9.4-bo'lim : Parametrik tenglamalar bilan yoy uzunligi
5. Quyidagi parametrik tenglamalar to‘plami orqali berilgan parametrik egri chiziq uzunligini beradigan integralni o‘rnating, lekin baholamang. Egri chiziq \(t\) ning berilgan diapazoni uchun aynan bir marta chiqadi deb taxmin qilishingiz mumkin.
\[x = {\cos ^3}\left( {2t} \right)\hspace{0.25in}y = \sin \left( {1 - {t^2}} \right)\hspace{0.25in} - \frac{3}{2} \le t \le 0\]Yechim
Bu erda bizga kerak bo'lgan birinchi narsa quyidagi ikkita hosiladir.
\[\frac{{dx}}{{dt}} = - 6{\cos ^2}\left( {2t} \right)\sin \left( {2t} \right)\hspace{0.5in}\frac{{dy}}{{dt}} = - 2t\cos \left( {1 - {t^2}} \right)\]2-qadam
Ushbu muammo uchun bizga \(ds\) kerak bo'ladi.
\[\begin{align*}ds & = \sqrt {{{\left[ { - 6{{\cos }^2}\left( {2t} \right)\sin \left( {2t} \right)} \right]}^2} + {{\left[ { - 2t\cos \left( {1 - {t^2}} \right)} \right]}^2}} \,dt\\ & = \sqrt {36{{\cos }^4}\left( {2t} \right){{\sin }^2}\left( {2t} \right) + 4{t^2}{{\cos }^2}\left( {1 - {t^2}} \right)} \,dt\end{align*}\]3-qadam
U holda yoy uzunligi uchun integrali,
\[L = \int_{{}}^{{}}{{ds}} = \int_{{ - \,\frac{3}{2}}}^{0}{{\sqrt {36{{\cos }^4}\left( {2t} \right){{\sin }^2}\left( {2t} \right) + 4{t^2}{{\cos }^2}\left( {1 - {t^2}} \right)} \,dt}}\]