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Question-203115

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Question Number 203115 by Mathstar last updated on 10/Jan/24
Commented by Mathstar last updated on 10/Jan/24
Send an explanatory solution, thank you
$$\mathrm{Send}\:\mathrm{an}\:\mathrm{explanatory}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 10/Jan/24
y=x^2 e^x y′=e^x (x^2 +2x) center of circle (0,r) say the circle tangents the curve additionally at P(p,q). q=p^2 e^p tan θ=y′=e^p (p^2 +2p) p=r sin θ=2r sin (θ/2) cos (θ/2) q=r(1−cos θ)=2r sin^2 (θ/2) tan (θ/2)=(q/p)=((p^2 e^p )/p)=pe^p tan θ=((2 tan (θ/2))/(1−tan^2 (θ/2))) e^p (p^2 +2p)=((2pe^p )/(1−p^2 e^(2p) )) ⇒(p+2)(1−p^2 e^(2p) )=2 ⇒p=0 ∨ p≈0.2619 r_(max) =(p/(sin θ))=(√(p^2 +(p^2 /(tan^2 θ)))) =(√(p^2 +(1/(e^(2p) (p+2)^2 ))))≈0.4294
$${y}={x}^{\mathrm{2}} {e}^{{x}} \\ $$$${y}'={e}^{{x}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}\right) \\ $$$${center}\:{of}\:{circle}\:\left(\mathrm{0},{r}\right) \\ $$$${say}\:{the}\:{circle}\:{tangents}\:{the}\:{curve} \\ $$$${additionally}\:{at}\:{P}\left({p},{q}\right). \\ $$$${q}={p}^{\mathrm{2}} {e}^{{p}} \\ $$$$\mathrm{tan}\:\theta={y}'={e}^{{p}} \left({p}^{\mathrm{2}} +\mathrm{2}{p}\right) \\ $$$${p}={r}\:\mathrm{sin}\:\theta=\mathrm{2}{r}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$${q}={r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\mathrm{2}{r}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{{q}}{{p}}=\frac{{p}^{\mathrm{2}} {e}^{{p}} }{{p}}={pe}^{{p}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}} \\ $$$${e}^{{p}} \left({p}^{\mathrm{2}} +\mathrm{2}{p}\right)=\frac{\mathrm{2}{pe}^{{p}} }{\mathrm{1}−{p}^{\mathrm{2}} {e}^{\mathrm{2}{p}} } \\ $$$$\Rightarrow\left({p}+\mathrm{2}\right)\left(\mathrm{1}−{p}^{\mathrm{2}} {e}^{\mathrm{2}{p}} \right)=\mathrm{2} \\ $$$$\Rightarrow{p}=\mathrm{0}\:\vee\:{p}\approx\mathrm{0}.\mathrm{2619} \\ $$$${r}_{{max}} =\frac{{p}}{\mathrm{sin}\:\theta}=\sqrt{{p}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \:\theta}} \\ $$$$\:\:=\sqrt{{p}^{\mathrm{2}} +\frac{\mathrm{1}}{{e}^{\mathrm{2}{p}} \left({p}+\mathrm{2}\right)^{\mathrm{2}} }}\approx\mathrm{0}.\mathrm{4294} \\ $$
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
$${Algebra} \\ $$$${Geometry} \\ $$$${Trigonometry} \\ $$$${Derivative} \\ $$$${Equations} \\ $$
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=∣(((1+y′^2 )^(3/2) )/(y′′))∣. in our case R=0.5. 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 (0,0) 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.
$${such}\:{that}\:{a}\:{circle}\:{only}\:{touches}\:{a} \\ $$$${curve}\:{at}\:{a}\:{point},\:{the}\:{radius}\:{of}\:{the} \\ $$$${circle}\:{may}\:{not}\:{be}\:{larger}\:{than}\:{the} \\ $$$${radius}\:{R}\:{of}\:{the}\:{osculating}\:{circle} \\ $$$${with}\:{R}=\mid\frac{\left(\mathrm{1}+{y}'^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{y}''}\mid. \\ $$$${in}\:{our}\:{case}\:{R}=\mathrm{0}.\mathrm{5}. \\ $$$${but}\:{this}\:{circle}\:{may}\:{intersect}\:{the} \\ $$$${curve}\:{at}\:{other}\:{points}.\:{such}\:{that} \\ $$$${this}\:{doesn}'{t}\:{happen},\:{we}\:{must}\: \\ $$$${ensure}\:{that}\:{the}\:{circle}\:{which}\:{touches} \\ $$$${the}\:{curce}\:{at}\:{point}\:\left(\mathrm{0},\mathrm{0}\right)\:{maximally} \\ $$$${touches}\:{the}\:{curve}\:{at}\:{an}\:{other}\:{point}. \\ $$$${this}\:{circle}\:{is}\:{then}\:{the}\:{largest}\:{circle} \\ $$$${which}\:{fulfills}\:{the}\:{requirement}\:{of} \\ $$$${the}\:{question}. \\ $$$${this}\:{is}\:{the}\:“{theory}''{behind}\:{my} \\ $$$${solution}. \\ $$
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 ��‍♂️

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