Question

8 a hot air balloon rising vertically is tracked by an observer located 2 miles from the lift - off point. at a certain moment, the angle between the observers line - of - sight and the horizontal is $\frac{pi}{4}$, and the angle is changing at a rate of $\frac{1}{10}$ rad/min. how fast is the balloon rising at this moment?

Explanation:

Step1: Establish a trigonometric relationship

Let $y$ be the height of the balloon and $\theta$ be the angle between the observer's line - of - sight and the horizontal. We know that $\tan\theta=\frac{y}{2}$ (since the horizontal distance from the lift - off point to the observer is 2 miles).

Step2: Differentiate both sides with respect to time $t$

Using the chain - rule, $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{2}\frac{dy}{dt}$.

Step3: Find the value of $\sec\theta$ at the given moment

When $\theta = \frac{\pi}{4}$, $\sec\theta=\sqrt{2}$. And we are given that $\frac{d\theta}{dt}=\frac{1}{10}$ rad/min.

Step4: Solve for $\frac{dy}{dt}$

Substitute $\sec\theta=\sqrt{2}$ and $\frac{d\theta}{dt}=\frac{1}{10}$ into the equation $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{2}\frac{dy}{dt}$.
$(\sqrt{2})^{2}\times\frac{1}{10}=\frac{1}{2}\frac{dy}{dt}$.
$2\times\frac{1}{10}=\frac{1}{2}\frac{dy}{dt}$.
$\frac{1}{5}=\frac{1}{2}\frac{dy}{dt}$.
$\frac{dy}{dt}=\frac{2}{5}$ miles/min.

Answer:

$\frac{2}{5}$ miles/min