In a straight alley, a cat is running from a dog at a speed of 5 m/s and maintains a lead of
7 m. The cat reached a corner and turned at a right angle. At what rate is the diagonal
distance between the cat and the dog changing when the cat is 4 m past the corner?

Let's set up the problem by defining the scenario and using the Pythagorean theorem to express the distance between the dog and the cat as they move.

1. **Define Variables**:
   - Let \( x \) be the distance the dog has traveled down the alley.
   - Let \( y \) be the distance the cat has traveled after turning the corner (perpendicular to the dog's path).
   - The diagonal distance between the dog and the cat, denoted as \( d \), can be expressed by the Pythagorean theorem:
     \[
     d = \sqrt{x^2 + y^2}
     \]

2. **Rates of Change**:
   - The dog is running at a speed of \( 5 \ \text{m/s} \), so \( \frac{dx}{dt} = 5 \ \text{m/s} \).
   - After the cat turns the corner, it continues to run at the same speed, \( 5 \ \text{m/s} \), so \( \frac{dy}{dt} = 5 \ \text{m/s} \).

3. **Initial Condition**:
   - The initial distance between the dog and the cat along the alley is 7 meters, so initially, \( x = 7 \ \text{m} \).

4. **When the Cat is 4 Meters Past the Corner**:
   - When the cat is 4 meters past the corner, \( y = 4 \ \text{m} \).

Now we need to find the rate at which the diagonal distance \( d \) between the cat and the dog is changing at this moment, \( \frac{dd}{dt} \).

5. **Differentiate the Distance Equation**:
   We start with the equation:
   \[
   d = \sqrt{x^2 + y^2}
   \]
   Taking the derivative with respect to time \( t \) on both sides, we get:
   \[
   \frac{dd}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt})
   \]
   Simplify:
   \[
   \frac{dd}{dt} = \frac{x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}}{\sqrt{x^2 + y^2}}
   \]

6. **Substitute Values**:
   - \( x = 7 \ \text{m} \)
   - \( y = 4 \ \text{m} \)
   - \( \frac{dx}{dt} = 5 \ \text{m/s} \)
   - \( \frac{dy}{dt} = 5 \ \text{m/s} \)

   Substitute these into the equation:
   \[
   \frac{dd}{dt} = \frac{(7)(5) + (4)(5)}{\sqrt{7^2 + 4^2}}
   \]

7. **Calculate**:
   - First, find \( \sqrt{7^2 + 4^2} \):
     \[
     \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65}
     \]
   - Then, evaluate \( \frac{dd}{dt} \):
     \[
     \frac{dd}{dt} = \frac{35 + 20}{\sqrt{65}} = \frac{55}{\sqrt{65}}
     \]

   Simplifying further, we approximate:
   \[
   \frac{55}{\sqrt{65}} \approx 6.82 \ \text{m/s}
   \]

### Answer:
The rate at which the diagonal distance between the cat and the dog is changing when the cat is 4 meters past the corner is approximately **6.82 m/s**.

Would you like more details or have questions about this solution?

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**Relative Questions**:
1. How would the rate change if the cat and dog had different speeds?
2. What if the initial distance between the cat and dog was greater?
3. How is this problem similar to or different from a typical related rates problem?
4. Can we generalize this problem for any initial distance and speed?
5. How does changing the angle at the corner affect the rate calculation?

**Tip**: When solving related rates problems, setting up clear variables and defining the relationships between them is key to finding the correct derivatives.

In a straight alley, a cat is running from a dog at a speed of 5 m/s and maintains a lead of 7 m. The cat reached a corner and turned at a right angle. At what rate is the diagonal distance between the cat and the dog changing when the cat is 4 m past the corner?

Explore a related rates problem involving a cat and dog in motion. Learn how to calculate the rate of change of the diagonal distance between them using the Pythagorean theorem and differentiation. Perfect for students in Grades 11-12.

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