Question

An Apache helicopter of the enemy is flying along the curve \[ y = x^2 + 7. \] A soldier placed at the point \( (3, 7) \), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.

Hint:

To find the nearest distance, minimize the square of the distance function, which is easier to differentiate.

Answer 1

The distance between the point \( (x, y) \) on the curve and the soldier at \( (3, 7) \) is given by the distance formula: \[ d = \sqrt{(x - 3)^2 + (y - 7)^2} \] Substitute \( y = x^2 + 7 \) into the distance formula: \[ d = \sqrt{(x - 3)^2 + (x^2 + 7 - 7)^2} = \sqrt{(x - 3)^2 + x^4} \] We want to minimize this distance, so we minimize \( d^2 \) to avoid dealing with the square root. Define the function: \[ f(x) = (x - 3)^2 + x^4 = x^2 - 6x + 9 + x^4 \] To minimize \( f(x) \), take the derivative and set it to zero: \[ \frac{d}{dx} f(x) = 4x^3 + 2x - 6 = 0 \] Factor the equation: \[ 2x(2x^2 + 1) = 6 $\Rightarrow$ x(2x^2 + 1) = 3 \] Solving this equation, we find that \( x = 1 \). Now substitute \( x = 1 \) into the distance formula: \[ y = 1^2 + 7 = 8 \] Thus, the point on the curve is \( (1, 8) \). The nearest distance is: \[ d = \sqrt{(1 - 3)^2 + (8 - 7)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \] Final Answer: The nearest distance is \( \boxed{\sqrt{5}} \).