Q.An Apache helicopter of enemy 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.

Concept: Distance Minimization – we minimise the squared distance to avoid the square root, then use calculus.

Let the helicopter’s position be $(x, x^2 + 7)$. The squared distance from $(3, 7)$ is

Differentiate:

Set to zero: $2x^3 + x - 3 = 0$. By inspection, $x = 1$ is a root. Factor: $(x - 1)(2x^2 + 2x + 3) = 0$. The quadratic has no real roots, so $x = 1$ is the only critical point. …

Why Distance Minimization Works

When a soldier wants to shoot a moving target at the moment it's closest, we're solving a pure geometry problem: find the point on the helicopter's path (the parabola $y = x^2 + 7$) that is nearest to the soldier's fixed position $(3, 7)$. The helicopter's motion along the curve is irrelevant — only the shape of the path matters.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Minimizing this is equivalent to minimizing its square (since the square root is a strictly increasing function for non-negative values), which avoids dealing with a square root in differentiation.

Step-by-Step Solution

1. Set up the distance function

Let the helicopter's position be $(x, y)$ on the curve $y = x^2 + 7$. The soldier is at $(3, 7)$. The squared distance is:

Substitute $y = x^2 + 7$:

So:

2. Differentiate to find critical points

Let $f(x) = (x - 3)^2 + x^4$. Differentiate with respect to $x$:

Set $f'(x) = 0$:

Divide through by 2:

So:

3. Solve the cubic equation

We need to solve $2x^3 + x - 3 = 0$. Try simple integer values:

• $x = 1$: $2(1)^3 + 1 - 3 = 2 + 1 - 3 = 0$. So $x = 1$ is a root.

Factor out $(x - 1)$ using polynomial division:

The quadratic $2x^2 + 2x + 3$ has discriminant $\Delta = 4 - 24 = -20 < 0$, so no real roots. Thus the only real critical point is $x = 1$.

Method: Minimizing Distance to a Curve

A general technique for finding the point on a curve closest to a fixed point, and the shortest distance between them.

Steps

Step 1: Write a general point on the curve and the distance to it

If the fixed point is $P=(a,b)$ and the curve is $y=f(x)$, a general point on the curve is $Q=(x, f(x))$. The distance between them is

Step 2: Minimize the squared distance instead

Since the square-root function is increasing, $D(x)$ and $D(x)^2$ are minimized at the same $x$-value. Work with $S(x) = D(x)^2$ instead — it avoids differentiating a square root and keeps the algebra polynomial.

Step 3: Differentiate, solve $S'(x)=0$, and verify it's a minimum

$S'(x) = 0$ …

Common Mistakes

Mistake 1: Differentiating the un-squared distance formula directly

Why it's wrong: working with $D(x)=\sqrt{(x-3)^2+x^4}$ and differentiating the square root directly leads to a messier expression and a higher chance of algebra errors, even though it eventually gives the same critical point. Correct approach: minimize the squared distance $D^2(x)=(x-3)^2+x^4$ instead — the $x$-value that minimizes $D^2$ is the same $x$ that minimizes $D$, and the derivative stays a clean polynomial.

Mistake 2: Forgetting to verify the critical point is actually a minimum …