Question Number 59814 by ANTARES VY last updated on 15/May/19
Commented by ANTARES VY last updated on 15/May/19
$$\boldsymbol{\mathrm{F}}\mathrm{ind}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{painted}}\:\:\boldsymbol{\mathrm{area}} \\ $$
Answered by mr W last updated on 15/May/19
$${y}={e}^{−{x}^{\mathrm{2}} } \\ $$$${y}'=−\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } =−\frac{{x}}{{y}}=−\frac{{x}}{{e}^{−{x}^{\mathrm{2}} } } \\ $$$$\Rightarrow{e}^{\mathrm{2}{x}^{\mathrm{2}} } =\mathrm{2} \\ $$$$\Rightarrow{x}=\pm\sqrt{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}={e}^{−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$${Area}\:{of}\:{circle}=\pi{r}^{\mathrm{2}} =\pi\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\pi\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}+{e}^{−\mathrm{ln}\:\mathrm{2}} \right) \\ $$
Commented by ANTARES VY last updated on 15/May/19
$$−\mathrm{2}\boldsymbol{\mathrm{xe}}^{−\boldsymbol{\mathrm{x}}^{\mathrm{2}} } =−\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\:\:? \\ $$
Commented by mr W last updated on 15/May/19
$${at}\:{point}\:\left({x},{y}\right)\:{the}\:{slope}\:{of}\:{tangent}\:{line} \\ $$$${is}\:{m}_{{t}} ={y}'=−\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } \\ $$$${the}\:{slope}\:{of}\:{normal}\:{line}\:{at}\:{this}\:{point} \\ $$$${is}\:{m}_{{n}} =\frac{{y}}{{x}}. \\ $$$${m}_{{t}} ×{m}_{{n}} =−\mathrm{1} \\ $$$$\Rightarrow{m}_{{t}} =−\frac{\mathrm{1}}{{m}_{{n}} } \\ $$$$\Rightarrow−\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } =−\frac{{x}}{{y}} \\ $$
Commented by ANTARES VY last updated on 15/May/19
$$? \\ $$
Commented by mr W last updated on 15/May/19
Commented by mr W last updated on 15/May/19
$$\mathrm{tan}\:\theta=−{y}'=\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\varphi=\frac{{y}}{{x}} \\ $$$$\mathrm{tan}\:\theta\:\mathrm{tan}\:\varphi=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } ×\frac{{y}}{{x}}=\mathrm{1} \\ $$$$…… \\ $$