Question

A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the balloon is 9 feet above the ground.

Answer 1

Answer:

$\displaystyle \theta' =0.24\ rad/s$

Explanation:

Rate Of Change

Let some variable y depend on time t. we can express y as a function of t as

$y=f(t)$

The instant rate of change of y respect to t is the first derivative, i.e.

$y'=f'(t)$

The balloon, the ground and the observer form a right triangle (shown below) where the height of the balloon y, the horizontal distance x, and the angle of elevation are related with the trigonometric formula

$\displaystyle tan\theta =\frac{y}{x}$

Since x is constant, we take the derivative with respect to time  by using the chain rule:

$\displaystyle sec^2\theta \ \theta' =\frac{y'}{x}$

Solving for $\theta'$

$\displaystyle \theta' =\frac{y'}{xsec^2\theta}$

Let's compute the actual angle with the initial conditions y=9 feet, x=12 feet

$\displaystyle tan\theta =\frac{y}{x}$

$\displaystyle tan\theta =\frac{9}{12}=\frac{3}{4}$

Knowing that

$\sec^2\theta=1+tan^2\theta$

$\displaystyle \sec^2\theta=1+\left(\frac{3}{4}\right)^2$

$\displaystyle \sec^2\theta=\frac{25}{16}$

The balloon is rising at y'=8 feet/sec, thus we compute the change of the angle of elevation:

$\displaystyle \theta' =\frac{8}{12\ \frac{25}{16}}$

$\displaystyle \theta' =\frac{32}{75}\ rad/s$

$\boxed{\displaystyle \theta' =0.43\ rad/s}$