A boat starts off 186 miles directly east from the city of Smithville. It travels due south at a speed of 27 miles per hour. After travelling 121 miles, how fast is the distance between the boat and Smithville changing?

To solve this problem, we need to determine how fast the distance between the boat and Smithville is changing when the boat has traveled 121 miles south.

### Step 1: Set Up the Problem
Let:
- \( x = 186 \) miles (the constant horizontal distance east of Smithville).
- \( y(t) \) be the vertical distance south of the starting point, where \( y(t) = 27t \) since the boat is traveling south at 27 miles per hour.
- \( z(t) \) be the distance between the boat and Smithville at time \( t \).

We can relate these using the Pythagorean theorem:
\[
z(t) = \sqrt{x^2 + y(t)^2}
\]
where
\[
z(t) = \sqrt{186^2 + (27t)^2}
\]

### Step 2: Differentiate with Respect to Time
To find how fast the distance \( z(t) \) is changing, differentiate both sides with respect to \( t \):
\[
\frac{dz}{dt} = \frac{1}{2\sqrt{186^2 + (27t)^2}} \cdot 2 \cdot 27t \cdot 27
\]
Simplifying this:
\[
\frac{dz}{dt} = \frac{27^2 t}{\sqrt{186^2 + (27t)^2}}
\]

### Step 3: Substitute the Given Values
The boat has traveled 121 miles, so:
\[
y(t) = 121 \text{ miles} \quad \text{and} \quad 121 = 27t
\]
Solve for \( t \):
\[
t = \frac{121}{27} \approx 4.481 \text{ hours}
\]

### Step 4: Compute \( \frac{dz}{dt} \)
Substitute \( t \) into the equation for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{27^2 \cdot 4.481}{\sqrt{186^2 + (27 \cdot 4.481)^2}}
\]

First, calculate the values:
- \( 27^2 = 729 \)
- \( 27 \cdot 4.481 \approx 121 \)
- \( 186^2 = 34596 \)
- \( 121^2 = 14641 \)

Then:
\[
\frac{dz}{dt} = \frac{729 \cdot 4.481}{\sqrt{34596 + 14641}} = \frac{3269.049}{\sqrt{49237}} \approx \frac{3269.049}{221.88} \approx 14.74 \text{ miles per hour}
\]

### Final Answer
The distance between the boat and Smithville is changing at a rate of approximately **14.74 miles per hour** after the boat has traveled 121 miles south.

Would you like more details on any step or have any questions?

Here are 5 related questions:
1. How does the Pythagorean theorem help in solving distance-related problems?
2. What would happen to the rate of change if the boat traveled at a different speed?
3. How can you calculate the time taken to travel a certain distance given a speed?
4. What is the significance of differentiating the distance function?
5. How would the problem change if the boat was traveling northwest instead?

**Tip:** In related rates problems, it's essential to clearly define all variables and their relationships before differentiating.

Learn how to calculate the rate of change in distance between a boat and Smithville using a related rates problem. This problem involves applying the Pythagorean theorem and differentiating with respect to time. Suitable for Grades 10-12.

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