Question

how fast is the angle of depression of the telescope changing when the boat is 260 meters from shore

Answer 1

Given:

The boat moves at a rate = 15 meters per second

The telescope is 40 meters above the water level

Let the distance between the boat and tower of the telescope = x

so,

$\begin{gathered} \tan \theta=\frac{x}{40} \\ \\ x=40\tan \theta \end{gathered}$

Differentiate both sides with respect to the time (t)

$\frac{dx}{dt}=40\cdot\sec ^2\theta\cdot\frac{d\theta}{dt}$

Where: (dx/dt) is the speed of the boat

(dθ/dt) is the change of the angle of the telescope

Substitute with (dx/dt = 15) and

When the boat is 260 meters from shore

$\begin{gathered} \tan \theta=\frac{260}{40}=6.5 \\ \theta=\tan ^{-1}6.5\approx81.254\degree \end{gathered}$

so,

$\begin{gathered} 15=40\cdot(\sec 81.254)^2\cdot\frac{d\theta}{dt} \\ \\ \frac{d\theta}{dt}=\frac{15}{40\cdot(\sec 81.254)^2} \end{gathered}$

Using the calculator:

$\frac{d\theta}{dt}=0.0087$

so, the answer will be 0.0087 degrees per seconds

Convert from to radians per second

So,

$0.0087\cdot(\frac{\pi}{180})=0.0002\text{ rad/s}$