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

Answer:

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

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:

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

$\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{r}$

$\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{\sqrt{x^{2}+y^{2}}}$ (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 $x = 40\,ft$, $y = 129\,ft$, $\dot x = 0\,\frac{ft}{s}$ and $\dot y = 53\,\frac{ft}{s}$, then the rate of change of the distance between man and helicopter is:

$\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{\sqrt{x^{2}+y^{2}}}$

$\dot r = 50.622\,\frac{ft}{s}$

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