9.4-bo'lim : Parametrik tenglamalar bilan yoy uzunligi
1. Quyidagi parametrik tenglamalar to‘plami orqali berilgan parametrik egri chiziq uzunligini aniqlang. \(t\) ning berilgan diapazoni uchun egri chiziq aynan bir marta chiqadi deb taxmin qilishingiz mumkin.
\[x = 8{t^{\frac{3}{2}}}\hspace{0.25in}y = 3 + {\left( {8 - t} \right)^{\frac{3}{2}}}\hspace{0.25in}0 \le t \le 4\]Yechim
Bu erda bizga kerak bo'lgan birinchi narsa quyidagi ikkita hosiladir.
\[\frac{{dx}}{{dt}} = 12{t^{\,\frac{1}{2}}}\hspace{0.5in}\frac{{dy}}{{dt}} = - \frac{3}{2}{\left( {8 - t} \right)^{\frac{1}{2}}}\]2-qadam
Ushbu muammo uchun bizga \(ds\) kerak bo'ladi.
\[ds = \sqrt {{{\left[ {12{t^{\,\frac{1}{2}}}} \right]}^2} + {{\left[ { - \frac{3}{2}{{\left( {8 - t} \right)}^{\frac{1}{2}}}} \right]}^2}} \,dt = \sqrt {144t + \frac{9}{4}\left( {8 - t} \right)} \,dt = \sqrt {\frac{{567}}{4}t + 18} \,dt\]3-qadam
U holda yoy uzunligi uchun integrali,
\[L = \int_{{}}^{{}}{{ds}} = \int_{0}^{4}{{\sqrt {\frac{{567}}{4}t + 18} \,dt}}\]4-qadam
Bu tez almashtirish bilan hisoblash uchun oddiy integraldir. Mana integral ish,
\[L = \int_{0}^{4}{{\sqrt {\frac{{567}}{4}t + 18} \,dt}} = \left. {\frac{4}{{567}}\left( {\frac{2}{3}} \right){{\left( {\frac{{567}}{4}t + 18} \right)}^{\frac{3}{2}}}} \right|_0^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{8}{{1701}}\left( {{{585}^{\,\frac{3}{2}}} - {{18}^{\,\frac{3}{2}}}} \right) = 66.1865}}\]