Rectifiable functions A continuous function $\alpha: [a,b] \to \mathbb{R}^k$ is called a curve. For each partition $P = \{t_0<t_1<....<t_n=b\}$, define $l(\alpha, P) = \sum_{i=1}^n \left|\alpha(t_i) - \alpha(t_{i-1})\right|$.
Let $l(\alpha) = \sup\{l(\alpha, P): P \text{ partitions } [a,b]\}$
If $l(\alpha) < \infty$, then $\alpha$ is called rectifiable.
a) Let $c < a < b < d.$ Suppose $\alpha'$ is continuous on $(c,d)$ for some curve $\alpha: [c,d] \to \mathbb{R}^k$. Show that the restriction $\alpha|_{[a,b]}$ of $\alpha$ to $[a,b]$ is rectifiable with $l(\alpha|_{[a,b]}) = \int_a^b \left|\alpha'(t)\right|\, dt$.
b) Let $c < a < b < d.$ Suppose $f: (c,d) \to \mathbb{R}$ is continuously differentiable. Let $\alpha: [a,b] \to \mathbb{R}^2$ be given by $\alpha(t) = (t,f(t))$. Find $l(\alpha)$ in terms of $f$.
Here's what I have so far. 
a) we can use the Cauchy-Schwarz inequality, unsure how to implement it, and not sure what else we can invoke
b) If $f$ is continuously differentiable, can I talk about uniform convergence, will it help? I am stuck on this one. 
 A: I'm expanding my original answer; so its no longer a hint anymore.
The inequality $l(\alpha)\leq\int_a^b|\alpha'(t)|dt\ $ is easy: One has $$|\alpha(t_i)-\alpha(t_{i-1})| = \Bigl |\int_{t_{i-1}}^{t_i} \alpha'(t)dt \Bigr| \leq \int_{t_{i-1}}^{t_i} |\alpha'(t)|dt \ ,$$
and by summation we obtain
$$l(\alpha,P)\leq \int_a^b |\alpha'(t)|dt\ .$$ 
Since this is true for all partitions $P$ the inequality follows.
For the other direction a trick is required. Let an $\epsilon>0$ be given and assume that the partition $P$ is fine enough to guarantee $|\alpha'(t)-\alpha'(t')|<\epsilon$ within each subinterval. Then for each $i$ one has
$$\int_{t_{i-1}}^{t_i}|\alpha'(t)|dt=|\alpha'(\tau)|(t_i-t_{i-1}) =\Bigl|\int_{t_{i-1}}^{t_i}\alpha'(\tau) dt\Bigr| \leq \Bigl|\int_{t_{i-1}}^{t_i}\alpha'(t) dt\Bigr| +\epsilon(t_i-t_{i-1})\ ,$$
where $\tau$ is a certain ${\it fixed}$ point in the interval $[t_{i-1},t_i]$. Summing over $i$ we get
$$\int_a^b|\alpha'(t)|dt\leq l(\alpha, P)+\epsilon(b-a)\leq l(\alpha)+\epsilon (b-a)\ .$$
As $\epsilon>0$ was arbitrary we conclude $\int_a^b|\alpha'(t)|dt\leq l(\alpha)\ .$
