9.9-bo'lim: Qutb koordinatalari bilan yoy uzunligi
Endi biz integrallarning Calculus II ilovalariga va ularni qutb koordinatalari nuqtai nazaridan qanday bajarishimizga o'tishimiz kerak. Ushbu bo'limda biz egri chiziqning yoy uzunligini ko'rib chiqamiz:
\[r = f\left( \theta \right)\hspace{0.5in}\alpha \le \theta \le \beta \]bu yerda egri chiziq aynan bir marta chizilgan deb ham faraz qilamiz. Xuddi biz bilan qilganimizdek qutb koordinatalaridagi tangens chiziqlar birinchi navbatda parametrik tenglamalar to'plamida egri chiziqni yozamiz,
\[\begin{align*}x & = r\cos \theta & \hspace{0.75in} y & = r\sin \theta \\ & = f\left( \theta \right)\cos \theta & \hspace{0.75in} & = f\left( \theta \right)\sin \theta \end{align*}\]va endi yoy uzunligini topish uchun parametrik formuladan foydalanishimiz mumkin.
Ushbu hisob-kitoblar uchun bizga quyidagi hosilalar kerak bo'ladi.
\[\begin{align*}\frac{{dx}}{{d\theta }} & = f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta & \hspace{0.75in}\frac{{dy}}{{d\theta }} & = f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta \\ & = \frac{{dr}}{{d\theta }}\cos \theta - r\sin \theta & \hspace{0.75in}& = \frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta \end{align*}\]\(ds\) uchun bizga quyidagilar kerak bo'ladi.
\[\begin{align*}{\left( {\frac{{dx}}{{d\theta }}} \right)^2} + {\left( {\frac{{dy}}{{d\theta }}} \right)^2} & = {\left( {\frac{{dr}}{{d\theta }}\cos \theta - r\sin \theta } \right)^2} + {\left( {\frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta } \right)^2}\\ & = {\left( {\frac{{dr}}{{d\theta }}} \right)^2}{\cos ^2}\theta - 2r\frac{{dr}}{{d\theta }}\cos \theta \sin \theta + {r^2}{\sin ^2}\theta \\ & \hspace{0.75in} + {\left( {\frac{{dr}}{{d\theta }}} \right)^2}{\sin ^2}\theta + 2r\frac{{dr}}{{d\theta }}\cos \theta \sin \theta + {r^2}{\cos ^2}\theta \\ & = {\left( {\frac{{dr}}{{d\theta }}} \right)^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\\ & = {r^2} + {\left( {\frac{{dr}}{{d\theta }}} \right)^2}\end{align*}\]Qutb koordinatalari uchun yoy uzunligi formulasi quyidagicha:
Qutb koordinatalari uchun yoy uzunligi
qayerda,
\[ds = \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} \,d\theta \]Keling, buning tezkor misolini ko'rib chiqaylik.
Yechim
Mayli, keling, to'g'ridan-to'g'ri formulaga o'taylik, chunki bu juda oddiy funktsiya.
\[L = \int_{{\,0}}^{{\,1}}{{\sqrt {{\theta ^2} + 1} \,d\theta }}\]Bu yerda trigni almashtirishdan foydalanishimiz kerak.
\[\theta = \tan x\hspace{0.5in}d\theta = {\sec ^2}x\,dx\] \[\begin{align*}\theta & = 0\hspace{0.5in}0 = \tan x\hspace{0.5in}x = 0\\ \theta & = 1\hspace{0.5in}1 = \tan x\hspace{0.5in}x = \frac{\pi }{4}\end{align*}\] \[\sqrt {{\theta ^2} + 1} = \sqrt {{{\tan }^2}x + 1} = \sqrt {{{\sec }^2}x} = \left| {\sec x} \right| = \sec x\]Keyin yoy uzunligi,
\[\begin{align*}L &= \int_{{\,0}}^{{\,1}}{{\sqrt {{\theta ^2} + 1} \,d\theta }}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{4}}}{{{{\sec }^3}x\,dx}}\\ & = \left. {\frac{1}{2}\left( {\sec x\tan x + \ln \left| {\sec x + \tan x} \right|} \right)} \right|_0^{\frac{\pi }{4}}\\ & = \frac{1}{2}\left( {\sqrt 2 + \ln \left( {1 + \sqrt 2 } \right)} \right)\end{align*}\]Biz ushbu bobni tark etishdan oldin bir chetga surib qo'ydik. \(r = \theta \) qutb tenglamasi spiral tenglamasidir. Mana \(0 \le \theta \le 4\pi \) uchun \(r = \theta \) ning qisqacha eskizi.
