Question
Can someone explain this problem? I need a step by step explanation of how to get the entire answer.
A lighthouse L is located on a small island 2 km from the nearest point A on a long, straight shoreline. If the lighthouse lamp rotates at 3 revolutions per minute, how fast is the illuminated spot P on the shoreline moving along the shoreline when it is 4 km from A? Referring to Figure 4.3, let x be the distance A P. and let theta be the angle PLA. Then x = 2 tan theta and dx / dt = 2 sec2 theta d theta / dt. Now d theta / dt = (3 rev/min)(2 pi radians/rev) = 6 pi radians/min. When x = 4, we have tan theta = 2 and sec2 theta = 1+ tan2 theta = 5. Thus dx / dt = (2)(5)(6 pi) = 60 pi 188.5. The spot of light is moving along the shoreline at a rate of about 189 km/min when it is 4 km from A. (Note that it was essential to convert the rate of change of theta from revolutions per minute to radians per minute. If theta were not measured in radians we could not assert that (d / d theta) tan theta = sec2 theta.) Figure 4.3
Explanation / Answer
Here, we know the speed at which light is rotating at teh shore.
Thus, AL = radius = 2km
Let, AP = x (we want to find the rate of change of x i.e. dx/dt )
Now, using trigonometry,
tan theta = x/2
Thus, x = 2 * tan theta
Differentiating both sides with respect ti time:
dx/dt = 2 * sec^2theta * d(theta)/dt
Now, speed of rotation = d(theta)/dt = 3 * 2pi = 6 pi rad/min
When x=4,
tan theta = 2
sec^2 theta = 1 + 4 = 5
Thus, dx/dt = 2(5)*(6pi) = 60 pi rad/min = 188.5
Thus, the speed is 188.5 km/min