 # A hot air balloon is attached to a spool ofrope that is 125 feet away from the balloonwhen it is on the ground. The hot airballoon rises straight up in such a way thatthe length of rope increases at a rate of15 ft/sec. How fast is the hot air balloonrising 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.

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