A​ hot-air balloon is 130 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes directly beneath it going 30 mi divided by hr ​(44 ft divided by s​). If the balloon rises vertically at a rate of 13 ft divided by s​, what is the rate of change of the distance between the motorcycle and the balloon 6 seconds ​later?

We are tasked with finding the rate of change of the distance between the motorcycle and the balloon, 6 seconds after the motorcycle passes directly beneath the balloon.

### Problem Breakdown

- The motorcycle is moving horizontally at a rate of 44 ft/s.
- The balloon is rising vertically at a rate of 13 ft/s.
- The balloon is initially 130 feet above the ground when the motorcycle passes directly underneath it.

### Step 1: Set up the coordinates

Let:
- \( x(t) \) be the horizontal distance (in feet) of the motorcycle from the point directly beneath the balloon at time \( t \).
- \( y(t) \) be the height of the balloon at time \( t \), in feet.
- \( D(t) \) be the distance between the motorcycle and the balloon at time \( t \).

At \( t = 0 \):
- The motorcycle is directly beneath the balloon, so \( x(0) = 0 \) and \( y(0) = 130 \).

### Step 2: Find expressions for \( x(t) \) and \( y(t) \)

- Since the motorcycle is moving horizontally at 44 ft/s, we have:
  \[
  x(t) = 44t
  \]
- Since the balloon is rising vertically at 13 ft/s, we have:
  \[
  y(t) = 130 + 13t
  \]

### Step 3: Distance between the motorcycle and the balloon

The distance \( D(t) \) can be found using the Pythagorean theorem, since the motorcycle's horizontal distance from the point beneath the balloon and the balloon's height form a right triangle:
\[
D(t) = \sqrt{x(t)^2 + y(t)^2}
\]
Substituting the expressions for \( x(t) \) and \( y(t) \):
\[
D(t) = \sqrt{(44t)^2 + (130 + 13t)^2}
\]
Simplify the equation:
\[
D(t) = \sqrt{1936t^2 + (130 + 13t)^2}
\]
\[
D(t) = \sqrt{1936t^2 + 16900 + 3380t + 169t^2}
\]
\[
D(t) = \sqrt{2105t^2 + 3380t + 16900}
\]

### Step 4: Find the rate of change of the distance \( \frac{dD}{dt} \)

To find the rate of change of the distance, we differentiate \( D(t) \) with respect to time \( t \):

\[
\frac{dD}{dt} = \frac{d}{dt} \left( \sqrt{2105t^2 + 3380t + 16900} \right)
\]
Using the chain rule:
\[
\frac{dD}{dt} = \frac{1}{2 \sqrt{2105t^2 + 3380t + 16900}} \cdot (2 \cdot 2105t + 3380)
\]
Simplify:
\[
\frac{dD}{dt} = \frac{2105t + 1690}{\sqrt{2105t^2 + 3380t + 16900}}
\]

### Step 5: Calculate the rate at \( t = 6 \) seconds

Substitute \( t = 6 \) into the equation:

First, calculate the values in the numerator and the denominator:
- \( 2105(6) + 1690 = 12630 + 1690 = 14320 \)
- For the denominator, we substitute \( t = 6 \) into \( 2105t^2 + 3380t + 16900 \):
  \[
  2105(6)^2 + 3380(6) + 16900 = 2105(36) + 3380(6) + 16900
  \]
  \[
  = 75690 + 20280 + 16900 = 112870
  \]
Thus:
\[
\frac{dD}{dt} = \frac{14320}{\sqrt{112870}} = \frac{14320}{336.29} \approx 42.5 \text{ ft/s}
\]

### Final Answer:
The rate of change of the distance between the motorcycle and the balloon 6 seconds later is approximately **42.5 ft/s**.

---

Would you like further details or explanations on any of the steps? Here are some related questions for you:

1. How does the rate of change of distance change if the motorcycle’s speed were different?
2. What would happen if the balloon rose at a slower or faster rate?
3. How can the Pythagorean theorem be applied to more complex motion scenarios?
4. How would the distance rate change if the motorcycle took a curved path instead of a straight one?
5. Can you apply this concept to objects moving in 3D space instead of just 2D?

**Tip**: When dealing with rates of change in geometry, always think of how the quantities relate to each other, often using the Pythagorean theorem for 2D motion or distance calculations.

A hot-air balloon is 130 ft above the ground when a motorcycle traveling in a straight line on a horizontal road passes directly beneath it going 30 mi/h (44 ft/s). If the balloon rises vertically at a rate of 13 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 6 seconds later?

This problem demonstrates how to calculate the rate of change in distance between a motorcycle and a rising hot-air balloon. The motorcycle moves horizontally while the balloon rises vertically, with specific rates provided. The solution involves applying the Pythagorean Theorem and differentiation techniques. Suitable for high school students in Grades 11-12.

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