Question Number 203115 by Mathstar last updated on 10/Jan/24
Commented by Mathstar last updated on 10/Jan/24
$$\mathrm{Send}\:\mathrm{an}\:\mathrm{explanatory}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 10/Jan/24
$${y}={x}^{\mathrm{2}} {e}^{{x}} \\ $$$${y}'={e}^{{x}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}\right) \\ $$$${center}\:{of}\:{circle}\:\left(\mathrm{0},{r}\right) \\ $$$${say}\:{the}\:{circle}\:{tangents}\:{the}\:{curve} \\ $$$${additionally}\:{at}\:{P}\left({p},{q}\right). \\ $$$${q}={p}^{\mathrm{2}} {e}^{{p}} \\ $$$$\mathrm{tan}\:\theta={y}'={e}^{{p}} \left({p}^{\mathrm{2}} +\mathrm{2}{p}\right) \\ $$$${p}={r}\:\mathrm{sin}\:\theta=\mathrm{2}{r}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$${q}={r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\mathrm{2}{r}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{{q}}{{p}}=\frac{{p}^{\mathrm{2}} {e}^{{p}} }{{p}}={pe}^{{p}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}} \\ $$$${e}^{{p}} \left({p}^{\mathrm{2}} +\mathrm{2}{p}\right)=\frac{\mathrm{2}{pe}^{{p}} }{\mathrm{1}−{p}^{\mathrm{2}} {e}^{\mathrm{2}{p}} } \\ $$$$\Rightarrow\left({p}+\mathrm{2}\right)\left(\mathrm{1}−{p}^{\mathrm{2}} {e}^{\mathrm{2}{p}} \right)=\mathrm{2} \\ $$$$\Rightarrow{p}=\mathrm{0}\:\vee\:{p}\approx\mathrm{0}.\mathrm{2619} \\ $$$${r}_{{max}} =\frac{{p}}{\mathrm{sin}\:\theta}=\sqrt{{p}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \:\theta}} \\ $$$$\:\:=\sqrt{{p}^{\mathrm{2}} +\frac{\mathrm{1}}{{e}^{\mathrm{2}{p}} \left({p}+\mathrm{2}\right)^{\mathrm{2}} }}\approx\mathrm{0}.\mathrm{4294} \\ $$
Commented by Mathstar last updated on 10/Jan/24
Thank you. I will look out for a video that can explain better.
Commented by mr W last updated on 10/Jan/24
Commented by mr W last updated on 10/Jan/24
$${Algebra} \\ $$$${Geometry} \\ $$$${Trigonometry} \\ $$$${Derivative} \\ $$$${Equations} \\ $$
Commented by Mathstar last updated on 10/Jan/24
Thank you so much Mr W. Please how can I understand something like this?
Commented by Mathstar last updated on 10/Jan/24
Could you please enumerate the scope in this type of question?
Commented by mr W last updated on 11/Jan/24
Commented by mr W last updated on 11/Jan/24
$${such}\:{that}\:{a}\:{circle}\:{only}\:{touches}\:{a} \\ $$$${curve}\:{at}\:{a}\:{point},\:{the}\:{radius}\:{of}\:{the} \\ $$$${circle}\:{may}\:{not}\:{be}\:{larger}\:{than}\:{the} \\ $$$${radius}\:{R}\:{of}\:{the}\:{osculating}\:{circle} \\ $$$${with}\:{R}=\mid\frac{\left(\mathrm{1}+{y}'^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{y}''}\mid. \\ $$$${in}\:{our}\:{case}\:{R}=\mathrm{0}.\mathrm{5}. \\ $$$${but}\:{this}\:{circle}\:{may}\:{intersect}\:{the} \\ $$$${curve}\:{at}\:{other}\:{points}.\:{such}\:{that} \\ $$$${this}\:{doesn}'{t}\:{happen},\:{we}\:{must}\: \\ $$$${ensure}\:{that}\:{the}\:{circle}\:{which}\:{touches} \\ $$$${the}\:{curce}\:{at}\:{point}\:\left(\mathrm{0},\mathrm{0}\right)\:{maximally} \\ $$$${touches}\:{the}\:{curve}\:{at}\:{an}\:{other}\:{point}. \\ $$$${this}\:{circle}\:{is}\:{then}\:{the}\:{largest}\:{circle} \\ $$$${which}\:{fulfills}\:{the}\:{requirement}\:{of} \\ $$$${the}\:{question}. \\ $$$${this}\:{is}\:{the}\:“{theory}''{behind}\:{my} \\ $$$${solution}. \\ $$
Commented by mr W last updated on 11/Jan/24
Commented by Mathstar last updated on 11/Jan/24
Thank you so much, Mr W.
You've been of great support to me ever since I joined this platform ��♂️