a valuable question about differential geometry, the curve$\beta(s)=\alpha(s)-rn(s)$ Let $\alpha(s)$,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.$k_{\beta}(s)={k_{\alpha}(s)\over {1+rk_{\alpha}(s))}}$.
(A( ) means the area bounded by the curve,the curvature is $k_{\alpha}$ and $k_{\beta}$)
This question is in my textbook. I have finished the first question, but I have no idea how to solve the rest of the questions. In terms of part(b), I tried to use${1\over 2}\int_{0}^{l} xdy-ydx$,but I couldn't get the result of (b), maybe my method is wrong. Can someone tell me how to solve part(b) and part(c)? 
 A: For (b), you can use your solution part (a) and Fubini's theorem, in the same way that you can derive that the area of a circle is $\pi r^2$ from the fact that the circumference is $2 \pi r$. To be specific, for each $t \in [0, r]$ let $\beta_t = \alpha - tn$, and compute $A(\beta) - A(\alpha)$ as an iterated integral
$$
   A(\beta) - A(\alpha) = \int_0^r \text{length}(\beta_t) dt = \int_0^r (l + 2\pi t) dt,
$$
which gives you your answer. 
For (c), I'd like to first note that $n'(s) = -k_{\alpha}(s) \alpha'(s)$. This follows from the fact that $||\alpha'(s)|| = 1$ and therefore that $n$ is obtained from $\alpha$ via a counter-clockwise $\pi/2$ rotation. It follows easily that
$$
 ||\beta'(s) || = (1 + r k_\alpha(s)) \alpha'(s).
$$
To compute $k_\beta$, we have to begin by parameterizing $\beta$ by arclength. Define $l(s)$ by
$$
  l(s) = \int_0^s ||\beta'(u)|| du,
$$ 
so that $l(s)$ is the length of $\beta$ from $0$ to $s$. We remark that
$$
   \frac{ds}{dl} = \frac{1}{||\beta'(s)||}.
$$
Let $\hat{\beta}$ be defined by $\hat{\beta}(l(s)) = \beta(s)$, so that $\hat{\beta}$ is the reparameterization of $\beta$ by arc-length. Then $\hat{\beta}'(l) = \alpha'(s)$ and
$$
 \hat{\beta}''(l) = \frac{\alpha''(s)}{||\beta'(s)||} = \frac{\alpha''(s)}{1 + rk_\alpha(s)},
$$
so that
$$
 || \hat{\beta}''(l) || = \frac{k_\alpha(s)}{1 + rk_{\alpha}(s)}
$$
as desired.
