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This is a related rates problem, involving the vertical motion of a balloon and the horizontal motion of a bicycle.

### Problem Summary:
- A balloon is rising vertically at 1 ft/sec.
- A bicycle is moving horizontally at 17 ft/sec.
- At the moment when the bicycle is directly under the balloon, the balloon is 65 feet above the ground.
- We are asked to find how fast the distance \( s(t) \) between the balloon and the bicycle is increasing 3 seconds after the bicycle passes under the balloon.

### Step-by-Step Solution:

1. **Define the variables:**
   - Let \( y(t) \) represent the height of the balloon (in feet) at time \( t \).
   - Let \( x(t) \) represent the horizontal distance of the bicycle from the point directly under the balloon (in feet).
   - The distance between the balloon and the bicycle is \( s(t) \), where \( s(t) \) is the hypotenuse of the right triangle formed by \( y(t) \) and \( x(t) \).

2. **Given rates:**
   - The balloon's rate of ascent is \( \frac{dy}{dt} = 1 \) ft/sec.
   - The bicycle's speed is \( \frac{dx}{dt} = 17 \) ft/sec.
   - At \( t = 0 \), the height of the balloon is \( y(0) = 65 \) feet, and the bicycle is directly under the balloon, meaning \( x(0) = 0 \).

3. **Equations:**
   The distance between the balloon and the bicycle is given by the Pythagorean theorem:
   \[
   s^2(t) = x^2(t) + y^2(t)
   \]
   Differentiating both sides with respect to \( t \):
   \[
   2s(t) \frac{ds}{dt} = 2x(t) \frac{dx}{dt} + 2y(t) \frac{dy}{dt}
   \]
   Simplifying:
   \[
   s(t) \frac{ds}{dt} = x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt}
   \]

4. **Find values at \( t = 3 \) seconds:**
   - At \( t = 3 \), the horizontal distance is \( x(3) = 17 \times 3 = 51 \) feet.
   - The height of the balloon is \( y(3) = 65 + 1 \times 3 = 68 \) feet.
   - The distance \( s(3) \) is:
     \[
     s(3) = \sqrt{x^2(3) + y^2(3)} = \sqrt{51^2 + 68^2} = \sqrt{2601 + 4624} = \sqrt{7225} = 85 \, \text{feet}.
     \]

5. **Find the rate of change of \( s(t) \):**
   Now, substitute into the differentiated equation:
   \[
   85 \frac{ds}{dt} = 51 \times 17 + 68 \times 1
   \]
   Simplifying:
   \[
   85 \frac{ds}{dt} = 867 + 68 = 935
   \]
   Solving for \( \frac{ds}{dt} \):
   \[
   \frac{ds}{dt} = \frac{935}{85} = 11 \, \text{ft/sec}.
   \]

### Final Answer:
The distance between the bicycle and the balloon is increasing at a rate of **11 ft/sec** after 3 seconds.

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Would you like a deeper explanation on any of the steps? Here are 5 related questions for further understanding:

1. What is the relationship between related rates and the Pythagorean theorem in this problem?
2. How does differentiating both sides of an equation help solve for related rates?
3. Can this problem be extended to three-dimensional motion involving more than two coordinates?
4. What is the significance of evaluating values at a specific time (e.g., 3 seconds)?
5. How would this problem change if the balloon's rate of ascent were variable?

### Tip:
In related rates problems, always start by clearly defining all variables and their relationships, and then differentiate the equation with respect to time.

A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?

Solve a related rates problem involving a balloon rising vertically and a bicycle moving horizontally. Learn how the Pythagorean theorem and differentiation help determine the rate of change in distance after 3 seconds. Suitable for advanced high school students.

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