A helicopter is flying along the curve given by y - x³/² = 7, (x ≥ 0). A soldier positioned at the p

Q: Why can we minimize $D^2$ instead of $D$? Isn't that changing the problem?

Great question! The distance $D$ is always non-negative. The function $g(D) = D^2$ is a strictly increasing function for $D \ge 0$. Therefore, the value of $x$ that minimizes $D^2$ must also minimize $D$. It is a powerful shortcut that saves you from the 'chain rule nightmare' of differentiating square roots.

Q: Why do we reject $x = -1$?

Look closely at the problem statement: $x \ge 0$. The helicopter's path is only defined for non-negative $x$. Mathematically, $x = -1$ is a valid critical point for the polynomial, but physically, it lies outside the domain of the helicopter's flight path. Always check your constraints before finalizing your answer!

Q: Is there a geometric way to solve this without calculus?

Yes! The shortest distance from a point to a curve is always along the normal to the curve at that point. If you were to find the slope of the tangent at $Q$ and ensure the line $PQ$ is perpendicular to it (i.e., the product of slopes is $-1$), you would arrive at the same result. Calculus is just the systematic way of finding that normal.

Question

A helicopter is flying along the curve given by $y - x^{3/2} = 7, (x \ge 0)$. A soldier positioned at the point $(1/2, 7)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

Options:

(A) 1/2

(B) $\frac{1}{3}\sqrt{7/3}$

(C) $\frac{1}{6}\sqrt{7}$

(D) $\frac{\sqrt{5}}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, open field. Above you, a helicopter is tracing a precise, mathematical trajectory through the sky. Its path is governed by the equation $y - x^{3/2} = 7$, where $x \ge 0$.

You are a soldier stationed at a fixed coordinate $P(\frac{1}{2}, 7)$. Your mission is to calculate the exact moment the helicopter is closest to you to ensure your shot is accurate. This is a problem of optimization, a dance between algebra and geometry.

### Defining the Target

To solve this, we define a point $Q$ on the curve. Since $Q$ must satisfy the equation of the path, its coordinates are linked. If we choose an arbitrary $x$-coordinate, the $y$-coordinate is forced by the equation $y = x^{3/2} + 7$.

Thus, our point $Q$ is defined as $(x, x^{3/2} + 7)$. We invoke the distance formula to find the distance $D$ between the soldier at $P(\frac{1}{2}, 7)$ and the helicopter at $Q(x, x^{3/2} + 7)$:

$$D = \sqrt{(x - \frac{1}{2})^2 + (x^{3/2} + 7 - 7)^2}$$

The constant $7$ in the $y$-coordinates cancels out perfectly. We are left with:

$$D = \sqrt{(x - \frac{1}{2})^2 + (x^{3/2})^2}$$

Simplifying the second term, $(x^{3/2})^2$ becomes $x^3$. Our distance function is $D = \sqrt{(x - \frac{1}{2})^2 + x^3}$.

### The Optimization Strategy

Differentiating $D$ directly involves the chain rule and a messy square root. Instead, we recognize that the value of $x$ that minimizes $D$ will also minimize $D^2$. We define a new function, $f(x)$, representing the squared distance:

$$f(x) = D^2 = (x - \frac{1}{2})^2 + x^3$$

Expanding this polynomial, we get:

$$f(x) = x^2 - x + \frac{1}{4} + x^3$$

Rearranging it into standard form, we have:

$$f(x) = x^3 + x^2 - x + \frac{1}{4}$$

### Finding the Critical Moment

To find the minimum, we differentiate $f(x)$ with respect to $x$:

$$f'(x) = \frac{d}{dx}(x^3 + x^2 - x + \frac{1}{4}) = 3x^2 + 2x - 1$$

For the distance to be at its minimum, we set the derivative to zero:

$$3x^2 + 2x - 1 = 0$$

We factor the quadratic by splitting the middle term:

$$3x^2 + 3x - x - 1 = 0$$
$$3x(x + 1) - 1(x + 1) = 0$$
$$(3x - 1)(x + 1) = 0$$

This yields two critical points: $x = \frac{1}{3}$ and $x = -1$.

### The Final Decision

The problem explicitly states that $x \ge 0$. The value $x = -1$ is physically impossible for the helicopter's path, so we discard it. The helicopter is closest when $x = \frac{1}{3}$.

Now, we substitute $x = \frac{1}{3}$ back into our squared distance function $f(x)$:

$$D^2 = (\frac{1}{3} - \frac{1}{2})^2 + (\frac{1}{3})^3$$
$$D^2 = (-\frac{1}{6})^2 + \frac{1}{27} = \frac{1}{36} + \frac{1}{27}$$

Finding the least common multiple of $36$ and $27$, which is $108$, we get:

$$D^2 = \frac{3}{108} + \frac{4}{108} = \frac{7}{108}$$

Finally, we take the square root to find the actual distance $D$:

$$D = \sqrt{\frac{7}{108}} = \sqrt{\frac{7}{36 	imes 3}} = \mathbf{\frac{1}{6}\sqrt{\frac{7}{3}}}$$

Animated Solution for Mathematics - Differentiation: A helicopter is flying along the curve given by $y - x^{3/2} = 7, (x \geq 0)$ . A soldier positioned at the point $(1/2, 7)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

Select Answer:

Visualized Solution

Visualize the Problem

Define a General Point $Q$

Set up the Distance Formula

Simplify the Distance Expression

Minimize the Square of Distance

Differentiate to find Critical Points

Set Derivative to Zero

Solve the Quadratic Equation

Select the Valid Root

Calculate the Minimum Distance Squared

Evaluate the Minimum Distance Squared

Final Calculation

The Way Forward

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Defining the Target

The Optimization Strategy

Finding the Critical Moment

The Final Decision

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