Question Number 50357 by ajfour last updated on 16/Dec/18
Commented by ajfour last updated on 16/Dec/18
$${Find}\:{maximum}\:{R}\:{as}\:{a}\:{function}\: \\ $$$${of}\:\theta\:.\:{Also}\:{find}\:{minimum}\:{R}_{{max}} . \\ $$
Answered by mr W last updated on 16/Dec/18
![y=(1/x) y′=−(1/x^2 ) y′′=(2/x^3 ) P(h,(1/h)) (1/h^2 )=tan θ=m ⇒h=(1/( (√(tan θ))))=(1/( (√m))) at point P (1/R)=((2/h^3 )/([1+(−(1/h^2 ))^2 ]^(3/2) ))=2((1/h^2 ))^(3/2) (1/([1+((1/h^2 ))^2 ]^(3/2) )) (1/R)=2((m/(1+m^2 )))^(3/2) =(2/((m+(1/m))^(3/2) )) ⇒R=(1/2)(m+(1/m))^(3/2) ≥(1/2)(2×(√(m×(1/m))))^(3/2) =(√2) min. R is when m=(1/m) or m=1 or θ=45° with R_(min) =(√2)](https://www.tinkutara.com/question/Q50382.png)
$${y}=\frac{\mathrm{1}}{{x}} \\ $$$${y}'=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${y}''=\frac{\mathrm{2}}{{x}^{\mathrm{3}} } \\ $$$${P}\left({h},\frac{\mathrm{1}}{{h}}\right) \\ $$$$\frac{\mathrm{1}}{{h}^{\mathrm{2}} }=\mathrm{tan}\:\theta={m} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\theta}}=\frac{\mathrm{1}}{\:\sqrt{{m}}} \\ $$$${at}\:{point}\:{P} \\ $$$$\frac{\mathrm{1}}{{R}}=\frac{\frac{\mathrm{2}}{{h}^{\mathrm{3}} }}{\left[\mathrm{1}+\left(−\frac{\mathrm{1}}{{h}^{\mathrm{2}} }\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{2}\left(\frac{\mathrm{1}}{{h}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \frac{\mathrm{1}}{\left[\mathrm{1}+\left(\frac{\mathrm{1}}{{h}^{\mathrm{2}} }\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{\mathrm{1}}{{R}}=\mathrm{2}\left(\frac{{m}}{\mathrm{1}+{m}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{2}}{\left({m}+\frac{\mathrm{1}}{{m}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\Rightarrow{R}=\frac{\mathrm{1}}{\mathrm{2}}\left({m}+\frac{\mathrm{1}}{{m}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}×\sqrt{{m}×\frac{\mathrm{1}}{{m}}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\sqrt{\mathrm{2}} \\ $$$${min}.\:{R}\:{is}\:{when}\:{m}=\frac{\mathrm{1}}{{m}}\:{or}\:{m}=\mathrm{1}\:{or}\:\theta=\mathrm{45}° \\ $$$${with}\:{R}_{{min}} =\sqrt{\mathrm{2}} \\ $$
Commented by ajfour last updated on 16/Dec/18
$$\mathbb{EXCELLENT}\:\mathcal{S}{i}\mathcal{R}\:!\:{Thank}\:{you}. \\ $$