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 9 seconds ​later?
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Part 1
Let x be the horizontal distance from the balloon to the​ motorcycle, y be the vertical distance from the balloon to the​ road, and z be the distance between the motorcycle and the balloon. Write an equation relating​ x, y, and z.
  
z squared equals x squared plus y squared
Part 2
Differentiate both sides of the equation with respect to t.
left parenthesis 2 x right parenthesis StartFraction dx Over dt EndFraction plus left parenthesis 2 y right parenthesis StartFraction dy Over dt EndFraction equals left parenthesis 2 z right parenthesis StartFraction dz Over dt EndFraction
  
  
  
Part 3
The rate of change of the distance between the motorcycle and the balloon after 9 seconds is about 
  
enter your response here 
▼ 
ft divided by s.
ft cubed divided by s.
ft squared divided by s.
ft.
​(Round to two decimal places as​ needed.)

### Part 1: Relating \( x \), \( y \), and \( z \)

We are given that:
- \( x \) is the horizontal distance between the motorcycle and the balloon,
- \( y \) is the vertical distance from the balloon to the ground (which changes because the balloon is rising), and
- \( z \) is the straight-line distance between the motorcycle and the balloon.

Using the Pythagorean theorem, we can relate these quantities by:

\[
z^2 = x^2 + y^2
\]

### Part 2: Differentiating with respect to time \( t \)

Now, we differentiate both sides of the equation with respect to time \( t \) to find the rates of change.

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

Applying the chain rule to both sides:

\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]

Simplifying:

\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]

### Part 3: Plugging in values

We are asked to find the rate of change of the distance between the balloon and the motorcycle after 9 seconds.

#### Step 1: Values at \( t = 9 \) seconds
- The horizontal speed of the motorcycle is \( \frac{dx}{dt} = 44 \, \text{ft/s} \).
- The vertical speed of the balloon is \( \frac{dy}{dt} = 13 \, \text{ft/s} \).
- Initially, \( y(0) = 130 \, \text{ft} \), so after 9 seconds, the vertical distance is:

\[
y(9) = 130 + 13 \times 9 = 130 + 117 = 247 \, \text{ft}
\]

- The horizontal distance after 9 seconds is:

\[
x(9) = 44 \times 9 = 396 \, \text{ft}
\]

#### Step 2: Finding \( z(9) \)

Now, we find \( z(9) \), the straight-line distance between the motorcycle and the balloon at \( t = 9 \) seconds:

\[
z(9) = \sqrt{x(9)^2 + y(9)^2} = \sqrt{396^2 + 247^2} = \sqrt{156,816 + 61,009} = \sqrt{217,825} \approx 466.69 \, \text{ft}
\]

#### Step 3: Solving for \( \frac{dz}{dt} \)

Now, we use the differentiated equation to find \( \frac{dz}{dt} \):

\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]

Substitute the known values:

\[
466.69 \frac{dz}{dt} = (396)(44) + (247)(13)
\]

First, calculate the right-hand side:

\[
466.69 \frac{dz}{dt} = 17,424 + 3,211 = 20,635
\]

Now, solve for \( \frac{dz}{dt} \):

\[
\frac{dz}{dt} = \frac{20,635}{466.69} \approx 44.23 \, \text{ft/s}
\]

### Final Answer:
The rate of change of the distance between the motorcycle and the balloon after 9 seconds is approximately \( \boxed{44.23 \, \text{ft/s}} \).

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Would you like more details or have any questions? 

Here are 5 questions you might consider next:
1. How would the answer change if the balloon's ascent rate were different?
2. What would happen to \( z \) over time if the motorcycle slowed down?
3. Can this same approach be applied if the motorcycle were moving in a curved path?
4. How does the rate of change of \( z \) evolve as the balloon gets higher and higher?
5. How does the differentiation process change if the motorcycle accelerates?

**Tip:** Always keep track of units carefully when working with rates of change in related rates problems.

Solve a related rates problem involving a hot-air balloon and a motorcycle. Learn how to calculate the rate of change of the distance between them using differentiation and the Pythagorean theorem. Suitable for students in Grades 11-12.

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