a balloon is rising vertically above a level, straight road at a constant rate of 4 ft/sec. just when the…

Explanation:

Step1: Define variables and relations

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, and $s(t)$ be the distance between the balloon and the bicycle. By the Pythagorean theorem, $s^{2}=x^{2}+y^{2}$.

Step2: Differentiate the equation with respect to time $t$

Differentiating both sides of $s^{2}=x^{2}+y^{2}$ 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 $s\frac{ds}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$.

Step3: Find the values of $x$, $y$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$ at $t = 6$ seconds

We know that $\frac{dy}{dt}=4$ ft/sec and $\frac{dx}{dt}=13$ ft/sec. Initially, $y(0)=80$ ft. After $t = 6$ seconds, $y=y(0)+\frac{dy}{dt}\times t=80 + 4\times6=80 + 24=104$ ft. And $x=\frac{dx}{dt}\times t=13\times6 = 78$ ft.

Step4: Calculate $s$ at $t = 6$ seconds

Using the Pythagorean theorem $s=\sqrt{x^{2}+y^{2}}=\sqrt{78^{2}+104^{2}}=\sqrt{(26\times3)^{2}+(26\times4)^{2}}=\sqrt{26^{2}(3^{2}+4^{2})}=\sqrt{26^{2}\times25}=26\times5 = 130$ ft.

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

Substitute $x = 78$, $\frac{dx}{dt}=13$, $y = 104$, $\frac{dy}{dt}=4$, and $s = 130$ into $s\frac{ds}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$. We have $130\frac{ds}{dt}=78\times13+104\times4$. First, calculate the right - hand side: $78\times13+104\times4=1014 + 416=1430$. Then $\frac{ds}{dt}=\frac{1430}{130}=11$ ft/sec.

Answer:

11 ft/sec