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

We have to find the shortest distance between the fixed point $$P\left(\dfrac12,\,7\right)$$ and any point $$Q(x,y)$$ that lies on the helicopter’s path given by the curve $$y-x^{3/2}=7$$ with $$x\ge 0$$.

First, from the equation of the curve we can express $$y$$ in terms of $$x$$:

$$y-x^{3/2}=7\; \Longrightarrow\; y = 7 + x^{3/2}.$$

The square of the distance between two points $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ is, by the distance formula,

$$D^2 = (x_2-x_1)^2 + (y_2-y_1)^2.$$

Here $$x_1=\dfrac12,\; y_1=7,\; x_2=x,\; y_2=y=7+x^{3/2}.$$ Substituting these coordinates gives

$$D^2 = (x-\tfrac12)^2 +\bigl[(7+x^{3/2})-7\bigr]^2 = (x-\tfrac12)^2 + (x^{3/2})^2.$$

Simplifying the second term $$\bigl(x^{3/2}\bigr)^2$$ yields $$x^3,$$ so we can define

$$f(x)=D^2=(x-\tfrac12)^2 + x^3,\qquad x\ge 0.$$

To minimise the distance it is enough to minimise its square. Hence we differentiate $$f(x)$$ with respect to $$x$$ and set the derivative equal to zero:

$$\frac{d}{dx}\bigl[(x-\tfrac12)^2\bigr] = 2(x-\tfrac12),$$ $$\frac{d}{dx}\bigl[x^3\bigr] = 3x^2,$$

so

$$f'(x)=2(x-\tfrac12)+3x^2=2x-1+3x^2.$$

Setting $$f'(x)=0$$ gives the quadratic equation

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

Using the quadratic formula $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with $$a=3,\; b=2,\; c=-1,$$ we obtain

$$x=\frac{-2\pm\sqrt{4-4\cdot3\cdot(-1)}}{2\cdot3} =\frac{-2\pm\sqrt{4+12}}{6} =\frac{-2\pm4}{6}.$$

This yields two roots:

$$x=\frac{-2+4}{6}=\frac13,\qquad x=\frac{-2-4}{6}=-1.$$

Because $$x\ge 0,$$ we keep $$x=\dfrac13$$ and discard $$x=-1.$

To confirm that this critical point gives a minimum, we check the second derivative:

$$f''(x)=\frac{d}{dx}\bigl[2x-1+3x^2\bigr]=2+6x.$$

For $$x\ge 0,$$ clearly $$f''(x)>0,$$ so $$x=\dfrac13$$ indeed corresponds to the minimum.

Now we substitute $$x=\dfrac13$$ back into $$f(x)$$ to obtain the minimum value of $$D^2$$:

$$f\!\left(\frac13\right)=\Bigl(\frac13-\frac12\Bigr)^2+\Bigl(\frac13\Bigr)^3.$$ Compute each part:

$$\bigl(\tfrac13-\tfrac12\bigr)=\frac{2-3}{6}=-\frac16,$$ so $$\Bigl(-\frac16\Bigr)^2=\frac1{36}.$$

Also $$(\tfrac13)^3=\frac1{27}.$$

Adding them,

$$f\!\left(\frac13\right)=\frac1{36}+\frac1{27} =\frac3{108}+\frac4{108} =\frac7{108}.$$

Therefore the least possible squared distance is $$\dfrac7{108},$$ and the least distance itself is

$$D_{\min}=\sqrt{\frac7{108}} =\frac{\sqrt7}{\sqrt{108}} =\frac{\sqrt7}{\sqrt{36\cdot3}} =\frac{\sqrt7}{6\sqrt3} =\frac16\sqrt{\frac73}\,.$$

Hence, the correct answer is Option A.