How do you find a point on the curve y = x^2 that is closest to the point? (16, 1/2)


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Question Number 139255 by bobhans last updated on 25/Apr/21

How do you find a point on the curve y = x^2 that is closest to the point? (16, 1/2)
	Answered by john_santu last updated on 25/Apr/21

	$T h e t a n g e n t t o p a r a b o l a a t i t s p o i n t$ $(r, r^{2}) h a s e q u a t i o n y = 2 r x - r^{2}$ $T h e n o r m a l f o r r \neq 0 w i l l h a v e$ $g r a d i e n t m = - \frac{1}{2 r} a n d i t w i l l h a v e$ $e q u a t i o n y = - \frac{x}{2 r} + r^{2} + \frac{1}{2}$ $s u c h a l i n e m u s t p a s s e s t h r o u g h$ $a t p o i n t (16, \frac{1}{2}) w e g e t$ $\frac{1}{2} = - \frac{16}{2 r} + r^{2} + \frac{1}{2} t h a t s i m p l i f i e s$ $t o r^{3} = 8 s o r = 2 a n d t h e p o i n t$ $o f m i n i m u m d i s t a n c e i s (2, 4)$
	Answered by mr W last updated on 25/Apr/21

	$d i s t a n c e f r o m p o i n t (16, \frac{1}{2}) t o p o i n t$ $(x, y) o n t h e c u r v e y = x^{2} i s d .$ $D = d^{2} = {(x - 16)}^{2} + {(y - \frac{1}{2})}^{2}$ $D = {(x - 16)}^{2} + {(x^{2} - \frac{1}{2})}^{2}$ $\frac{d D}{d x} = 2 (x - 16) + 2 (x^{2} - \frac{1}{2}) (2 x) = 0$ $x^{3} = 8$ $x = 2$ $y = 2^{2} = 4$ $\Rightarrow p o i n t (2, 4) i s c l o s e s t t o (16, \frac{1}{2}) .$
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