Question

Determinate the minimal value of the module of vector:


`\vec{u}=(m-1)\vec{i}+(2+m)\vec{j}`

Answer 1

`\text{We know that }`
`||u(m)||= √((m-1)^2+(2+m)^2)`


`\text{If we want the minimum of this}`,
`\text{we can take the minimum of the square}`


`√(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= (-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)||= (3)/(2) √(2)`

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`\text{Proof that} √(f(x)) \ \text{and} \ f(x) \ \text{have the same extrema}`


`√((f'(x))')= (f'(x))/(2√(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.}`


`(d)/(dm)(u*u)=0`


`u'*u=0`

Answer 2

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