A​ hot-air balloon is 160 ft160 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes directly beneath it going 60 mi divided by hr60 mi/hr ​(88 ft divided by s88 ft/s​). If the balloon rises vertically at a rate of 6 ft divided by s6 ft/s​, what is the rate of change of the distance between the motorcycle and the balloon 5 seconds5 seconds ​later? 2 decimals

We are tasked with finding the rate of change of the distance between the hot-air balloon and the motorcycle at a given time. To approach this, we'll use related rates and apply the Pythagorean Theorem, as the problem involves two perpendicular motions: the balloon's vertical rise and the motorcycle's horizontal movement.

### Step-by-Step Solution

1. **Let the variables be:**
   - \( x(t) \) = the horizontal distance of the motorcycle from the point directly below the balloon at time \( t \).
   - \( y(t) \) = the vertical height of the balloon at time \( t \).
   - \( z(t) \) = the distance between the balloon and the motorcycle at time \( t \).

2. **Given information:**
   - The motorcycle travels at a constant speed of \( 88 \, \text{ft/s} \), so \( \frac{dx}{dt} = 88 \, \text{ft/s} \).
   - The balloon rises at a constant rate of \( 6 \, \text{ft/s} \), so \( \frac{dy}{dt} = 6 \, \text{ft/s} \).
   - Initially, the balloon is \( 160 \, \text{ft} \) above the ground, so \( y(0) = 160 \, \text{ft} \).

3. **Find the values of \( x \), \( y \), and \( z \) at \( t = 5 \, \text{seconds} \):**
   - The horizontal distance after 5 seconds is:
     \[
     x(5) = 88 \times 5 = 440 \, \text{ft}
     \]
   - The vertical distance of the balloon at \( t = 5 \) seconds is:
     \[
     y(5) = 160 + 6 \times 5 = 160 + 30 = 190 \, \text{ft}
     \]
   - Using the Pythagorean Theorem to find \( z(5) \), the distance between the balloon and the motorcycle:
     \[
     z(5) = \sqrt{x(5)^2 + y(5)^2} = \sqrt{440^2 + 190^2} = \sqrt{193600 + 36100} = \sqrt{229700} \approx 479.33 \, \text{ft}
     \]

4. **Differentiate using related rates:**
   The Pythagorean relation is:
   \[
   z^2 = x^2 + y^2
   \]
   Differentiating both sides with respect to \( t \):
   \[
   2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
   \]
   Simplifying:
   \[
   z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
   \]
   Solving for \( \frac{dz}{dt} \):
   \[
   \frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}
   \]

5. **Substitute known values:**
   - At \( t = 5 \):
     - \( x = 440 \, \text{ft} \)
     - \( y = 190 \, \text{ft} \)
     - \( z \approx 479.33 \, \text{ft} \)
     - \( \frac{dx}{dt} = 88 \, \text{ft/s} \)
     - \( \frac{dy}{dt} = 6 \, \text{ft/s} \)

   Substituting these into the equation:
   \[
   \frac{dz}{dt} = \frac{440 \times 88 + 190 \times 6}{479.33} = \frac{38720 + 1140}{479.33} = \frac{39860}{479.33} \approx 83.15 \, \text{ft/s}
   \]

### Final Answer:
The rate of change of the distance between the balloon and the motorcycle 5 seconds later is approximately \( 83.15 \, \text{ft/s} \).

Would you like a more detailed breakdown of any part of the solution?

Here are 5 related questions to explore further:
1. How would the rate of change of distance vary if the motorcycle accelerated instead of moving at a constant speed?
2. What if the balloon was descending instead of rising? How would that affect the related rates?
3. How would the rate change if the balloon started at a different height above the ground?
4. What happens to the rate of change if the motorcycle moves at a different angle, not directly underneath the balloon?
5. Can we calculate the instantaneous rate of change if the speeds were functions of time instead of constant?

**Tip**: For any related rates problem, identify which variables change over time and how they are geometrically or algebraically related. This helps in setting up the equations correctly.

A hot-air balloon is 160 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 60 mi/hr (88 ft/s). If the balloon rises vertically at a rate of 6 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 5 seconds later? (Answer to 2 decimals)

Solve the problem of finding the rate of change of the distance between a hot-air balloon and a motorcycle using related rates and the Pythagorean Theorem. Given the vertical rise of the balloon at 6 ft/s and the motorcycle's speed of 88 ft/s, calculate the rate of change 5 seconds later. Suitable for Grades 10-12.

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