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

The answer is
`34.37 deg/h` .

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:

`(sin \theta)/(18t)=(sin(\pi /2 - \theta))/(3)=(cos\theta)/(3)`

`tan(\theta)=(18t)/(3) = 6t`.

`\theta=arctan(6t)`.

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

`(d \theta)/(dt)=6(1)/(1+(6t)^2)`.

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

`(d \theta)/(dt)(t=1/2)=6(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(180deg)/(\pi)=34.37 deg/h`