A radar antenna is located on a ship that is 4km from a straight shore. It is rotating at 32rev/min. How fast does the radar beam sweep along the shore when the angle between the beam and the shortest distance to the shore is pi/4?

Question

Let's say that the shoreline is y, are we looking for dy/dt?

tanθ=y/4
y=4tan(pi/4)
y=4
r=radar beam
Using pythagarus, I found out that r=4root2

Would the equation be dy/dt=(dθ/dt)(dy/dθ)?

Answer 1

We don't really need r, because of the nice angle.

From
y = 4tanØ
dy/dt = 4sec^2 Ø (dØ/dt)

so we need sec^2 (π/4) and dØ/dt
but we are given dØ/dt = 32(2π) rpm , (one rotation is 2π radians)
and cos π/4 = 1/√2
then sec π/4 = √2
and sec^2 π/4 = 2

so dy/dt = 4(2)(32)(2π) km/min

I will let you finish and convert to any unit you need.

Answer 2

Ohh, so that's how you find dy/dt, thanks!