To solve this problem, let’s break down the situation and apply the Pythagorean theorem to relate the positions of the police car and the truck.

1. Define Variables:

   • Let $x(t)$ represent the distance of the truck east of Sioux Falls at time $t$.
   • Let $y(t)$ represent the distance of the police car north of Sioux Falls at time $t$.
   • Let $z(t)$ represent the distance between the police car and the truck at time $t$.

2. Initial Conditions:

   • At $t = 0$, the police car is 60 km north of Sioux Falls, so $y(0) = 60$ km.
   • At $t = 0$, the truck is 50 km east of Sioux Falls, so $x(0) = 50$ km.

3. Rates of Change:

   • The truck moves east at 140 km/h, so $\frac{dx}{dt} = 140$ km/h.
   • The police car moves south toward Sioux Falls at 160 km/h, so $\frac{dy}{dt} = -160$ km/h (negative because it’s moving south).

4. Relationship between $x$, $y$, and $z$: Using the Pythagorean theorem: $z^2 = x^2 + y^2$ Differentiate both sides with respect to $t$: $2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$ Simplify by dividing by 2: $z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}$ Solving for $\frac{dz}{dt}$: $\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}$

5. Find $x$, $y$, and $z$ at $t = 10$ minutes:

   • Since $t = 10$ minutes is $\frac{1}{6}$ hours, we calculate $x$ and $y$ at this time.
   • $x(10) = 50 + 140 \times \frac{1}{6} = 50 + \frac{140}{6} = 50 + 23.333 = 73.333$ km
   • $y(10) = 60 - 160 \times \frac{1}{6} = 60 - \frac{160}{6} = 60 - 26.667 = 33.333$ km
   • Now, find $z(10)$: $z(10) = \sqrt{x(10)^2 + y(10)^2} = \sqrt{(73.333)^2 + (33.333)^2}$ $z(10) = \sqrt{5377.78 + 1111.11} = \sqrt{6488.89} \approx 80.599 \text{ km}$

6. Calculate $\frac{dz}{dt}$ at $t = 10$ minutes: Substitute the values: $\frac{dz}{dt} = \frac{(73.333)(140) + (33.333)(-160)}{80.599}$ $= \frac{10266.62 - 5333.28}{80.599} = \frac{4933.34}{80.599} \approx 61.205 \text{ km/h}$

Answer:

The rate at which the distance between the vehicles is changing at $t = 10$ minutes is approximately 61.205 km/h.

Would you like further details or have questions on specific steps?

Here are five related questions that expand on the concepts:

1. How would the rate of change in distance differ if the truck traveled in a different direction?
2. What if both vehicles were moving away from Sioux Falls at the same angle instead of perpendicular directions?
3. How would the solution change if we considered different initial distances from Sioux Falls for each vehicle?
4. Can we apply similar methods to determine when the police car would catch up to the truck?
5. How would this problem be solved if the speeds of the police car and truck were not constant?

Tip: In related rates problems, always start by identifying how quantities relate and differentiating with respect to time.