9.2-bo'lim : Parametrik tenglamalar bilan tangentlar
1. Quyidagi parametrik tenglamalar to‘plami uchun \(\frac{{dy}}{{dx}}\) va \(\frac{{{d^2}y}}{{d{x^2}}}\) ni hisoblang.
\[x = 4{t^3} - {t^2} + 7t\hspace{0.5in}\,\,y = {t^4} - 6\]Yechim
Bu erda bizga kerak bo'lgan birinchi narsa quyidagi ikkita hosiladir.
\[\frac{{dx}}{{dt}} = 12{t^2} - 2t + 7\hspace{0.5in}\frac{{dy}}{{dt}} = 4{t^3}\]2-qadam
Birinchi hosila shundan iboratki,
\[\frac{{dy}}{{dx}} = \frac{{\displaystyle \,\,\frac{{dy}}{{dt}}\,\,}}{{\displaystyle \,\,\frac{{dx}}{{dt}}\,\,}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4{t^3}}}{{12{t^2} - 2t + 7}}}}\]3-qadam
Ikkinchi hosila uchun bizga endi kerak bo'ladi,
\[\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dt}}\left( {\frac{{4{t^3}}}{{12{t^2} - 2t + 7}}} \right) = \frac{{\left( {12{t^2}} \right)\left( {12{t^2} - 2t + 7} \right) - 4{t^3}\left( {24t - 2} \right)}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}}}}\]4-qadam
Ikkinchi hosila esa,
\[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}} = \frac{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}}}}{{12{t^2} - 2t + 7}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^3}}}}}\]