Let's break down the solution step by step for part D of the problem, which involves related rates.

Step 1: Understand the problem

• The red car is moving north at 62 mph, and after 2 hours, it has traveled 124 miles.
• The blue car is moving east at 66 mph, and after 2 hours, it has traveled 132 miles.
• We already found that the distance between the two cars after 2 hours is 181.11 miles using the Pythagorean theorem.

Now we need to find how fast the distance between the cars is changing after 2 hours. This is a related rates problem, where the rate of change of the distance between the cars is a function of time.

Step 2: Set up the variables

• Let $x(t)$ be the distance the blue car has traveled (in the east direction) at time $t$.
• Let $y(t)$ be the distance the red car has traveled (in the north direction) at time $t$.
• Let $z(t)$ be the distance between the two cars at time $t$.

From the Pythagorean theorem, the relationship between $x(t)$, $y(t)$, and $z(t)$ is:

$z^2(t) = x^2(t) + y^2(t)$

Step 3: Differentiate with respect to time

To find the rate at which the distance between the cars is changing, we differentiate both sides of the equation with respect to time $t$:

[ \frac{d}{dt}[z^2(t)] = \frac{d}{dt}[x^2(t) + y^2(t)] ]

Using the chain rule, this becomes:

$2z(t) \frac{dz}{dt} = 2x(t) \frac{dx}{dt} + 2y(t) \frac{dy}{dt}$

Simplifying:

$z(t) \frac{dz}{dt} = x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt}$

Step 4: Plug in the values

From the given information:

• $x(t) = 132$ miles (distance traveled by the blue car),
• $y(t) = 124$ miles (distance traveled by the red car),
• $z(t) = 181.11$ miles (the distance between the two cars after 2 hours),
• $\frac{dx}{dt} = 66$ mph (speed of the blue car),
• $\frac{dy}{dt} = 62$ mph (speed of the red car).

Substitute these values into the equation:

$181.11 \cdot \frac{dz}{dt} = 132 \cdot 66 + 124 \cdot 62$

Simplify the right-hand side:

$181.11 \cdot \frac{dz}{dt} = 8712 + 7688 = 16400$

Now, solve for $\frac{dz}{dt}$ (the rate at which the distance between the cars is changing):

$\frac{dz}{dt} = \frac{16400}{181.11} \approx 90.55 \text{ mph}$

Final Answer:

The distance between the two cars is changing at approximately 90.55 miles per hour.

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Would you like more details on any part of this solution or further clarification?

Here are five related questions you could explore:

1. How would the answer change if the blue car were traveling west instead of east?
2. What is the distance between the cars after 3 hours?
3. How does the rate of change of the distance between the cars behave as time increases?
4. How can you generalize this approach for cars moving in any direction?
5. What if the speeds of the cars were time-dependent functions?

Tip: When solving related rates problems, always differentiate the given equation with respect to time, even if time isn't explicitly in the equation.