Ismael Molina

2022-07-24

Show that ${v}_{2}$ is the smaller root of the quadratic equation
${v}_{2}^{2}-\frac{2\sigma M}{{v}_{1}{l}_{1}}{v}_{2}-\left[{v}_{1}^{2}-\frac{2\sigma M}{{l}_{1}}\right]=0$

gardapati5u

Expert

${v}_{2}^{2}-\frac{2\sigma M}{{v}_{1}{l}_{1}}{v}_{2}-\left[{v}_{1}^{2}-\frac{2\sigma M}{{l}_{1}}\right]=0$
By using roots method
${v}_{2}=\frac{-\left(\frac{2\sigma M}{{v}_{1}{l}_{1}}\right)±\sqrt{\left(\frac{-2\sigma M}{{v}_{1}{l}_{1}}{\right)}^{2}}-4\left[1\right]\left[-{v}_{1}^{2}+\frac{2\sigma M}{{l}_{1}}\right]}{2\cdot 1}\phantom{\rule{0ex}{0ex}}{v}_{2}=\frac{+\frac{2\sigma M}{{v}_{1}{l}_{1}}±\sqrt{\left(\frac{2\sigma M}{{v}_{1}{l}_{1}}{\right)}^{2}+4{v}_{1}^{2}-4\cdot \frac{2\sigma M}{{l}_{1}}}}{2}\phantom{\rule{0ex}{0ex}}{v}_{2}=\frac{\frac{2\sigma M}{{v}_{1}{l}_{1}}±\left(2{v}_{1}-\frac{2\sigma M}{{v}_{1}{l}_{1}}\right)}{2}\phantom{\rule{0ex}{0ex}}{v}_{2}^{\prime }=\frac{2\sigma M}{{v}_{1}{l}_{1}}+2{v}_{1}-\frac{2\sigma M}{{l}_{1}{v}_{1}}\phantom{\rule{0ex}{0ex}}{v}_{2}^{\prime }=2{v}_{1}\phantom{\rule{0ex}{0ex}}{v}_{2}^{″}=\frac{2\sigma M}{{v}_{1}{l}_{1}}+2{v}_{1}-\frac{2\sigma M}{{l}_{1}{v}_{1}}\phantom{\rule{0ex}{0ex}}{v}_{2}^{″}=2{v}_{1}\phantom{\rule{0ex}{0ex}}{v}_{2}^{″}=\frac{2\sigma M}{{V}_{1}{l}_{1}}-2{V}_{1}+\frac{2\sigma M}{{v}_{1}{l}_{1}}\phantom{\rule{0ex}{0ex}}{v}_{2}^{″}=2\cdot \frac{2\sigma m}{{v}_{1}{l}_{1}}-2{v}_{1}$
Therefore for the second roots ${v}_{2}^{″}$
as we increases the value of ${v}_{1}$
the ${v}_{2}^{″}$ will decreases due to two factor
1. negative stop " $-2{v}_{1}$ "
2. exponential decreasing function " $\alpha \frac{1}{{v}_{1}}$ "Therefore ${v}_{2}^{″}<{v}_{1}$ proved.

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