $\gamma: \mathbb{R} \to \mathbb{R}^2$ of class $C^2$. Assume that $|\gamma'(t)| = 1$ and the curvature $\kappa(t) := |\gamma''(t)|$ is positive for every $t$.
- Derive the equality $\kappa(t) = |\gamma''_1(t) \cdot \gamma'_2(t) - \gamma'_1(t) \cdot \gamma''_2(t)|$.
- Justify that the integral from $0$ to $1$ of $\kappa(t) \, dt = 2\pi$.
- For a given (you can assume it's small enough) $r > 0$, determine the area of the region $\Gamma_r = \{\gamma(t) + s \cdot \gamma'(t) : t \in [0, 1], s \in [0, r]\}$.
SOLUTION:
(1)
First, differentiate the condition $|\gamma'(t)|^2 = 1$ with respect to $t$. This gives:
$$ 2\gamma'(t) \cdot \gamma''(t) = 0 $$
This implies that $\gamma'(t)$ and $\gamma''(t)$ are orthogonal. Now, compute the cross product of $\gamma'(t)$ and $\gamma''(t)$ and take its magnitude to obtain $\kappa(t)$:
$$ \kappa(t) = |\gamma''_1(t) \cdot \gamma'_2(t) - \gamma'_1(t) \cdot \gamma''_2(t)| $$
(2)
Consider the derivative of the angle $\alpha(t) := \arctan(\frac{\gamma'_1(t)}{\gamma'_2(t)})$. Using the chain rule, we have:
$$ \frac{d\alpha}{dt} = \frac{\gamma'_2(t)\gamma''_1(t) - \gamma'_1(t)\gamma''_2(t)}{(\gamma'_1(t))^2 + (\gamma'_2(t))^2} $$
Since $|\gamma'(t)|^2 = 1$, the denominator is equal to 1. Thus, we have:
$$ \frac{d\alpha}{dt} = \gamma'_2(t)\gamma''_1(t) - \gamma'_1(t)\gamma''_2(t) $$
Integrate the derivative of $\alpha(t)$ over the interval $[0, 1]$:
$$ \int_0^1 \frac{d\alpha}{dt} dt = \int_0^1 (\gamma'_2(t)\gamma''_1(t) - \gamma'_1(t)\gamma''_2(t)) dt $$
Using the fact that the unit vector $\gamma'(t)$ makes a rotation of $2\pi$ in this interval, we have:
$$ \int_0^1 \kappa(t) dt = 2\pi $$
(3)
Parameterize the region Gamma_r using the given definition of Gamma_r:
$$ \Gamma_r(u, v) = \gamma(u) + r \cdot \nu(\gamma(u)) \cdot v $$
where $u \in [0, 1]$ and $v \in [0, r]$.
Compute the Jacobian determinant of the transformation from the parameter space to the region:
$$ J(u, v) = \det \begin{pmatrix} \frac{\partial \Gamma_{r1}}{\partial u} & \frac{\partial \Gamma_{r1}}{\partial v} \\ \frac{\partial \Gamma_{r2}}{\partial u} & \frac{\partial \Gamma_{r2}}{\partial v} \end{pmatrix} $$
Finally, integrate the Jacobian determinant over the parameter space $[0, 1] \times [0, r]$ to obtain the area of the region:
$$ \text{Area}(\Gamma_r) = \int_0^1 \int_0^r J(u, v) \, dv \, du $$