We can solve this problem using related rates and the Law of Sines. Here's a step-by-step solution:

Step 1: Define the variables

• Let $t$ represent time in hours after noon.
• Let $x(t)$ be the distance between the ship and the port at time $t$. Since the ship travels at a constant speed of 17 mi/h, we have: $x(t) = 17t$
• The distance between the radar station and the port is a constant 5 miles.
• Let $\theta(t)$ be the angle between the shore (connecting the radar station to the port) and the line connecting the radar station to the ship.

Step 2: Set up the situation at 12:30 PM

At 12:30 PM, the time $t = 0.5$ hours since noon. The distance the ship has traveled is: $x(0.5) = 17 \times 0.5 = 8.5 \, \text{miles}$ Thus, the ship is 8.5 miles from the port.

Now, we have a triangle with:

• One side (the distance from the port to the ship) = 8.5 miles,
• Another side (the distance between the port and the radar station) = 5 miles,
• An angle $\theta(t)$ between the shore and the line connecting the radar station to the ship.

Step 3: Apply the Law of Sines

The Law of Sines states: $\frac{\sin(\theta)}{x} = \frac{\sin(\pi/2)}{d}$ where:

• $x$ is the distance from the ship to the port (8.5 miles at 12:30 PM),
• $d = 5$ miles is the distance between the radar station and the port,
• $\pi/2$ radians is the right angle between the shore and the port-radar station line.

Thus, we get: $\sin(\theta) = \frac{5}{8.5}$ This simplifies to: $\sin(\theta) = 0.588$

Step 4: Differentiate with respect to time

We now differentiate both sides of the Law of Sines equation with respect to $t$, using implicit differentiation. We will use the chain rule to differentiate $\sin(\theta(t))$, keeping in mind that $\frac{d}{dt}[\sin(\theta)] = \cos(\theta) \frac{d\theta}{dt}$.

The Law of Sines gives us: $\frac{\sin(\theta)}{x} = \frac{1}{5}$

Differentiating both sides with respect to $t$ gives: $\frac{d}{dt} \left( \frac{\sin(\theta)}{x} \right) = 0$ Using the product rule, we get: $\frac{\cos(\theta) \cdot \frac{d\theta}{dt}}{x} - \frac{\sin(\theta) \cdot \frac{dx}{dt}}{x^2} = 0$ Simplifying: $\frac{\cos(\theta) \cdot \frac{d\theta}{dt}}{x} = \frac{\sin(\theta) \cdot \frac{dx}{dt}}{x^2}$

Step 5: Substitute known values

At 12:30 PM:

• $\sin(\theta) = 0.588$,
• $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - 0.588^2} = \sqrt{1 - 0.345} = 0.809$,
• $\frac{dx}{dt} = 17 \, \text{mi/hr}$,
• $x = 8.5 \, \text{miles}$.

Substitute these into the equation: $\frac{0.809 \cdot \frac{d\theta}{dt}}{8.5} = \frac{0.588 \cdot 17}{(8.5)^2}$ Simplifying: $\frac{0.809 \cdot \frac{d\theta}{dt}}{8.5} = \frac{9.996}{72.25}$ $\frac{d\theta}{dt} = \frac{9.996}{72.25} \times 8.5 \div 0.809$ $\frac{d\theta}{dt} = \frac{84.966}{58.819}$ $\frac{d\theta}{dt} \approx 1.445 \, \text{radians per hour}.$

Final Answer:

The rate of change of the tracking angle $\theta$ at 12:30 PM is approximately $1.445 \, \text{radians per hour}$.

Do you need more details or have any questions?

Related Questions:

1. What happens if the ship changes its speed?
2. How does the rate of change of the angle vary with different distances?
3. What if the distance between the radar station and the port changes?
4. How can this problem be solved using Cartesian coordinates?
5. What would the tracking angle rate be if the ship was moving slower?

Tip:

When solving related rates problems, always differentiate with respect to time and use the chain rule to handle implicitly defined variables.