curvesgeometryintuition
We call a curve admitting a parameterization $t\to z(t)$, $t\in[0,1]$ differentiable if the vector function $z$ is differentiable. We call the curve smooth if it is differentiable and its derivative is non-zero.
What are examples of differentiable curves that are not smooth? What are the geometrical differences between the two types? Are there curves, which are not smooth at a point and still have a unique tangent at that point? Which are some differentiable curves do not have a tangent.
This is in response to tipshoni’s request above.
Let us first introduce some conventions and definitions.
For each $ n \in \mathbb{N} $, we define $ [n] \stackrel{\text{df}}{=} \{ i \in \mathbb{N}_{0} \mid 0 \leq i \leq n - 1 \} $.
Fix a metric space $ (X,d) $.
Let $ \gamma: [0,1] \to X $ be a (continuous) curve in $ X $.
A partition of $ [0,1] $ is defined as a finite subset of $ [0,1] $ containing $ 0 $ and $ 1 $.
We denote the set of all partitions of $ [0,1] $ by $ \mathbf{P} $. Observe that $ (\mathbf{P},\subseteq) $ is a directed set.
If $ \mathcal{P} \in \mathbf{P} $ has cardinality $ n $ and $ t^{\mathcal{P}}: [n] \to \mathcal{P} $ denotes the order-preserving enumeration of $ \mathcal{P} $, then we define $ {L_{\gamma}}(\mathcal{P}) \in [0,\infty] $ by $$ {L_{\gamma}}(\mathcal{P}) \stackrel{\text{df}}{=} \sum_{i \in [n]} d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))). $$
If $ \mathcal{P},\mathcal{Q} \in \mathbf{P} $ and $ \mathcal{P} \subseteq \mathcal{Q} $, then by the Triangle Inequality applied to the metric $ d $, we obtain $$ {L_{\gamma}}(\mathcal{P}) \leq {L_{\gamma}}(\mathcal{Q}). $$
Finally, we say that $ \gamma $ is rectifiable if and only if $$ L_{\gamma} \stackrel{\text{df}}{=} \sup(\{ {L_{\gamma}}(\mathcal{P}) \in [0,\infty] \mid \mathcal{P} \in \mathbf{P} \}) < \infty. $$
Let $ \gamma $ be a rectifiable curve and $ F: X \to \mathbb{R} $ a bounded function (not assumed to be continuous yet).
For each $ \mathcal{P} \in \mathbf{P} $, define \begin{align} \mathcal{L}(\gamma,\mathcal{P};F) & \stackrel{\text{df}}{=} \sum_{i \in [\mathsf{card}(\mathcal{P})]} \left[ \inf_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))); \\ \mathcal{U}(\gamma,\mathcal{P};F) & \stackrel{\text{df}}{=} \sum_{i \in [\mathsf{card}(\mathcal{P})]} \left[ \sup_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))). \end{align}
Define also \begin{align} \mathcal{L}(\gamma;F) & \stackrel{\text{df}}{=} \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F), \quad \text{if the limit exists in $ \mathbb{R} $}; \\ \mathcal{U}(\gamma;F) & \stackrel{\text{df}}{=} \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{U}(\gamma,\mathcal{P};F), \quad \text{if the limit exists in $ \mathbb{R} $}. \end{align}
Here, the limits are to be taken in the sense of a net. For example, if $ A \in \mathbb{R} $, then we write $$ A = \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) $$ if and only if for every $ \epsilon > 0 $, there exists a $ \mathcal{P}_{0} \in \mathbf{P} $ such that for all $ \mathcal{P} \in \mathbf{P} $ with $ \mathcal{P}_{0} \subseteq \mathcal{P} $, we have $$ |A - \mathcal{L}(\gamma,\mathcal{P};F)| < \epsilon. $$
If $ \mathcal{L}(\gamma;F) $ and $ \mathcal{U}(\gamma;F) $ both exist and are equal, then we define the line integral of $ F $ along $ \gamma $ by $$ \int_{\gamma} F \stackrel{\text{df}}{=} \mathcal{L}(\gamma;F) = \mathcal{U}(\gamma;F). $$
If $ X = \mathbb{R}^{2} $ and $ \gamma $ is a Jordan curve, then we usually denote the line integral of $ F $ along $ \gamma $ by $ \displaystyle \oint_{\gamma} F $, if it exists.
Now, assume that $ \gamma $ is a rectifiable curve as before and that $ F $ is continuous.
Claim: The line integral of $ F $ along $ \gamma $ exists.
Proof of Claim
As $ F \circ \gamma $ is a continuous function defined on the closed and bounded interval $ [0,1] $, its range lies in the interval $ [- M,M] $ for some finite $ M > 0 $. Then we clearly have $$ \forall \mathcal{P} \in \mathbf{P}: \quad |\mathcal{L}(\gamma,\mathcal{P};F)| \leq \sum_{i \in [\mathsf{card}(\mathcal{P})]} M \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))) \leq M \cdot L_{\gamma}. $$
Furthermore, observe that if $ \mathcal{P},\mathcal{Q} \in \mathbf{P} $ and $ \mathcal{P} \subseteq \mathcal{Q} $, then by the Triangle Inequality, we get $$ \mathcal{L}(\gamma,\mathcal{P};F) \leq \mathcal{L}(\gamma,\mathcal{Q};F). $$ Hence, the net $ \{ \mathbf{P} \to \mathbb{R}; \mathcal{P} \mapsto \mathcal{L}(\gamma,\mathcal{P};F) \} $ is bounded and monotone, which implies that $$ \mathcal{L}(\gamma;F) = \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) = \sup_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) \in \mathbb{R}. $$
Next, we will show that $$ \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{U}(\gamma,\mathcal{P};F) = \mathcal{L}(\gamma;F). $$
Let $ \epsilon > 0 $. As $ F \circ \gamma $ is uniformly continuous on $ [0,1] $, we can find a $ \mathcal{P}_{0} \in \mathbf{P} $ sufficiently fine so that for all $ \mathcal{P} \in \mathbf{P} $ with $ \mathcal{P}_{0} \subseteq \mathcal{P} $, we have $$ \mathcal{L}(\gamma;F) - \epsilon < \mathcal{L}(\gamma,\mathcal{P};F) \leq \mathcal{L}(\gamma;F) \qquad (\clubsuit) $$ and $$ \forall i \in [\mathsf{card}(\mathcal{P})]: \quad \left[ \sup_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] - \left[ \inf_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] < \frac{\epsilon}{L_{\gamma} + 1}. $$ Let $ \mathcal{P} \in \mathbf{P} $ satisfy $ \mathcal{P}_{0} \subseteq \mathcal{P} $. It follows immediately from the preceding inequality that $$ 0 \leq \mathcal{U}(\gamma,\mathcal{P};F) - \mathcal{L}(\gamma,\mathcal{P};F) \leq \frac{\epsilon}{L_{\gamma} + 1} \cdot L_{\gamma} < \epsilon. \qquad (\spadesuit) $$ Adding the inequalities $ (\clubsuit) $ and $ (\spadesuit) $ yields $$ \mathcal{L}(\gamma;F) - \epsilon < \mathcal{U}(\gamma,\mathcal{P};F) < \mathcal{L}(\gamma;F) + \epsilon, $$ or equivalently, $$ |\mathcal{U}(\gamma,\mathcal{P};F) - \mathcal{L}(\gamma;F)| < \epsilon. $$ Therefore, as $ \epsilon $ is arbitrary, we obtain $ \mathcal{U}(\gamma;F) = \mathcal{L}(\gamma;F) $. This concludes the proof. $ \quad \blacksquare $
Concluding remarks: As it is not required for $ X $ to be equipped with a smooth structure, it is not necessary to invoke the notion of ‘smoothness’ in order to discuss line integrals. However, in the context of $ X = \mathbb{R}^{2} $, where a canonical smooth structure exists, it would certainly be profitable to restrict one’s attention to the class of positively oriented and piecewise-smooth Jordan curves so as to exploit powerful results like Green’s Theorem.
Best Answer
the smoothness that you mean is what we call geometric continuity of a curve , in other words a curve is sad to be geometric continuous if it has at each point a tangent vector and varie continuously along the curve. I m interseting at such field of geometry, i will be happy if you give me any open question in .