Respuesta :

[tex]\displaystyle Note~that~|\overrightarrow{u}|= \sqrt{(m-1)^2+(2+m)^2}=\sqrt{2m^2+2m+5}. \\ \\ Thus~the~minmal~value~of~|\overrightarrow{u}|~holds~when~2m^2+2m+5~is~ \\ \\minimum. \\ \\ 2m^2+2m+5= \frac{4m^2+4m+10}{2}= \frac{(2m+1)^2+9}{2} \ge \frac{0+9}{2}= \frac{9}{2}. \\ \\ Hence,~|\overrightarrow{u}|_{min}= \sqrt{\frac{9}{2}}= \frac{3 \sqrt{2}}{2}~for~2m+1=0 \Leftrightarrow m= -\frac{1}{2}.[/tex]
[tex]\text{We know that }[/tex] [tex]||u(m)||= \sqrt{(m-1)^2+(2+m)^2} [/tex] 

[tex]\text{If we want the minimum of this}[/tex], [tex]\text{we can take the minimum of the square}[/tex] 

[tex] \sqrt{f(x)} \ \text{and} \ f(x) \ \text{have the same minimum}[/tex] 

[tex](||u(m)||^2) '=2(m-1)+2(2+m)[/tex] 

[tex]\text{Now set it, so it will equal to zero}[/tex] 

[tex]2(m-1)+2(2+m)=0[/tex] 

[tex]m= \dfrac{-1}{2} [/tex]  [tex]\text{Solution}[/tex]

[tex]\text{The particular value for m minimizes}[/tex] ∥u(m)∥,  [tex]\text{using the second derivative test.}[/tex] 

[tex]\text{Now plug m back into the norm. }[/tex] 

[tex]||u(-12)||= \dfrac{3}{2} \sqrt{2} [/tex] 

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

[tex]\text{Proof that} \sqrt{f(x)} \ \text{and} \ f(x) \ \text{have the same extrema}[/tex] 

[tex] \sqrt{(f'(x))'}= \dfrac{f'(x)}{2\sqrt{f(x)}}=0[/tex] 

[tex] f'(x)=0\ \text{because both have the same solution.}[/tex] 

[tex]\text{It's minimized when the velocity is perpendicular to the position. }[/tex].[tex]\text{That makes sense cause rotational motion has a constant}[/tex][tex]\text{ length throughout its motion}[/tex][tex]\text{Whereas if you don't have a locally circular motion,}[/tex] [tex]\text{you must either be moving further or closer away so }[/tex][tex]\text{you're no at a local min or max.}[/tex] 

[tex] \dfrac{d}{dm}(u*u)=0[/tex] 

[tex]u'*u=0[/tex] 

Otras preguntas