# Does this limit of arc length exist?

We have a parametrized curve $\gamma: \mathbb{R} \rightarrow \mathbb{R^2}$ given by $\gamma (t) = \langle e^t\cos (t), e^t\sin(t)\rangle$. I want to compute the arc-length of this curve on $[a,b]$ in general.

Is this always possible? For example, I am not sure I can do it when $a\rightarrow - \infty$. Thank you

## 1 Answer

The arc length for a $C^1$ curve on $[a,b]$ is given by $l(\gamma) = \int_a^b \|\dot{\gamma}(t)\| dt$. Since $\|\dot{\gamma}(t) \| = \sqrt{2}e^{t}$, we have $l(\gamma) = \sqrt{2}(e^{b}-e^{a})$.

• Probably there is $sqrt(2)$ in front but this is what I got as well. My question is more like what happens when a is very very small? The formula for arclen holds as well? – Whats My Name Feb 18 '14 at 17:05
• Why would there be a $\sqrt{2}$ in front??? You can see that if you let $a \to -\infty$ then the length tends towards $e^b$. – copper.hat Feb 18 '14 at 17:08
• Isn't the norm defined a square root of the squares of the individual components of the derivative in this case $(2e^{2t})^{\frac{1}{2}$ – Whats My Name Feb 18 '14 at 17:12
• @WhatsMyName: Oops, I made a mistake. Thanks for catching that. – copper.hat Feb 18 '14 at 17:14
• Well, generally for a curve, the length is given by $l(\gamma) = \sup \{ \sum_k |\gamma(t_k)-\gamma(t_{k-1}) | a = t_0 < t_1< \cdots < t_{n-1} < t_n = b \}$. If the length is finite, the curve is called rectifiable. A sufficient condition is that the curve be $C^1$. – copper.hat Feb 18 '14 at 17:29