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

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

$$Q45.$$$$\frac{dx}{dt}=3t^2$$$$\frac{dy}{dt}=2t$$$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t}{3t^2}=\frac{2}{3t}$$$${\left(\frac{dy}{dx}\right)}^4={\left(\frac{2}{3t}\right)}^4=\frac{16}{81t^4}$$$$\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{-\frac{1}{3t^2}}{3t^2}=-\frac{1}{9t^4}$$$$=-\frac{1}{9}\times \frac{81}{16}\left(\frac{16}{81t^4}\right)=-\frac{9}{16}{\left(\frac{dy}{dx}\right)}^4$$$$=k{\left(\frac{dy}{dx}\right)}^{4}withk=-\frac{9}{16}=constant$$$$$$$$Q44.$$$$similarly$$

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

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

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