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


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 $l(\gamma) = \sqrt{2}(e^{b}-e^{a})$.

  • $\begingroup$ 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? $\endgroup$ – Whats My Name Feb 18 '14 at 17:05
  • $\begingroup$ Why would there be a $\sqrt{2}$ in front??? You can see that if you let $a \to -\infty$ then the length tends towards $e^b$. $\endgroup$ – copper.hat Feb 18 '14 at 17:08
  • $\begingroup$ Isn't the norm defined a square root of the squares of the individual components of the derivative in this case $(2e^{2t})^{\frac{1}{2}$ $\endgroup$ – Whats My Name Feb 18 '14 at 17:12
  • $\begingroup$ @WhatsMyName: Oops, I made a mistake. Thanks for catching that. $\endgroup$ – copper.hat Feb 18 '14 at 17:14
  • 1
    $\begingroup$ 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 $C^1$. $\endgroup$ – copper.hat Feb 18 '14 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.