8.1-bo'lim : Ark uzunligi
1. \(y = \sqrt {x + 2} \) , \(1 \le x \le 7\) 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
Bizga birinchi navbatda funktsiyaning hosilasi kerak bo'ladi.
\[\frac{{dy}}{{dx}} = \frac{1}{2}{\left( {x + 2} \right)^{ - \,\frac{1}{2}}} = \frac{1}{{2{{\left( {x + 2} \right)}^{\,\frac{1}{2}}}}}\]2-qadam
Buni \(\displaystyle ds\) formulasiga kiritish,
\[ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ {\frac{1}{{2{{\left( {x + 2} \right)}^{\,\frac{1}{2}}}}}} \right]}^2}} \,dx = \sqrt {1 + \frac{1}{{4\left( {x + 2} \right)}}} \,dx = \sqrt {\frac{{4x + 9}}{{4\left( {x + 2} \right)}}} \,dx\]3-qadam
Endi qilishimiz kerak bo'lgan yagona narsa yoy uzunligi uchun integralni o'rnatishdir. Shuni ham yodda tutingki, bizda \(\displaystyle ds\) formulasida \(dx\) bor va shuning uchun bizga muammo bayonida berilgan \(x\) integratsiya chegaralari kerakligini bilamiz.
\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{1}^{7}{{\sqrt {\frac{{4x + 9}}{{4x + 8}}} \,dx}}}}\]b \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy\)
Yechim
Bu holda biz birinchi navbatda \(x\) funksiyasini hal qilishimiz kerak, shunda biz \(\displaystyle ds\) da hosilani hisoblashimiz mumkin.
\[y = \sqrt {x + 2} \hspace{0.5in} \to \hspace{0.5in}x = {y^2} - 2\]Buning hosilasi,
\[\frac{{dx}}{{dy}} = 2y\]2-qadam
Buni \(\displaystyle ds\) formulasiga ulash, beradi,
\[ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy = \sqrt {1 + {{\left[ {2y} \right]}^2}} \,dy = \sqrt {1 + 4{y^2}} \,dy\]3-qadam
Keyin, \(\displaystyle ds\)-da \(dy\) borligini unutmang va shuning uchun bizga \(y\) integratsiya chegaralari kerak bo'ladi.
Muammo bayonotida bizga faqat \(x\) chegaralari berilgan. Biroq, biz ularni \(y\) chegaralariga aylantirish uchun muammo bayonotida berilgan funksiyaga ulashimiz mumkin. Shunday qilib,
\[x = 1:y = \sqrt 3 \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = 7:y = \sqrt 9 = 3\]Shunday qilib, mos keladigan \(y\) chegaralari: \(\sqrt 3 \le y \le 3\).
4-qadam
Nihoyat, biz qilishimiz kerak bo'lgan yagona narsa integralni o'rnatishdir.
\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{{\sqrt 3 }}^{3}{{\sqrt {1 + 4{y^2}} \,dy}}}}\]