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)?
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.