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 break down the problem and set up the necessary equations.

### Problem Setup:
1. The cat is running from a dog at a speed of \( 5 \, \text{m/s} \) and has a head start of \( 7 \, \text{m} \).
2. The cat reaches a corner and turns at a right angle.
3. We are asked to find the rate at which the diagonal distance between the cat and the dog is changing when the cat is \( 4 \, \text{m} \) past the corner.

Let's define the following variables:
- \( x \): the distance the cat has traveled past the corner (at the moment we are interested in, \( x = 4 \, \text{m} \)).
- \( y \): the distance between the dog and the corner.
- \( z \): the diagonal distance between the cat and the dog.

### Step 1: Set up the Relationship between \( x \), \( y \), and \( z \)

Since the cat turns at a right angle, \( x \), \( y \), and \( z \) form a right triangle. By the Pythagorean theorem:
\[
z^2 = x^2 + y^2
\]

### Step 2: Differentiate with Respect to Time \( t \)

We are interested in the rate of change of the diagonal distance \( z \), so let's differentiate both sides of the equation with respect to \( t \):
\[
\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)
\]
Using the chain rule, this becomes:
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Dividing by 2:
\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]

Solving for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}
\]

### Step 3: Substitute Known Values

#### 1. Speeds \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
- The cat's speed is \( \frac{dx}{dt} = 5 \, \text{m/s} \).
- Since the dog is always \( 7 \, \text{m} \) behind, the dog’s distance from the corner (along \( y \)) also changes at \( \frac{dy}{dt} = 5 \, \text{m/s} \).

#### 2. Current Values of \( x \), \( y \), and \( z \):
- \( x = 4 \, \text{m} \) (the cat is \( 4 \, \text{m} \) past the corner).
- \( y = 7 \, \text{m} \) (the dog’s initial lead distance).

Calculate \( z \) using the Pythagorean theorem:
\[
z = \sqrt{x^2 + y^2} = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65} \approx 8.06 \, \text{m}
\]

#### 3. Substitute into the Equation for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}
\]
\[
\frac{dz}{dt} = \frac{(4)(5) + (7)(5)}{8.06}
\]
\[
\frac{dz}{dt} = \frac{20 + 35}{8.06} = \frac{55}{8.06} \approx 6.82 \, \text{m/s}
\]

### Final Answer:
The rate at which the diagonal distance between the cat and the dog is changing when the cat is \( 4 \, \text{m} \) past the corner is approximately:
\[
\frac{dz}{dt} \approx 6.82 \, \text{m/s}
\]

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?

Learn how to determine the rate at which the diagonal distance between a cat and a dog is changing using related rates and the Pythagorean theorem. This problem involves key calculus concepts, including differentiation and the chain rule, suitable for students in Grades 10-12.

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