Question

At a distance of 40 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 53 ft/sec when it is at an altitude of 129 ft, how fast is the distance between the helicopter and the man changing at that instant

Answer 1

The distance between the helicopter and the man is increasing at a rate of 50.622 feet per second.

Step-by-step explanation:

First, we create a geometric diagram representing the situation between man (point O), helicopter (point P) and helipad (point H). The distance between man and helicopter is represented by Pythagorean Theorem:

`x^(2)+y^(2) = r^(2)` (1)

Where:

`x` - Distance between man and helipad, in feet.

`y` - Distance between helipad and helicopter, in feet.

`r` - Distance between man and helicopter, in feet.

By Differential Calculus, we derive an expression for the rate of change of the distance between man and helicopter (
`\dot r`), in feet per second:

`2\cdot x\cdot \dot x + 2\cdot y\cdot \dot y = 2\cdot r\cdot \dot r`

`\dot r = (x\cdot \dot x + y\cdot \dot y)/(r)`

`\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{\sqrt{x^(2)+y^(2)}}` (2)

Where:

`\dot x` - Rate of change of the distance between man and helipad, in feet per second.

`\dot y` - Rate of change of the distance between helicopter and helipad, in feet per second.

If we know that
`x = 40\,ft`,
`y = 129\,ft`,
`\dot x = 0\,(ft)/(s)` and
`\dot y = 53\,(ft)/(s)`, then the rate of change of the distance between man and helicopter is:

`\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{\sqrt{x^(2)+y^(2)}}`

`\dot r = 50.622\,(ft)/(s)`

The distance between the helicopter and the man is increasing at a rate of 50.622 feet per second.