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(A) For each value of $x$,the helicopter's position is at point $(x, x^{2} + 7)$.Therefore,the distance $d$ between the helicopter and the soldier at

Vedclass · Accepted Answer

$\sqrt{5}$

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(A) For each value of $x$,the helicopter's position is at point $(x, x^{2} + 7)$.Therefore,the distance $d$ between the helicopter and the soldier at $(3, 7)$ is given by $d = \sqrt{(x - 3)^{2} + (x^{2} + 7 - 7)^{2}} = \sqrt{(x - 3)^{2} + x^{4}}$.Let $f(x) = (x - 3)^{2} + x^{4}$. To minimize the distance,we minimize $f(x)$.Taking the derivative,$f'(x) = 2(x - 3) + 4x^{3} = 2(x - 1)(2x^{2} + 2x + 3)$.Setting $f'(x) = 0$,we get $x = 1$ (since $2x^{2} + 2x + 3 = 0$ has no real roots as its discriminant $D = 4 - 24 = -20 The value of $f(x)$ at $x = 1$ is $f(1) = (1 - 3)^{2} + (1)^{4} = 4 + 1 = 5$.Thus,the minimum distance is $\sqrt{f(1)} = \sqrt{5}$.

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

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