Let consider γ :I→R^2 a parametric curve 1)Prove that if a<b and γ(a)≠γ(b) then there exist t_0 ∈]a,b[ such as γ′(t_0 ) is colinear to γ(b)−γ(a) 2)Show that if γ is regular and the function f :I→R t→f(t)=∣∣γ(t)−O(0,0) ∣∣ is maximal in t_0 ∈I Then ∣K_γ (t_0 )∣≥(1/(f(t


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Question Number 74300 by ~blr237~ last updated on 21/Nov/19

$L e t c o n s i d e r γ : I \to R^{2} a p a r a m e t r i c c u r v e$ $1) P r o v e t h a t i f a < b a n d γ (a) \neq γ (b) t h e n t h e r e e x i s t t_{0} \in] a, b [$ $s u c h a s γ^{'} (t_{0}) i s c o l i n e a r t o γ (b) - γ (a)$ $2) S h o w t h a t i f γ i s r e g u l a r a n d t h e f u n c t i o n f : I \to R t \to f (t) =∣∣ γ (t) - O (0, 0) ∣∣ i s m a x i m a l i n t_{0} \in I$ $T h e n ∣ K_{γ} (t_{0}) ∣⩾ \frac{1}{f (t_{0})}$
	Answered by mind is power last updated on 21/Nov/19

	$γ (t) = (x (t), y (t))$ $γ (a) \neq γ (b)$ $\Rightarrow x (a) \neq x (b) o r y (a) \neq y (b)$ $x (a) \neq x (b)$ $g (t) = y (t) (x (a) - x (b) - x (t) (y (a) - y (b))$ $g (a) = - y (a) x (b) + x (a) y (b)$ $g (b) = y (b) x (a) - x (b) y (a) = g (a)$ $g (a) = g (b) m e a n v a l u e s \Rightarrow \exists t_{0} \in] a, b [∣ g^{'} (t_{0}) = 0$ $g^{'} (t_{0}) = y^{'} (t_{0}) (x (a) - x (b)) - x^{'} (t_{0}) . (y (a) - y (b)) = 0$ $s i n c e γ (a) \neq γ (b) w e h a v e 3 c a s e s$ $i f x (a) = x (b) \Rightarrow x^{'} (t_{0}) = 0$ $x (a) \neq x (b) & y (a) - y (b) = 0 \Rightarrow y^{'} (t_{0}) = 0$ $You can't use 'macro parameter character #' in math mode$ $\Leftrightarrow \frac{y^{'} (t_{0})}{x^{'} (t_{0})} = \frac{y (a) - y (b)}{x (a) - x (b)}$ $\frac{y^{'} (t_{0})}{x^{'} (t_{0})} i s c o e f i c u e n t o f t a n g e n t i n t_{o}$ $\Rightarrow i n t_{0}, γ^{'} (t_{0}) i s c o l i n e a r t o γ (b) - γ (a)$ $2) c a l c u l e c o r b u r$
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