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Question Number 131121 by abdurehime last updated on 01/Feb/21

Answered by mathmax by abdo last updated on 01/Feb/21

$$\mathrm{A}){\mathrm{x}}^2+{\mathrm{y}}^2=25\Rightarrow {\mathrm{y}}^2=25-{\mathrm{x}}^2\Rightarrow \mathrm{y}=\xi \sqrt{25-{\mathrm{x}}^2}(\xi =\stackrel{-}{+}1)\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\mathrm{x}\xi }{2\sqrt{25-{\mathrm{x}}^2}}=-\frac{\xi \mathrm{x}}{\sqrt{25-{\mathrm{x}}^2}}\mathrm{also}\mathrm{due}\mathrm{to}\mathrm{symetrie}\mathrm{we}\mathrm{get}$$$$\frac{\mathrm{dx}}{\mathrm{dy}}=-\frac{\xi \mathrm{y}}{\sqrt{25-{\mathrm{y}}^2}}$$$$\mathrm{c})\mathrm{x}+{\mathrm{xy}}^2-\mathrm{y}=\mathrm{xy}\Rightarrow {\mathrm{xy}}^2-\mathrm{y}-\mathrm{xy}+\mathrm{x}=0\Rightarrow {\mathrm{xy}}^2-(\mathrm{x}+1)\mathrm{y}+\mathrm{x}=0$$$$\mathrm{\Delta }={(\mathrm{x}+1)}^2-{4\mathrm{x}}^2={\mathrm{x}}^2+2\mathrm{x}+1-{4\mathrm{x}}^2={(\mathrm{x}-1)}^2\Rightarrow$$$${\mathrm{y}}_1=\frac{\mathrm{x}+1+\mid \mathrm{x}-1\mid }{2\mathrm{x}}\mathrm{and}{\mathrm{y}}_2=\frac{\mathrm{x}+1-\mid \mathrm{x}-1\mid }{2\mathrm{x}}$$$$\mathrm{x}⩾1\Rightarrow {\mathrm{y}}_1=1\mathrm{and}{\mathrm{y}}_2=\frac{1}{\mathrm{x}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=0\mathrm{or}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{{\mathrm{x}}^2}$$$$\mathrm{x}⩽1\Rightarrow {\mathrm{y}}_1=1\mathrm{and}{\mathrm{y}}_2=\frac{1}{\mathrm{x}}\Rightarrow \mathrm{same}\mathrm{result}$$$$\mathrm{we}\mathrm{have}\mathrm{x}+{\mathrm{xy}}^2-\mathrm{xy}=\mathrm{y}\Rightarrow (1+{\mathrm{y}}^2-\mathrm{y})\mathrm{x}=\mathrm{y}\Rightarrow \mathrm{x}=\frac{\mathrm{y}}{{\mathrm{y}}^2-\mathrm{y}+1}\Rightarrow$$$$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{{\mathrm{y}}^2-\mathrm{y}+1-\mathrm{y}(2\mathrm{y}-1)}{{({\mathrm{y}}^2-\mathrm{y}+1)}^2}=\frac{{\mathrm{y}}^2-\mathrm{y}+1-{2\mathrm{y}}^2+\mathrm{y}}{{({\mathrm{y}}^2-\mathrm{y}+1)}^2}=\frac{1-{\mathrm{y}}^2}{{({\mathrm{y}}^2-\mathrm{y}+1)}^2}$$

Commented by abdurehime last updated on 02/Feb/21

$$\mathrm{thank}\mathrm{u}$$

Answered by liberty last updated on 02/Feb/21

$$\left(\mathrm{b}\right)\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}$$$$\Rightarrow 3\mathrm{y}+{3\mathrm{xy}}^{\prime }-(2\mathrm{xy}+{\mathrm{x}}^2{\mathrm{y}}^{\prime })=0$$$$\Rightarrow 3\mathrm{y}+{3\mathrm{xy}}^{\prime }-2\mathrm{xy}-{\mathrm{x}}^2{\mathrm{y}}^{\prime }=0$$$$\Rightarrow (3\mathrm{x}-{\mathrm{x}}^2){\mathrm{y}}^{\prime }=2\mathrm{xy}-3\mathrm{y}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{xy}-3\mathrm{y}}{3\mathrm{x}-{\mathrm{x}}^2};\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{\mathrm{dy}/\mathrm{dx}}=\frac{3\mathrm{x}-{\mathrm{x}}^2}{2\mathrm{xy}-3\mathrm{y}}$$

Answered by liberty last updated on 02/Feb/21

$$\left(\mathrm{a}\right)\frac{\mathrm{d}}{\mathrm{dx}}[{\mathrm{x}}^2+{\mathrm{y}}^2]=0$$$$\Rightarrow 2\mathrm{x}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=0;\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}\wedge \frac{\mathrm{dx}}{\mathrm{dy}}=-\frac{\mathrm{y}}{\mathrm{x}}$$$$\mathrm{where}\mathrm{y}=\pm \sqrt{25-{\mathrm{x}}^2}$$$$$$

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