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: $\frac{dz}{dt} \approx 6.82 \, \text{m/s}$