To solve this problem, we need to calculate the rate of change of the tracking angle $\theta$ at 12:30 PM.

Step 1: Define the variables

Let’s define the following:

• The ship's speed is 10 miles per hour, so in $t$ hours, it travels a distance of $x = 10t$ miles.
• The distance between the port and the radar station is 3 miles.
• The tracking angle $\theta$ is the angle between the line from the radar station to the ship and the shore.
• The ship starts moving at noon, and we need to find the rate of change of $\theta$ at 12:30 PM (i.e., when $t = 0.5$ hours).

We can model this problem using trigonometry and related rates. Let $x(t)$ represent the horizontal distance the ship has traveled after $t$ hours. The total distance from the radar station to the ship at any time $t$ forms a right triangle with the shore. The base of this triangle is $x(t) = 10t$, and the other leg is the fixed distance of 3 miles (the distance between the radar station and the port).

Step 2: Set up the equation

The tangent of the angle $\theta$ is the ratio of the ship's distance from the port to the fixed distance between the radar station and the port:

$\tan(\theta) = \frac{x(t)}{3}$ Since $x(t) = 10t$, we substitute:

$\tan(\theta) = \frac{10t}{3}$

Step 3: Differentiate with respect to time

To find the rate of change of $\theta$ with respect to time, we differentiate both sides of the equation with respect to $t$:

$\frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{10t}{3} \right)$

Using the chain rule, the derivative of $\tan(\theta)$ is:

$\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{10}{3}$

Step 4: Solve for $\frac{d\theta}{dt}$

Now we solve for $\frac{d\theta}{dt}$, which is the rate of change of the tracking angle:

$\frac{d\theta}{dt} = \frac{\frac{10}{3}}{\sec^2(\theta)}$

Next, we need to find $\sec^2(\theta)$. At 12:30 PM, $t = 0.5$ hours, so the ship has traveled:

$x(0.5) = 10 \times 0.5 = 5 \text{ miles}$

At this time, $\tan(\theta) = \frac{5}{3}$, so:

$\theta = \tan^{-1} \left( \frac{5}{3} \right)$

Now, use the identity $\sec^2(\theta) = 1 + \tan^2(\theta)$:

$\sec^2(\theta) = 1 + \left( \frac{5}{3} \right)^2 = 1 + \frac{25}{9} = \frac{34}{9}$

Step 5: Final calculation

Substitute $\sec^2(\theta) = \frac{34}{9}$ into the equation for $\frac{d\theta}{dt}$:

$\frac{d\theta}{dt} = \frac{\frac{10}{3}}{\frac{34}{9}} = \frac{10}{3} \times \frac{9}{34} = \frac{90}{102} = \frac{45}{51} = \frac{15}{17}$

Thus, the rate of change of the tracking angle at 12:30 PM is $\frac{15}{17}$ radians per hour.

Conclusion:

At 12:30 PM, the rate of change of the tracking angle $\theta$ is $\frac{15}{17}$ radians per hour.

Would you like more details on this, or have any questions?

Here are 5 related questions to expand your understanding:

1. How does the rate of change of $\theta$ vary as the ship gets farther from the shore?
2. How would the problem change if the ship's speed were not constant?
3. What happens to $\frac{d\theta}{dt}$ as $t$ approaches infinity?
4. How does the calculation change if the radar station is not 3 miles from the port?
5. What is the ship's distance from the radar station at any time $t$?

Tip: When working with related rates, always set up your equation based on known relationships (like trigonometric identities) before differentiating.