Two runner A and B start at the origin and run along positive x axis ,with B running three times as fast as A. An obsever, standeing one unit above the origin , keeps A and B in view.Then the maximum angle theta of sight between the observer's view of A and B is

Options

A. `pi//8`
B. `pi//6`
C. `pi//3`
D. `pi//4`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Define Variables
Let the speed of runner A be \( v_A \). Then, the speed of runner B, who runs three times as fast, will be \( v_B = 3v_A \).

### Step 2: Distance Traveled by Runners
At any time \( t \):
- The distance traveled by runner A is \( x = v_A t \).
- The distance traveled by runner B is \( 3x = 3v_A t \).

### Step 3: Set Up the Coordinate System
We can represent the positions of the runners in a coordinate system:
- Runner A's position at time \( t \) is \( (x, 0) \).
- Runner B's position at time \( t \) is \( (3x, 0) \).
- The observer is located at point \( O(0, 1) \).

### Step 4: Calculate Angles of Sight
We need to find the angles of sight from the observer to each runner:
- For runner A, the angle \( \theta_1 \):
\[ \tan(\theta_1) = \frac{y}{x} = \frac{1}{x} \]
- For runner B, the angle \( \theta_2 \):
\[ \tan(\theta_2) = \frac{y}{3x} = \frac{1}{3x} \]

### Step 5: Find the Angle Between the Lines of Sight
The angle \( \theta \) between the lines of sight to runners A and B can be expressed as:
\[ \tan(\theta) = \tan(\theta_2 - \theta_1) = \frac{\tan(\theta_2) - \tan(\theta_1)}{1 + \tan(\theta_1) \tan(\theta_2)} \]

### Step 6: Substitute the Values
Substituting the values of \( \tan(\theta_1) \) and \( \tan(\theta_2) \):
\[ \tan(\theta) = \frac{\frac{1}{3x} - \frac{1}{x}}{1 + \left(\frac{1}{x}\right)\left(\frac{1}{3x}\right)} \]
This simplifies to:
\[ \tan(\theta) = \frac{\frac{1 - 3}{3x}}{1 + \frac{1}{3x^2}} = \frac{-\frac{2}{3x}}{1 + \frac{1}{3x^2}} = \frac{-2}{3x + \frac{1}{x}} \]

### Step 7: Maximize the Angle
To find the maximum value of \( \tan(\theta) \), we need to maximize:
\[ y = \frac{-2}{3x + \frac{1}{x}} \]
We can differentiate this with respect to \( x \) and set the derivative to zero to find critical points.

### Step 8: Differentiate and Solve
Differentiating \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{-2}{3x + \frac{1}{x}}\right) \]
Using the quotient rule, we find:
\[ \frac{dy}{dx} = \frac{0 \cdot (3x + \frac{1}{x}) - (-2)(3 - \frac{1}{x^2})}{(3x + \frac{1}{x})^2} \]
Setting the numerator equal to zero gives:
\[ 2(3 - \frac{1}{x^2}) = 0 \implies 3 = \frac{1}{x^2} \implies x^2 = \frac{1}{3} \implies x = \frac{1}{\sqrt{3}} \]

### Step 9: Determine the Maximum Angle
Substituting \( x = \frac{1}{\sqrt{3}} \) back into the expression for \( \tan(\theta) \):
\[ \tan(\theta) = \frac{2\sqrt{3}}{3 + 3} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \]
Thus, the maximum angle \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \text{ radians} = 30^\circ \]

### Final Answer
The maximum angle \( \theta \) of sight between the observer's view of A and B is \( 30^\circ \).

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