A port and a radar station are 3 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of 10 ​mi/hr. If the ship maintains its speed and​ course, what is the rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at​ 12:30 PM?

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.

Learn how to calculate the rate of change of the tracking angle θ as a ship moves away from a port. This problem involves related rates, trigonometry, and calculus concepts, using formulas like tan(θ) = x(t)/3 and sec²(θ) to find dθ/dt. Ideal for advanced high school students studying calculus.

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