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.

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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.