a balloon is rising vertically above a level, straight road at a constant rate of 5 ft/sec. just when the…

Explanation:

Step1: Define variables and equations

Let $y(t)$ be the height of the balloon, $x(t)$ be the horizontal distance of the bicycle from the point directly below the balloon. By the Pythagorean theorem, $s(t)^2=x(t)^2 + y(t)^2$.

Step2: Find $y(t)$ and $x(t)$ at $t = 3$

The initial height of the balloon $y(0)=33$ ft and it rises at a rate of $\frac{dy}{dt}=5$ ft/sec, so $y(t)=33 + 5t$. At $t = 3$, $y(3)=33+5\times3=33 + 15=48$ ft. The bicycle moves at a rate of $\frac{dx}{dt}=12$ ft/sec, so $x(t)=12t$. At $t = 3$, $x(3)=12\times3 = 36$ ft.

Step3: Differentiate the Pythagorean - equation

Differentiate $s(t)^2=x(t)^2 + y(t)^2$ with respect to $t$ using the chain - rule. We get $2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, which simplifies to $s\frac{ds}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$.

Step4: Calculate $s$ at $t = 3$

At $t = 3$, $x = 36$, $y = 48$. Using the Pythagorean theorem, $s=\sqrt{x^{2}+y^{2}}=\sqrt{36^{2}+48^{2}}=\sqrt{1296 + 2304}=\sqrt{3600}=60$ ft.

Step5: Solve for $\frac{ds}{dt}$

We know $x = 36$, $\frac{dx}{dt}=12$, $y = 48$, $\frac{dy}{dt}=5$, and $s = 60$. Substitute these values into $s\frac{ds}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$: $60\frac{ds}{dt}=36\times12+48\times5$. First, calculate the right - hand side: $36\times12+48\times5=432+240 = 672$. Then, $\frac{ds}{dt}=\frac{672}{60}=11.2$ ft/sec.

Answer:

$11.2$ ft/sec