A helicopter flying along the path `y=7+x^((3)/(2))`, A soldier standint at point `((1)/(2),7)` wants to hit the helicopter when it is closest from him, then minimum distance is equal to

Options

A. a. `1/2`
B. b. `1/3sqrt(7/3)`
C. c. `1/6sqrt(7/3)`
D. d. `sqrt5/6`

Answer

To solve the problem, we need to find the minimum distance between the soldier standing at point \((\frac{1}{2}, 7)\) and the helicopter flying along the path given by the equation \(y = 7 + x^{\frac{3}{2}}\).

### Step 1: Define the coordinates of the helicopter
Let the coordinates of the helicopter be \(P(\alpha, 7 + \alpha^{\frac{3}{2}})\), where \(\alpha\) is the x-coordinate of the helicopter's position.

### Step 2: Write the distance formula
The distance \(D\) between the soldier at point \(A(\frac{1}{2}, 7)\) and the helicopter at point \(P(\alpha, 7 + \alpha^{\frac{3}{2}})\) can be expressed using the distance formula:
\[ D = \sqrt{(\alpha - \frac{1}{2})^2 + (7 + \alpha^{\frac{3}{2}} - 7)^2} \]
This simplifies to:
\[ D = \sqrt{(\alpha - \frac{1}{2})^2 + (\alpha^{\frac{3}{2}})^2} \]
\[ D = \sqrt{(\alpha - \frac{1}{2})^2 + \alpha^3} \]

### Step 3: Minimize the distance
To minimize \(D\), we can minimize \(D^2\) (since the square root function is monotonically increasing):
\[ D^2 = (\alpha - \frac{1}{2})^2 + \alpha^3 \]

### Step 4: Differentiate \(D^2\)
Now, we differentiate \(D^2\) with respect to \(\alpha\):
\[ \frac{d(D^2)}{d\alpha} = 2(\alpha - \frac{1}{2}) + 3\alpha^2 \]

### Step 5: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 2(\alpha - \frac{1}{2}) + 3\alpha^2 = 0 \]
\[ 3\alpha^2 + 2\alpha - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 3\), \(b = 2\), and \(c = -1\):
\[ \alpha = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ = \frac{-2 \pm \sqrt{4 + 12}}{6} \]
\[ = \frac{-2 \pm \sqrt{16}}{6} \]
\[ = \frac{-2 \pm 4}{6} \]
This gives us two possible values for \(\alpha\):
\[ \alpha_1 = \frac{2}{6} = \frac{1}{3}, \quad \alpha_2 = \frac{-6}{6} = -1 \quad (\text{not valid since } \alpha \geq 0) \]

### Step 7: Find the coordinates of point P
Using \(\alpha = \frac{1}{3}\):
\[ y = 7 + \left(\frac{1}{3}\right)^{\frac{3}{2}} = 7 + \frac{1}{\sqrt{27}} = 7 + \frac{1}{3\sqrt{3}} \]
So, the coordinates of point \(P\) are \(\left(\frac{1}{3}, 7 + \frac{1}{3\sqrt{3}}\right)\).

### Step 8: Calculate the minimum distance
Now, we can calculate the minimum distance \(D\) using the distance formula:
\[ D = \sqrt{\left(\frac{1}{3} - \frac{1}{2}\right)^2 + \left(7 + \frac{1}{3\sqrt{3}} - 7\right)^2} \]
\[ = \sqrt{\left(-\frac{1}{6}\right)^2 + \left(\frac{1}{3\sqrt{3}}\right)^2} \]
\[ = \sqrt{\frac{1}{36} + \frac{1}{27}} \]
Finding a common denominator (which is 108):
\[ = \sqrt{\frac{3}{108} + \frac{4}{108}} = \sqrt{\frac{7}{108}} = \frac{\sqrt{7}}{6\sqrt{3}} = \frac{\sqrt{21}}{18} \]

Thus, the minimum distance is \(\frac{\sqrt{21}}{18}\).