Question Number 74300 by ~blr237~ last updated on 21/Nov/19 | ||
$${Let}\:{consider}\:\:\gamma\:\::{I}\rightarrow\mathbb{R}^{\mathrm{2}} \:\:{a}\:{parametric}\:{curve}\: \\ $$ $$\left.\mathrm{1}\left.\right){Prove}\:{that}\:{if}\:\:{a}<{b}\:\:{and}\:\:\gamma\left({a}\right)\neq\gamma\left({b}\right)\:{then}\:{there}\:{exist}\:\:{t}_{\mathrm{0}} \in\right]{a},{b}\left[\:\:\right. \\ $$ $${such}\:{as}\:\:\gamma'\left({t}_{\mathrm{0}} \right)\:\:{is}\:{colinear}\:{to}\:\gamma\left({b}\right)−\gamma\left({a}\right)\: \\ $$ $$\left.\mathrm{2}\right){Show}\:{that}\:{if}\:\:\gamma\:{is}\:{regular}\:{and}\:{the}\:\:{function}\:{f}\::{I}\rightarrow\mathbb{R}\:\:\:\:{t}\rightarrow{f}\left({t}\right)=\mid\mid\gamma\left({t}\right)−{O}\left(\mathrm{0},\mathrm{0}\right)\:\mid\mid\:\:{is}\:{maximal}\:{in}\:{t}_{\mathrm{0}} \in{I} \\ $$ $${Then}\:\:\mid{K}_{\gamma} \left({t}_{\mathrm{0}} \right)\mid\geqslant\frac{\mathrm{1}}{{f}\left({t}_{\mathrm{0}} \right)} \\ $$ | ||
Answered by mind is power last updated on 21/Nov/19 | ||
$$\gamma\left({t}\right)=\left({x}\left({t}\right),{y}\left({t}\right)\right) \\ $$ $$\gamma\left({a}\right)\neq\gamma\left({b}\right) \\ $$ $$\Rightarrow{x}\left({a}\right)\neq{x}\left({b}\right)\:\:{or}\:{y}\left({a}\right)\neq{y}\left({b}\right) \\ $$ $${x}\left({a}\right)\neq{x}\left({b}\right) \\ $$ $${g}\left({t}\right)={y}\left({t}\right)\left({x}\left({a}\right)−{x}\left({b}\right)−{x}\left({t}\right)\left({y}\left({a}\right)−{y}\left({b}\right)\right)\right. \\ $$ $${g}\left({a}\right)=−{y}\left({a}\right){x}\left({b}\right)+{x}\left({a}\right){y}\left({b}\right) \\ $$ $${g}\left({b}\right)={y}\left({b}\right){x}\left({a}\right)−{x}\left({b}\right){y}\left({a}\right)={g}\left({a}\right) \\ $$ $$\left.{g}\left({a}\right)={g}\left({b}\right)\:{mean}\:{values}\:\Rightarrow\exists{t}_{\mathrm{0}} \in\right]{a},{b}\left[\mid{g}'\left({t}_{\mathrm{0}} \right)=\mathrm{0}\right. \\ $$ $${g}'\left({t}_{\mathrm{0}} \right)={y}'\left({t}_{\mathrm{0}} \right)\left({x}\left({a}\right)−{x}\left({b}\right)\right)−{x}'\left({t}_{\mathrm{0}} \right).\left({y}\left({a}\right)−{y}\left({b}\right)\right)=\mathrm{0} \\ $$ $${since}\:\gamma\left({a}\right)\neq\gamma\left({b}\right)\:{we}\:{have}\:\mathrm{3}\:{cases} \\ $$ $${if}\:{x}\left({a}\right)={x}\left({b}\right)\Rightarrow{x}'\left({t}_{\mathrm{0}} \right)=\mathrm{0} \\ $$ $${x}\left({a}\right)\neq{x}\left({b}\right)\:\&\:{y}\left({a}\right)−{y}\left({b}\right)=\mathrm{0}\Rightarrow{y}'\left({t}_{\mathrm{0}} \right)=\mathrm{0} \\ $$ $${x}\left({a}\right)#{x}\left({b}\right)\&{y}\left({a}\right)\neq{y}\left({b}\right)\Rightarrow \\ $$ $$\Leftrightarrow\frac{{y}'\left({t}_{\mathrm{0}} \right)}{{x}'\left({t}_{\mathrm{0}} \right)}=\frac{{y}\left({a}\right)−{y}\left({b}\right)}{{x}\left({a}\right)−{x}\left({b}\right)} \\ $$ $$\frac{{y}'\left({t}_{\mathrm{0}} \right)}{{x}'\left({t}_{\mathrm{0}} \right)}\:\:{is}\:{coeficuent}\:{of}\:{tangent}\:{in}\:{t}_{{o}} \\ $$ $$\Rightarrow{in}\:{t}_{\mathrm{0}} ,\gamma'\left({t}_{\mathrm{0}} \right)\:\:{is}\:{colinear}\:{to}\:\gamma\left({b}\right)−\gamma\left({a}\right) \\ $$ $$\left.\mathrm{2}\right)\:{calcule}\:\:{corbur} \\ $$ $$ \\ $$ $$ \\ $$ | ||