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Question-165168




Question Number 165168 by saboorhalimi last updated on 26/Jan/22
Answered by mahdipoor last updated on 26/Jan/22
(dy/dx)=((dy/dt)/(dx/dt))=((3t^2 )/(3t^2 −4))=1+(4/(3t^2 −4))  (d^2 y/dx^2 )=((d(dy/dx))/dx)=(((d(dy/dx))/dt)/(dx/dt))=(((−4(6t))/((3t^2 −4)^2 ))/(3t^2 −4))=  −((24t)/((3t^2 −4)^3 ))  ⇒^(t=1) =24
$$\frac{{dy}}{{dx}}=\frac{{dy}/{dt}}{{dx}/{dt}}=\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}\left({dy}/{dx}\right)}{{dx}}=\frac{\frac{{d}\left({dy}/{dx}\right)}{{dt}}}{\frac{{dx}}{{dt}}}=\frac{\frac{−\mathrm{4}\left(\mathrm{6}{t}\right)}{\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}= \\ $$$$−\frac{\mathrm{24}{t}}{\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{3}} }\:\:\overset{{t}=\mathrm{1}} {\Rightarrow}=\mathrm{24} \\ $$

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