8.1-bo'lim : Ark uzunligi
2. \(x = \cos \left( y \right)\) , \(0 \le x \le \frac{1}{2}\) uzunligi uchun integralni sozlang, lekin baholamang,
- \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx\)
- \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy\)
Yechim
Bu holda biz birinchi navbatda \(y\) funksiyasini hal qilishimiz kerak, shunda biz \(ds\) da hosilani hisoblashimiz mumkin.
\[x = \cos \left( y \right)\hspace{0.5in} \to \hspace{0.5in}\,\,\,y = {\cos ^{ - 1}}\left( x \right) = \arccos \left( x \right)\]Teskari tangens uchun qaysi belgidan foydalanish muhim emas, chunki hosila olingandan keyin u “yo'qoladi”.
Bu haqda gapiradigan bo'lsak, bu erda funktsiyaning hosilasi.
\[\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}\]2-qadam
Buni \(ds\) formulasiga kiritish,
\[ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ { - \frac{1}{{\sqrt {1 - {x^2}} }}} \right]}^2}} \,dx = \sqrt {1 + \frac{1}{{1 - {x^2}}}} \,dx = \sqrt {\frac{{2 - {x^2}}}{{1 - {x^2}}}} \,dx\]3-qadam
Endi qilishimiz kerak bo'lgan yagona narsa yoy uzunligi uchun integralni o'rnatishdir. Shuni ham yodda tutingki, bizda \( ds\) formulasida \(dx\) bor va shuning uchun bizga muammo bayonida berilgan \(x\) integratsiya chegaralari kerakligini bilamiz.
\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{0}^{{\frac{1}{2}}}{{\sqrt {\frac{{2 - {x^2}}}{{1 - {x^2}}}} \,dx}}}}\]b \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy\)
Yechim
Bizga birinchi navbatda funktsiyaning hosilasi kerak bo'ladi.
\[\frac{{dx}}{{dy}} = - \sin \left( y \right)\]2-qadam
Buni \(ds\) formulasiga ulash, beradi,
\[ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy = \sqrt {1 + {{\left[ { - \sin \left( y \right)} \right]}^2}} \,dy = \sqrt {1 + {{\sin }^2}\left( y \right)} \,dy\]3-qadam
Keyin, \(ds\)-da \(dy\) borligini unutmang va shuning uchun bizga \(y\) integratsiya chegaralari kerak bo'ladi.
Muammo bayonotida bizga faqat \(x\) chegaralari berilgan. Biroq, qisman (a) olish uchun \(y\) funksiyasini hal qildik,
\[y = {\cos ^{ - 1}}\left( x \right) = \arccos \left( x \right)\]va biz qilishimiz kerak bo'lgan yagona narsa, ularni \(y\) chegaralariga aylantirish uchun bizga berilgan \(x\) chegaralarini ulashdir. Shunday qilib,
\[\begin{align*}x = & 0: \hspace{0.25in}y = {\cos ^{ - 1}}\left( 0 \right) = \arccos \left( 0 \right) = \frac{\pi }{2}\\ x = & \frac{1}{2}: \hspace{0.25in} y = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \arccos \left( {\frac{1}{2}} \right) = \frac{\pi }{3}\end{align*}\]Shunday qilib, mos keladigan \(y\) chegaralari: \(\frac{\pi }{3} \le y \le \frac{\pi }{2}\).
E'tibor bering, biz bu erda teskari kosinus uchun ikkala belgidan ham foydalandik, lekin siz faqat o'zingizga qulay bo'lganidan foydalanishingiz kerak. Shuni ham eslaylikki, teskari kosinus funksiyasining diapazoni,
\[0 \le {\cos ^{ - 1}}\left( x \right) \le \pi \]Shuning uchun \(y\) ning faqat bitta mumkin bo'lgan qiymati mavjud bo'lib, biz \(x\) ning har bir qiymatidan chiqa olamiz.
4-qadam
Nihoyat, biz qilishimiz kerak bo'lgan yagona narsa integralni o'rnatishdir.
\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int\limits_{{\frac{\pi }{3}}}^{{\frac{\pi }{2}}}{{\sqrt {1 + {{\sin }^2}\left( y \right)} \,dy}}}}\]