A police car is approaching a right-angled intersection from the north at 35 mph, and a speeding car is heading east at 30 mph. When the police car is 0.75 miles north of the intersection and the speeding car is 1 mile to the east, what is the rate that the distance between the police car and car is changing?

To solve this problem, we can use the concept of related rates in calculus. Here's the setup:

### Problem Setup:
Let:
- \( x(t) \) be the distance of the speeding car from the intersection, where \( x(t) = 1 \) mile initially.
- \( y(t) \) be the distance of the police car from the intersection, where \( y(t) = 0.75 \) miles initially.
- \( z(t) \) be the distance between the two cars, which we want to find the rate of change of.

The rates of change:
- The speeding car is moving east, so \( \frac{dx}{dt} = -30 \) mph (negative because the distance \( x(t) \) is decreasing).
- The police car is moving south towards the intersection, so \( \frac{dy}{dt} = -35 \) mph (negative because the distance \( y(t) \) is decreasing).

### Applying the Pythagorean Theorem:
Since the cars form a right triangle with the intersection:
\[
z^2(t) = x^2(t) + y^2(t)
\]

Taking the derivative with respect to time \( t \) on both sides:
\[
2z(t) \frac{dz}{dt} = 2x(t) \frac{dx}{dt} + 2y(t) \frac{dy}{dt}
\]
\[
z(t) \frac{dz}{dt} = x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt}
\]

### Plugging in the Values:
- \( x(t) = 1 \) mile
- \( y(t) = 0.75 \) miles
- \( \frac{dx}{dt} = -30 \) mph
- \( \frac{dy}{dt} = -35 \) mph

First, calculate \( z(t) \) using the Pythagorean theorem:
\[
z(t) = \sqrt{x^2(t) + y^2(t)} = \sqrt{1^2 + 0.75^2} = \sqrt{1 + 0.5625} = \sqrt{1.5625} \approx 1.25 \text{ miles}
\]

Now, substitute into the related rates equation:
\[
1.25 \frac{dz}{dt} = (1)(-30) + (0.75)(-35)
\]
\[
1.25 \frac{dz}{dt} = -30 - 26.25
\]
\[
1.25 \frac{dz}{dt} = -56.25
\]
\[
\frac{dz}{dt} = \frac{-56.25}{1.25} = -45 \text{ mph}
\]

### Conclusion:
The rate at which the distance between the two cars is changing is \(-45\) mph. The negative sign indicates that the distance between the two cars is decreasing at a rate of 45 mph.

Would you like more details or have any questions?

### Related Questions:
1. What if the speeding car was going faster, say at 50 mph?
2. How would the rate change if the police car was farther from the intersection?
3. What if the police car was moving south at 50 mph?
4. How does the distance between the cars change over time if both cars maintain their speeds?
5. What would happen if the cars were initially closer to each other?

**Tip:** When solving related rates problems, always start by identifying the relationship between the variables using geometry or physics, then differentiate with respect to time.

Explore how to calculate the rate of change between a police car and a speeding car approaching an intersection. Learn the application of calculus and the Pythagorean theorem in solving related rates problems. Suitable for advanced high school and college students.

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