Two rocket cars, A and B, start moving from the same point. Car A travels south and car B travels west.

At time t(in minutes), the distance(in meters) from the starting point to car A can be found by
a = 4t2 + 10t

At time t(in miutes), the distance(in meters) from the starting point to car B can be found by
b = 3t3 + 7

At what rate is the distance between the cars changing fifteen minutes after they start?
Give your answer to two decimal places.

 
Incorrect: Your answer is incorrect.
 m/min

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**.

---

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!

Two rocket cars, A and B, start moving from the same point. Car A travels south and car B travels west.

At time t (in minutes), the distance (in meters) from the starting point to car A can be found by
a = 4t^2 + 10t

At time t (in minutes), the distance (in meters) from the starting point to car B can be found by
b = 3t^3 + 7

At what rate is the distance between the cars changing fifteen minutes after they start? Give your answer to two decimal places.

Explore how to find the rate of change of distance between two rocket cars, A and B, as they move in perpendicular directions. Given the functions a(t) = 4t^2 + 10t and b(t) = 3t^3 + 7, learn how to apply the Pythagorean theorem and related rates to solve for the rate at t = 15 minutes. Suitable for students in Grades 11-12.

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