Hello, pantsaregood!

A weather balloon is rising vertically at a rate of 10 ft/s. An observer is standing

on the ground 500 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 in the air? Code:

B *
| *
| * r
y | *
| *
| *
* - - - - - - - - * A
500

The observer is at $\displaystyle A.$

The balloon is at $\displaystyle B$ . . . Its height is $\displaystyle y$

Let: $\displaystyle r \:=\:AB.$

We are told that: .$\displaystyle \frac{dy}{dt} = 10$ ft/sec.

Pythagorus says: .$\displaystyle r^2 \:=\:y^2 + 500^2$

Differentiate with respect to time: .$\displaystyle 2r\cdot\frac{dr}{dt} \:=\:2y\cdot\frac{dy}{dt} \quad\Rightarrow\quad \frac{dr}{dt} \:=\:\frac{y}{r}\cdot\frac{dy}{dt}$ .[1]

When $\displaystyle y = 400,\;r^2 \:=\:400^2 + 500^2 \:=\:410,000 \quad\Rightarrow\quad r \:=\:100\sqrt{41}$

Substitute into [1]: .$\displaystyle \frac{dr}{dt} \;=\;\frac{400}{100\sqrt{41}}(10) \;=\;\frac{40}{\sqrt{41}}$

Therefore: .$\displaystyle \frac{dr}{dt}\;\approx\;6.25$ ft/sec.