# Thread: Derivatives in relation to time.

1. ## Derivatives in relation to time.

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?

The answer I got seems a bit ridiculous. 8000/[2sqrt(410000)].

Is that correct, or did I mess up?

2. 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.

,

,

### calculus height of weather balloon with two observer

Click on a term to search for related topics.