- #1
Demon117
- 165
- 1
Homework Statement
(a) Let [itex]\alpha:I=[a,b]→R^2[/itex] be a differentiable curve. Assume the parametrization is arc length. Show that for [itex]s_{1},s_{2}\in I, |\alpha(s_{1})-\alpha(s_{2})|≤|s_{1}-s_{2}|[/itex] holds.
(b) Use the previous part to show that given [itex]\epsilon >0[/itex] there are finitely many two dimensional open discs [itex]B_{\epsilon}(x_{i}), i=1,..,n[/itex] such that [itex]\alpha (I)\subset \cup _{i=1..n}B_{\epsilon}(x_{i})[/itex] and [itex]\sum_{i}Area(B_{\epsilon}(x_{i}))<\epsilon[/itex].
2. The attempt at a solution
(a) For this I made an argument using the mean value theorem for equality. For all [itex]s\in I[/itex], we have [itex]|\alpha '(s)|=1[/itex] since the curve is parametrized by arc length. Then, given some [itex]s_{o}\in (s_{1},s_{2})[/itex] for some [itex]s_{1},s_{2}\in[a,b][/itex] we have by the mean value theorem
[itex]\frac{\alpha(s_{1})-\alpha(s_{2})}{s_{1}-s_{2}}=\alpha'(s_{o})[/itex]
whereby,
[itex]\frac{|\alpha(s_{1})-\alpha(s_{2})|}{|s_{1}-s_{2}|}=1[/itex], so equality holds.
I am unsure how I should show that it has to be less than. I will have to think of this more. What I am really stuck on is how to even attempt part (b). Any suggestions would be very appreciated.