Questions: A hot-air balloon is 150 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 30 mi / hr (44 ft / s). If the balloon rises vertically at a rate of 7 t / s, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later? The rate of change of the distance between the motorcycle and the balloon after 10 seconds is about ft / s. (Round to two decimal places as needed.)

A hot-air balloon is 150 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 30 mi / hr (44 ft / s). If the balloon rises vertically at a rate of 7 t / s, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later?

The rate of change of the distance between the motorcycle and the balloon after 10 seconds is about ft / s. (Round to two decimal places as needed.)

Transcript text: A hot-air balloon is 150 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going $30 \mathrm{mi} / \mathrm{hr}$ ( $44 \mathrm{ft} / \mathrm{s}$ ). If the balloon rises vertically at a rate of $7 \mathrm{t} / \mathrm{s}$, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later? The rate of change of the distance between the motorcycle and the balloon after 10 seconds is about $\square$ $\mathrm{ft} / \mathrm{s}$. (Round to two decimal places as needed.)

Solution

Solution Steps

Step 1: Set up the relationship using the Pythagorean theorem

Given the Pythagorean theorem, $z^2 = x^2 + y^2$.

Step 2: Express $x$ and $y$ as functions of time ($t$)

For $x(t) = v_x \cdot t$, we have $x(10) = 44 \cdot 10 = 440$. For $y(t) = y_0 + v_y \cdot t$, we have $y(10) = 150 + 7 \cdot 10 = 220$. Thus, $z(t) = \sqrt{x^2 + y^2} = \sqrt{440^2 + 220^2} = 491.935$.

Step 3: Differentiate $z^2 = x^2 + y^2$ with respect to time ($t$)

Using the chain rule, $2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$, substituting $ rac{dx}{dt} = v_x$ and $ rac{dy}{dt} = v_y$, we get $ rac{dz}{dt} = \frac{xv_x + yv_y}{z} = \frac{440 \cdot 44 + 220 \cdot 7}{491.935} = 42.49$.