\(\text{We know that }\) \(||u(m)||= \sqrt{(m-1)^2+(2+m)^2}\)
\(\text{If we want the minimum of this}\), \(\text{we can take the minimum of the square}\)
\(\sqrt{f(x)} \ \text{and} \ f(x) \ \text{have the same minimum}\)
\((||u(m)||^2) '=2(m-1)+2(2+m)\)
\(\text{Now set it, so it will equal to zero}\)
\(2(m-1)+2(2+m)=0\)
\(m= \dfrac{-1}{2}\) \(\text{Solution}\)
\(\text{The particular value for m minimizes}\) ∥u(m)∥, \(\text{using the second derivative test.}\)
\(\text{Now plug m back into the norm. }\)
\(||u(-12)||= \dfrac{3}{2} \sqrt{2}\)
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\(\text{Proof that} \sqrt{f(x)} \ \text{and} \ f(x) \ \text{have the same extrema}\)
\(\sqrt{(f'(x))'}= \dfrac{f'(x)}{2\sqrt{f(x)}}=0\)
\(f'(x)=0\ \text{because both have the same solution.}\)
\(\text{It's minimized when the velocity is perpendicular to the position. }\).\(\text{That makes sense cause rotational motion has a constant}\)\(\text{ length throughout its motion}\)\(\text{Whereas if you don't have a locally circular motion,}\) \(\text{you must either be moving further or closer away so }\)\(\text{you're no at a local min or max.}\)
\(\dfrac{d}{dm}(u*u)=0\)
\(u'*u=0\)
