Question Number 11413 by agni5 last updated on 24/Mar/17 | ||
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbolic} \\ $$$$\mathrm{spiral}\:\:\mathrm{r}\theta=\mathrm{a}\:\:\mathrm{lying}\:\mathrm{between}\:\:\mathrm{r}=\mathrm{a}\:\:\mathrm{and}\: \\ $$$$\mathrm{r}=\mathrm{2a}. \\ $$ | ||
Answered by mrW1 last updated on 26/Mar/17 | ||
$${r}=\frac{{a}}{\theta} \\ $$$$\frac{{dr}}{{d}\theta}=−\frac{{a}}{\theta^{\mathrm{2}} } \\ $$$$\sqrt{{r}^{\mathrm{2}} +\left(\frac{{dr}}{{d}\theta}\right)^{\mathrm{2}} }=\frac{{a}\sqrt{\mathrm{1}+\theta^{\mathrm{2}} }}{\theta^{\mathrm{2}} } \\ $$$${L}=\int_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \sqrt{{r}^{\mathrm{2}} +\left(\frac{{dr}}{{d}\theta}\right)^{\mathrm{2}} }{d}\theta={a}\int_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \frac{\sqrt{\mathrm{1}+\theta^{\mathrm{2}} }}{\theta^{\mathrm{2}} }{d}\theta \\ $$$$={a}\left[−\frac{\sqrt{\mathrm{1}+\theta^{\mathrm{2}} }}{\theta}+\mathrm{ln}\:\left(\theta+\sqrt{\mathrm{1}+\theta^{\mathrm{2}} }\right)\right]_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \\ $$$$={a}\left[\frac{\sqrt{\mathrm{1}+\theta_{\mathrm{1}} ^{\mathrm{2}} }}{\theta_{\mathrm{1}} }−\frac{\sqrt{\mathrm{1}+\theta_{\mathrm{2}} ^{\mathrm{2}} }}{\theta_{\mathrm{2}} }+\mathrm{ln}\:\frac{\theta_{\mathrm{2}} +\sqrt{\mathrm{1}+\theta_{\mathrm{2}} ^{\mathrm{2}} }}{\theta_{\mathrm{1}} +\sqrt{\mathrm{1}+\theta_{\mathrm{1}} ^{\mathrm{2}} }}\right] \\ $$$$ \\ $$$${with}\:\theta_{\mathrm{1}} =\frac{{a}}{{r}_{\mathrm{1}} }=\frac{{a}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\theta_{\mathrm{2}} =\frac{{a}}{{r}_{\mathrm{2}} }=\frac{{a}}{{a}}=\mathrm{1} \\ $$$${L}={a}\left[\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}}{\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\sqrt{\mathrm{1}+\mathrm{1}}}{\mathrm{1}}+\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{1}}}{\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}}\right] \\ $$$${L}={a}\left[\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}+\mathrm{ln}\:\frac{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{1}+\sqrt{\mathrm{5}}}\right] \\ $$ | ||
Commented by mrW1 last updated on 26/Mar/17 | ||
$${the}\:{answer}\:{is}\:{corrected}. \\ $$$${please}\:{see}\:{also}\:{Q}\mathrm{11433}. \\ $$ | ||