A hot-air balloon is 180 ft above the ground when a motorcycle passes directly beneath it (travelling in a straight line on a horizontal road) going at a constant speed of 60 ft/s. If the balloon is rising vertically at a rate of 15 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 20 seconds after the motorcycle was directly beneath the balloon.
This is what I understand.
d^2=h^2+x^2
(d)dd/dt= (h)dh/dt + (x)dx/dt
dd/dt= ((180)dh/dt + (60)dx/dt)/15
Other than that, I really have no idea what else to do. How do I find what dh/dt is as well as dx/dt.
Did you read the jproblem? dh/dt=15ft/sec
dx/dt= 60 ft/sec
To find the rate of change of the distance between the motorcycle and the balloon, we need to determine the values of dh/dt and dx/dt.
First, let's find dh/dt, which represents the rate at which the balloon is rising vertically. The problem states that the balloon is rising at a constant rate of 15 ft/s. Therefore, dh/dt = 15 ft/s.
Next, let's determine dx/dt, which represents the rate at which the motorcycle is moving horizontally. The problem states that the motorcycle is traveling at a constant speed of 60 ft/s. Since the motorcycle is traveling in a straight line on a horizontal road, dx/dt is also constant at 60 ft/s.
Now, we can substitute these values into the equation dd/dt = (180 * dh/dt + 60 * dx/dt) / 15.
Using the given values, we have:
dd/dt = (180 * 15 + 60 * 60) / 15
dd/dt = (2700 + 3600) / 15
dd/dt = 6300 / 15
dd/dt = 420 ft/s
Therefore, the rate of change of the distance between the motorcycle and the balloon 20 seconds after the motorcycle was directly beneath the balloon is 420 ft/s.