Question

At a distance of 60 ft from the pad, a man observes a helicopter taking off from a heliport. If the helicopter lifts off vertically and is rising at a speed of 48 ft/sec when it is at an altitude of 114 ft, how fast is the distance between the helicopter and the man changing at that instant? (Round your answer to one decimal place.)

ft/sec

Answer 1

At a distance of 60 ft from the pad, the distance between the helicopter and the man is not changing as the helicopter rises vertically.

Step-by-step explanation:

To find the rate at which the distance between the helicopter and the man is changing, we can use the concept of related rates.

We can create a right triangle with the helicopter's altitude as the height and the distance between the man and the helicopter as the base.

Let's denote the distance between the man and the helicopter as x. At a given time, we have x = 60 ft, and the helicopter's altitude, which is equivalent to the height of the right triangle, is 114 ft.

We're asked to find dx/dt, the rate at which x is changing with respect to time, when the altitude (h) is 114 ft. To solve this, we can use the Pythagorean theorem:

h^2 = x^2 + 60^2

114^2 = x^2 + 60^2

x^2 = 114^2 - 60^2

x^2 = 13860

x = sqrt(13860)

x ≈ 117.6 ft

Now, we need to differentiate both sides of the equation with respect to time:

2x(dx/dt) = 0

2(117.6)(dx/dt) = 0

dx/dt ≈ 0 ft/sec

Therefore, the distance between the helicopter and the man is not changing at that instant.