Question

A port and a radar station are 3 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of 18 ​mi/hr. If the ship maintains its speed and​ course, what is the rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at​ 12:30 PM?​ (Hint: Use the law of​ sines.)

Answer 1

Answer:

The answer is $34.37 deg/h$ .

Step-by-step explanation:

Use the image provided here for clarification purposes. It helps so much to make a drawing in this sort of questions. The tracking angle here is drawn as $\theta$.

Law of sines is applied to the vertical side (ship's trayectory) and the horizontal side (shore line). Here, time must be expressed in hours:

$\frac{sin \theta}{18t}=\frac{sin(\pi /2 - \theta)}{3}=\frac{cos\theta}{3}$

$tan(\theta)=\frac{18t}{3} = 6t$.

$\theta=arctan(6t)$.

So, the rate of change of our tracking angle, $\theta$, can be obtained by deriving this expression:

$\frac{d \theta}{dt}=6\frac{1}{1+(6t)^2}$.

At 12:30 PM, i.e., half an hour after departure, t = 1/2, so:

$\frac{d \theta}{dt}(t=1/2)=6\frac{1}{1+(3)^2}=6/10=0.6$ rads/h.

Please note this is the actual rate of change of the tracking angle at this very moment, but not for the rest of moments. It is a function, variable with time, applied to a specific instant. For a value in degrees,

$0.6\frac{180deg}{\pi}=34.37 deg/h$