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The problem you provided involves two cars moving in perpendicular directions, and we need to determine how fast the distance between the two cars is increasing at a specific moment.

### Given:
- **Car 1** is moving south at **45 miles per hour**.
- **Car 2** is moving west at **60 miles per hour**.
- When Car 1 is **3 miles** from Math-town and Car 2 is **4 miles** from Math-town, how fast is the distance between the two cars increasing?

### Step 1: Define variables
- Let \( x \) represent the distance of Car 2 (heading west) from Math-town. At the given moment, \( x = 4 \) miles.
- Let \( y \) represent the distance of Car 1 (heading south) from Math-town. At the given moment, \( y = 3 \) miles.
- Let \( z \) represent the distance between the two cars. We want to find \( \frac{dz}{dt} \), the rate at which the distance between the two cars is increasing.

### Step 2: Use the Pythagorean theorem
The two cars are moving in perpendicular directions, so their distances from Math-town form a right triangle with the distance between the two cars as the hypotenuse. The relationship between \( x \), \( y \), and \( z \) is given by:

\[
z^2 = x^2 + y^2
\]

### Step 3: Differentiate with respect to time
To find the rate at which the distance between the cars is increasing, we differentiate both sides of the equation with respect to time \( t \):

\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]

Simplifying:

\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]

### Step 4: Plug in known values
At the given moment:
- \( x = 4 \) miles,
- \( y = 3 \) miles,
- \( \frac{dx}{dt} = 60 \) miles per hour (speed of Car 2),
- \( \frac{dy}{dt} = 45 \) miles per hour (speed of Car 1).

First, we calculate \( z \), the distance between the two cars:

\[
z = \sqrt{x^2 + y^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ miles}.
\]

Now, substitute all the values into the differentiated equation:

\[
5 \frac{dz}{dt} = 4(60) + 3(45)
\]

\[
5 \frac{dz}{dt} = 240 + 135 = 375
\]

\[
\frac{dz}{dt} = \frac{375}{5} = 75 \text{ miles per hour}.
\]

### Conclusion:
The distance between the two cars is increasing at a rate of **75 miles per hour** when Car 1 is 3 miles and Car 2 is 4 miles from Math-town.

Would you like more details or have any other questions? Here are 5 related questions for deeper understanding:

1. What happens to the rate of distance change if Car 1 slows down to 30 miles per hour?
2. How would the scenario change if Car 2 was heading east instead of west?
3. Can you derive the equation if one of the cars was moving at a variable speed?
4. What would the total distance be after 1 hour of movement at the given speeds?
5. How would the solution change if the cars were not moving perpendicularly?

### Tip:
When solving related rates problems, always start by defining a geometric relationship (like the Pythagorean theorem in this case) and then differentiate with respect to time.

Car 1 is heading due south from Math-town at a rate of 45 miles per hour. Car 2 is heading due west from Math-town at a rate of 60 miles per hour. How fast is the distance between the cars increasing when Car 1 and Car 2 are 3 miles and 4 miles from Math-town, respectively?

This problem involves finding how fast the distance between two cars is increasing as they move perpendicularly from Math-town. The cars travel at 45 mph south and 60 mph west. Using related rates and the Pythagorean theorem, learn how to calculate the rate of change at the moment when one car is 3 miles and the other is 4 miles from the starting point.

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