A Helicopter is flying along the curve given by $y = x^{2} + 7$ . A solder placed at (3, 7) wants to shoot out the helicopter when it is nearest to him. Find the nearest distance

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Solution

$c u r v e i s g i v e n a s y = x^{2} + 7 l e t h e l i c o p t e r b e a t p o i n t (x, y) = (x, x^{2} + 7) l e t d b e t h e d i s t a n c e b e t w e e n h e l i c o p t e r s o l d i e r a t (3, 7) . d = \sqrt{{(x 2 - x 1)}^{2} + {(y 2 - y 1)}^{2}} d = \sqrt{{(x - 3)}^{2} + {((x^{2} + 7) - 7)}^{2}} d = \sqrt{{(x - 3)}^{2} + {(x)}^{4}} l e t f (x) i s m i n i m u m, d i s m i n i m u m . F i n d f^{^{'}} (x) f^{^{'}} (x) = 2 (x - 3) + 4 x^{3} = 2 x - 6 + 4 x^{3} = 4 x^{3} + 2 x - 6 = (x - 1) (4 x^{2} + 4 x + 6) x = 1, x = \frac{- 2 \pm \sqrt{4 - 4 (2) (3)}}{4} = - 2 \pm \frac{\sqrt{- 20}}{4} (N o t p o s s i b l e) ∴ x = 1$
This is either minima or maxima.
$f^{^{''}} (x) = 12 x^{2} + 2 ∴ f^{11} (1) = 12 + 2 = 14 x = 1 i s m i n i m a . v a l u e o f f (1) i s f (1) = {(1 - 3)}^{2} + 1^{4} = {(- 2)}^{2} + 1 = 5$
Minimum distance between soldier and helicopter
$d = \sqrt{f (1)} \Rightarrow d = \sqrt{5}$

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