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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