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?
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)?
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.
Ohh, so that's how you find dy/dt, thanks!
Yes, you are correct that we are looking for dy/dt, which represents the rate at which the radar beam sweeps along the shore.
To find dy/dt, we can use the equation dy/dt = (dθ/dt)(dy/dθ), which represents the chain rule in calculus.
In this equation:
- (dθ/dt) represents the rate at which the angle between the beam and the shortest distance to the shore is changing, which is given as 32 revolutions per minute.
- (dy/dθ) represents the rate of change of y with respect to θ.
To find (dy/dθ), we start with the equation tan(θ) = y/4 and differentiate both sides with respect to θ:
sec^2(θ) * dθ/dθ = (1/4) * dy/dθ
Since dθ/dθ is equal to 1, we can simplify the equation to:
sec^2(θ) = (1/4) * dy/dθ
Now, we substitute the given angle θ = π/4:
sec^2(π/4) = (1/4) * dy/dθ
Since sec^2(π/4) = 2, we can solve for (dy/dθ):
2 = (1/4) * dy/dθ
Multiplying both sides by 4 gives:
8 = dy/dθ
Now, we have found (dy/dθ) = 8.
Next, we substitute this value along with the given value of (dθ/dt) = 32 revolutions per minute into the equation dy/dt = (dθ/dt)(dy/dθ):
dy/dt = (32 rev/min) * (8)
Calculating this gives us:
dy/dt = 256 rev/min
Therefore, the radar beam sweeps along the shore at a rate of 256 revolutions per minute.
Yes, you are correct. You are looking for the rate at which the distance along the shore (y) changes with respect to time (t), so you are looking for dy/dt.
You have correctly determined that the tangent of the angle θ is equal to y/4, where y is the distance along the shore and 4 is the distance from the ship to the shore. Therefore, y = 4tan(θ).
You have also determined that when θ = π/4, y = 4.
Additionally, you have found that r, the distance traveled by the radar beam, is equal to 4√2, which can be deduced by using the Pythagorean theorem.
To find the equation relating dy/dt, dθ/dt, and dy/dθ, we can differentiate the equation y = 4tan(θ) with respect to time (t):
dy/dt = d(4tan(θ))/dt.
By applying the chain rule, this can be rewritten as:
dy/dt = 4(dtan(θ)/dθ)(dθ/dt).
Now, since tan(θ) = y/4, we can differentiate this equation with respect to θ:
sec^2(θ) = (1/4)(dy/dθ).
Now, we can substitute this back into the previous equation:
dy/dt = 4(sec^2(θ))(dθ/dt).
So, the equation you are looking for is:
dy/dt = 4(sec^2(θ))(dθ/dt),
where dy/dt represents the rate at which the radar beam sweeps along the shore, dθ/dt represents the rate at which the angle between the beam and the shortest distance to the shore is changing, and θ = π/4.