Question

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Answer 1

Given a figure represents the distance between an airplane and a radar station on the ground.

We will find the following:

Step 1: we will find a relation between x and θ

As shown we may consider the triangle which a right-angle triangle

With the height = 3000 ft

And the side (x) is the adjacent side to the angle θ

the side (3000) is the opposite side to the angle θ

so, we can write the following equation:

$\begin{gathered} tan\text{ }θ=\frac{opposite}{adjacent} \\ \\ tan\text{ }θ=\frac{3000}{x}\rightarrow x=\frac{3000}{tan\text{ }θ} \end{gathered}$

Step (2): at the instant x = 2000 ft, we will compute the following:

$\begin{gathered} x=2000 \\ θ=tan^{-1}(\frac{3000}{x})\approx56.31\degree \end{gathered}$

Also, we will find the first derivative of x and θ

$\begin{gathered} x=\frac{3000}{tanθ}=3000*cot\text{ }θ \\ \\ \frac{dx}{dt}=3000*(-csc^2θ)*\frac{dθ}{dt} \end{gathered}$

the value of (dx/dt) is given and equal to 200 ft/sec

Substitute dx/dt and θ to find dθ/dt

$\begin{gathered} 200=3000*(-csc^256.31)*\frac{dθ}{dt} \\ \\ \frac{dθ}{dt}=-\frac{200}{3000*csc^256.31}=-\frac{3}{65} \end{gathered}$

So, the answer to step (2) will be as follows:

$\begin{gathered} x=2000\text{ }ft \\ \frac{dx}{dt}=200\text{ ft/sec} \\ θ=56.31\degree \\ \frac{dθ}{dt}=-\frac{3}{65}\text{ deg/sec} \end{gathered}$

Step (3):

We will find how fast the radar is rotating

Which will be the angular velocity ω

$ω=\frac{d\theta}{dt}=-\frac{3}{65}\text{ deg/sec}$

convert from degree to radian

$ω=-\frac{3}{65}*\frac{\pi}{180}=0.001\text{ rad/sec}$