Question

A weather balloon is rising vertically at the rate of 10 ft s . An observer is standing on the ground 300 ft horizontally from the point where the balloon was released. At what rate is the distance between the observer and the balloon changing when the balloon is 400 ft height

Answer 1

Answer:

The distance between the observer and the balloon is increasing at a rate 8 feet per second.

Step-by-step explanation:

Let be A the point at which the observer is located, O the initial location of the  weather balloon and B is the current location above the ground. Since the weather balloon is rising vertically and the distance between its initial position and the position of the observer, we can represent all distances by the Pythagorean Theorem:

$r^{2} = h^{2} + (300\,ft)^{2}$ (1)

Where:

$r$ - Distance between observer and current position of the weather balloon, in feet.

$h$ - Current height of the weather balloon above ground, in feet.

By Differential Calculus, we derive an expression for the rate of change of the distance between observer and current position of the weather balloon ($\dot r$), in feet per second:

$2\cdot r \cdot \dot r = 2\cdot h \cdot \dot h$

$\dot r = \frac{h\cdot \dot h}{r}$

$\dot r = \frac{h\cdot \dot h}{\sqrt{h^{2}+ (300\,ft)^{2}}}$ (2)

Where $\dot h$ is the rate of change of the height of the weather balloon, in feet per second.

If we know that $h = 400\,ft$ and $\dot h = 10\,\frac{ft}{s}$, then the rate of change of the distance between the observer and the balloon is:

$\dot r = \frac{(400\,ft)\cdot \left(10\,\frac{ft}{s} \right)}{500\,ft}$

$\dot r = 8\,\frac{ft}{s}$

The distance between the observer and the balloon is increasing at a rate 8 feet per second.