Key Concepts and Formulas

• Distance Formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Derivative as Slope: The derivative of a function $y = f(x)$, denoted as $\frac{dy}{dx}$, represents the slope of the tangent line to the curve at a given point $(x, f(x))$.
• Perpendicular Lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 \cdot m_2 = -1$.

Step-by-Step Solution

Step 1: Define the curve and the fixed point. We are given the curve $y - x^{3/2} = 7$ or $y = 7 + x^{3/2}$, where $x \ge 0$. The soldier's position is at the point $P\left(\frac{1}{2}, 7\right)$. We want to find the point $Q(x, y)$ on the curve that minimizes the distance to $P$.

Step 2: Express the distance between the point P and a general point Q on the curve. The distance $d$ between $P\left(\frac{1}{2}, 7\right)$ and $Q(x, y)$ is given by $d = \sqrt{\left(x - \frac{1}{2}\right)^2 + (y - 7)^2}$. Since $y = 7 + x^{3/2}$, we can substitute this into the distance formula: $d = \sqrt{\left(x - \frac{1}{2}\right)^2 + (7 + x^{3/2} - 7)^2} = \sqrt{\left(x - \frac{1}{2}\right)^2 + (x^{3/2})^2} = \sqrt{\left(x - \frac{1}{2}\right)^2 + x^3}$ To minimize $d$, it suffices to minimize $d^2$. Let $D = d^2 = \left(x - \frac{1}{2}\right)^2 + x^3 = x^2 - x + \frac{1}{4} + x^3$.

Step 3: Find the critical points by differentiating D with respect to x. To find the minimum distance, we need to find the critical points of $D(x)$ by taking the derivative with respect to $x$ and setting it equal to zero: $\frac{dD}{dx} = \frac{d}{dx} \left(x^3 + x^2 - x + \frac{1}{4}\right) = 3x^2 + 2x - 1$ Setting $\frac{dD}{dx} = 0$, we get $3x^2 + 2x - 1 = 0$. Factoring the quadratic gives $(3x - 1)(x + 1) = 0$. Thus, $x = \frac{1}{3}$ or $x = -1$. Since $x \ge 0$, we have $x = \frac{1}{3}$.

Step 4: Verify that x = 1/3 corresponds to a minimum using the second derivative test. We compute the second derivative of $D$ with respect to $x$: $\frac{d^2D}{dx^2} = \frac{d}{dx}(3x^2 + 2x - 1) = 6x + 2$ Evaluating the second derivative at $x = \frac{1}{3}$, we get $\frac{d^2D}{dx^2}\Big|_{x = \frac{1}{3}} = 6\left(\frac{1}{3}\right) + 2 = 2 + 2 = 4 > 0$. Since the second derivative is positive, we have a local minimum at $x = \frac{1}{3}$.

Step 5: Calculate the y-coordinate of the point Q and then the minimum distance. When $x = \frac{1}{3}$, we have $y = 7 + x^{3/2} = 7 + \left(\frac{1}{3}\right)^{3/2} = 7 + \frac{1}{3\sqrt{3}}$. Now, we find the minimum distance $d$ using $x = \frac{1}{3}$: $d = \sqrt{\left(\frac{1}{3} - \frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^3} = \sqrt{\left(-\frac{1}{6}\right)^2 + \frac{1}{27}} = \sqrt{\frac{1}{36} + \frac{1}{27}} = \sqrt{\frac{3 + 4}{108}} = \sqrt{\frac{7}{108}} = \sqrt{\frac{7}{36 \cdot 3}} = \frac{1}{6}\sqrt{\frac{7}{3}}$

Common Mistakes & Tips

• Remember to minimize the square of the distance to simplify calculations and avoid dealing with square roots until the end.
• Always check the domain of the variable and discard any solutions that fall outside the domain.
• Use the second derivative test to confirm that you have found a minimum.

Summary We found the point on the curve $y = 7 + x^{3/2}$ closest to the point $(\frac{1}{2}, 7)$ by minimizing the squared distance between a general point on the curve and the given point. We found the critical point by taking the derivative of the squared distance and setting it equal to zero. We then used the second derivative test to confirm that this critical point corresponded to a minimum. Finally, we calculated the minimum distance.

Final Answer The final answer is $\boxed{\frac{1}{6}\sqrt{\frac{7}{3}}}$, which corresponds to option (A).