A boat is sailing directly toward a buoy at a constant speed of 10 meters per second. The boat's angle of approach to the buoy is θ, and θ is decreasing at a rate of 0.2 radians per second. How far is the boat from the buoy when θ is 30 degrees?
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First, let's convert 30 degrees to radians. We know that 2π radians is equal to 360 degrees, so to convert from degrees to radians, we divide by 360 and multiply by 2π.
θ = (30/360) * 2π = π/6 radians.
Since θ is decreasing at a rate of 0.2 radians per second, we can set up a differential equation to represent this relationship:
dθ/dt = -0.2.
To solve for the value of θ when it's 30 degrees, we need to integrate the differential equation with respect to time:
∫dθ = ∫-0.2 dt.
Δθ = -0.2t + C,
where Δθ is the change in θ and C is the constant of integration.
Since θ is initially at π/6 radians and t = 0 when θ is π/6 radians, we can substitute these values into the equation:
π/6 = -0.2*0 + C
C = π/6.
Now we can substitute the value of C into the equation and solve for t when θ is 30 degrees:
Δθ = -0.2t + π/6,
π/6 + π/6 = -0.2t + π/6,
π/3 = -0.2t,
t = -π/6.
Since t represents time and cannot be negative in this context, we discard the negative solution. Thus, t = π/6.
Now, we can use the speed of the boat and the value of t to find the distance from the boat to the buoy:
distance = speed * time,
distance = 10 * (π/6),
distance ≈ 5.24 meters.
When θ is 30 degrees, the boat is approximately 5.24 meters from the buoy.
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