Question-127490

Question Number 127490 by ZiYangLee last updated on 30/Dec/20

Commented by mr W last updated on 30/Dec/20

Q45. (dx/dt)=3t^2 (dy/dt)=2t (dy/dx)=((dy/dt)/(dx/dt))=((2t)/(3t^2 ))=(2/(3t)) ((dy/dx))^4 =((2/(3t)))^4 =((16)/(81t^4 )) (d^2 y/dx^2 )=(d/dx)((dy/dx))=(((d/dt)((dy/dx)))/(dx/dt))=((−(1/(3t^2 )))/(3t^2 ))=−(1/(9t^4 )) =−(1/9)×((81)/(16))(((16)/(81t^4 )))=−(9/(16))((dy/dx))^4 =k((dy/dx))^4 with k=−(9/(16))=constant Q44. similarly

$${Q}\mathrm{45}. \\ $$$$\frac{{dx}}{{dt}}=\mathrm{3}{t}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dt}}=\mathrm{2}{t} \\ $$$$\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{dt}}}{\frac{{dx}}{{dt}}}=\frac{\mathrm{2}{t}}{\mathrm{3}{t}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}{t}} \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{4}} =\left(\frac{\mathrm{2}}{\mathrm{3}{t}}\right)^{\mathrm{4}} =\frac{\mathrm{16}}{\mathrm{81}{t}^{\mathrm{4}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{\frac{{d}}{{dt}}\left(\frac{{dy}}{{dx}}\right)}{\frac{{dx}}{{dt}}}=\frac{−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }}{\mathrm{3}{t}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{9}{t}^{\mathrm{4}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{9}}×\frac{\mathrm{81}}{\mathrm{16}}\left(\frac{\mathrm{16}}{\mathrm{81}{t}^{\mathrm{4}} }\right)=−\frac{\mathrm{9}}{\mathrm{16}}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{4}} \\ $$$$={k}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{4}} \:{with}\:{k}=−\frac{\mathrm{9}}{\mathrm{16}}={constant} \\ $$$$ \\ $$$${Q}\mathrm{44}. \\ $$$${similarly} \\ $$

Answered by som(math1967) last updated on 30/Dec/20

x=tant⇒(dx/dt)=sec^2 t y=tanpt⇒(dy/dt)=psec^2 pt ∴(dy/dx)=((psec^2 pt)/(sec^2 t))=((p(1+tan^2 pt))/(1+tan^2 t)) ∴(dy/dx)=((p(1+y^2 ))/(1+x^2 )) [∵x=tant,y=tanpt] (1+x^2 )(dy/dx)=p+py^2 (1+x^2 )(d^2 y/dx^2 ) +2x(dy/dx)=2py(dy/dx) ∴(1+x^2 )(d^2 y/dx^2 )=2(py−x)(dy/dx)

$${x}={tant}\Rightarrow\frac{{dx}}{{dt}}={sec}^{\mathrm{2}} {t} \\ $$$${y}={tanpt}\Rightarrow\frac{{dy}}{{dt}}={psec}^{\mathrm{2}} {pt} \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{{psec}^{\mathrm{2}} {pt}}{{sec}^{\mathrm{2}} {t}}=\frac{{p}\left(\mathrm{1}+{tan}^{\mathrm{2}} {pt}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} {t}} \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{{p}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\left[\because{x}={tant},{y}={tanpt}\right] \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}={p}+{py}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\mathrm{2}{x}\frac{{dy}}{{dx}}=\mathrm{2}{py}\frac{{dy}}{{dx}} \\ $$$$\therefore\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{2}\left({py}−{x}\right)\frac{{dy}}{{dx}} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply