A hot air balloon is attached to a spool of
rope that is 125 feet away from the balloon
when it is on the ground. The hot air
balloon rises straight up in such a way that
the length of rope increases at a rate of
15 ft/sec. How fast is the hot air balloon
rising 20 seconds after it lifts off?
We know the rate of change of the rope's length with respect to time (dL/dt), and we want to find the rate of change of the balloon's height with respect to time (dh/dt) at t = 20 seconds.
Let's denote the length of the rope as L and the height of the balloon as h. The rope, the ground, and the vertical line from the balloon to the ground form a right triangle, so by the Pythagorean theorem, we have L² = h² 125².
We're given that dL/dt = 15 ft/sec. We're asked to find dh/dt when t = 20 sec.
The length of the rope at t = 20 sec is L = 125 ft 15 ft/sec * 20 sec = 425 ft.
Now, differentiate both sides of the equation L² = h² 125² with respect to time t:
2L * (dL/dt) = 2h * (dh/dt).
Then solve for dh/dt, the rate of change of the height with respect to time:
dh/dt = (L * dL/dt) / h.
Substitute the given values into this equation:
dh/dt = (425 ft * 15 ft/sec) / sqrt(425² - 125²) = 6375 / sqrt(425² - 125²) ft/sec ≈ 18.68 ft/sec.
So, the hot air balloon is rising at a speed of approximately 18.68 feet per second 20 seconds after it lifts off.
ExpertInStudy.com is a smart community of thousands of students and experts. It is thanks to them that even the most difficult questions get quick and good answers. Here you can become an expert and start making money!