We are given the following functions for the distances traveled by the rocket cars:

• Distance from the starting point to car A: $a(t) = 4t^2 + 10t$
• Distance from the starting point to car B: $b(t) = 3t^3 + 7$

We are asked to find the rate at which the distance between the two cars is changing at $t = 15$ minutes.

Step 1: Define the distance between the two cars

The distance $d(t)$ between the two cars at any time $t$ can be found using the Pythagorean theorem, as the two cars are moving in perpendicular directions (south and west). The formula for the distance is:

$d(t) = \sqrt{a(t)^2 + b(t)^2}$

Step 2: Differentiate $d(t)$ with respect to time $t$

To find the rate at which the distance between the two cars is changing, we need to differentiate $d(t)$ with respect to $t$:

$\frac{dd}{dt} = \frac{d}{dt} \left( \sqrt{a(t)^2 + b(t)^2} \right)$

Using the chain rule, we have:

$\frac{dd}{dt} = \frac{1}{2} \cdot \frac{1}{\sqrt{a(t)^2 + b(t)^2}} \cdot 2a(t) \frac{da}{dt} + 2b(t) \frac{db}{dt}$

This simplifies to:

$\frac{dd}{dt} = \frac{a(t) \frac{da}{dt} + b(t) \frac{db}{dt}}{\sqrt{a(t)^2 + b(t)^2}}$

Step 3: Find $\frac{da}{dt}$ and $\frac{db}{dt}$

Now, we differentiate $a(t)$ and $b(t)$ with respect to time $t$:

• $a(t) = 4t^2 + 10t$

$\frac{da}{dt} = 8t + 10$

• $b(t) = 3t^3 + 7$

$\frac{db}{dt} = 9t^2$

Step 4: Substitute $t = 15$

Now, substitute $t = 15$ into $a(t), b(t), \frac{da}{dt},$ and $\frac{db}{dt}$:

• $a(15) = 4(15)^2 + 10(15) = 4(225) + 150 = 900 + 150 = 1050$ meters
• $b(15) = 3(15)^3 + 7 = 3(3375) + 7 = 10125 + 7 = 10132$ meters
• $\frac{da}{dt} = 8(15) + 10 = 120 + 10 = 130$ m/min
• $\frac{db}{dt} = 9(15)^2 = 9(225) = 2025$ m/min

Step 5: Calculate $d(15)$

Now, calculate the distance $d(15)$:

$d(15) = \sqrt{(1050)^2 + (10132)^2} = \sqrt{1102500 + 102659824} = \sqrt{103762324} \approx 10186.38 \text{ meters}$

Step 6: Calculate $\frac{dd}{dt}$ at $t = 15$

Now, substitute the values into the derivative formula:

$\frac{dd}{dt} = \frac{1050(130) + 10132(2025)}{10186.38}$

First, calculate the numerator:

$1050(130) = 136500, \quad 10132(2025) = 20567100$

$136500 + 20567100 = 20703600$

Now, calculate the rate of change:

$\frac{20703600}{10186.38} \approx 2033.58 \text{ m/min}$

Final Answer:

The rate at which the distance between the cars is changing 15 minutes after they start is approximately 2033.58 m/min.

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Would you like any further details on the solution or clarifications?

Here are 5 related questions you might find interesting:

1. How would the rate change if the cars were traveling at different speeds?
2. How does the rate of change evolve over time in this problem?
3. What would the rate of change be if the distances involved functions of different degrees?
4. How would the situation change if the cars were moving at angles other than 90 degrees?
5. What is the significance of using the Pythagorean theorem in motion problems like this?

Tip: When differentiating functions involving square roots, always apply the chain rule carefully to avoid mistakes!