Let me writh down what I'm trying to prove.
It comes from Do carmo's differential geometry sec 1.7.
Lets $\alpha(s), s \in [0, l]$ be a closed convex plane curve positively oriented.
The curve
$\beta(s)=\alpha(s)-rn(s)$, where $r$ is a positive constant and $n$ is the normal vector, is called a parallel curve to $\alpha$.
Show that
a. Length of $\beta$ = length of $\alpha$+$2 \pi r$
b. $A(\beta)=A(\alpha)+rl+\pi r^2$
c. $k_\beta (s) = \frac{k_\alpha (s)}{1+r k_\alpha (s)} $.
For (a)-(c), A($~~$) denotes the area bounded by the corresponding curve, and $k_\alpha , k_\beta$ sare the curvatures of $\alpha$ and $\beta$, respectively.
And these are solutions to the above problems.
I have no difficulty with part (a), but part (b) bugges me.
The only formula I have been learned from this book so far is
If $\alpha(s)=(x(s), y(s))$ is the parametrized simple closed curve,
then the area bounded by the curve is given $\int xy'-yx' dt$, which can be written as a $\int \alpha \times \alpha' dt$.
And I think that the first approach of the second solution uses this method.
But I can't see how $\int \vert (\alpha - rn) \times (t - rn') \vert ds$ can be the integral $\frac{1}{2} \int \vert \alpha \times \alpha' \vert ds + r \int \vert b \vert ds + r^2 \int k_\alpha ds$ which follows it.
and the first solution and the second approach of the second solution argue in the same reasoning, which is relied on $A(\beta)-A(\alpha)=\int_0^r L(\alpha)+2 \pi t dt$. Why this is so?
Thank you for reading this and answering in advance.
