A surface ship is moving in a straight line (horizontally) at 8km/hr. At the same time, an enemy submarine maintains a position directly below the ship while driving at an angle that is 20 degrees below the horizontal. How fast is the submarines altitude decreasing?
To find the rate at which the submarine's altitude is decreasing, we can use trigonometry and differentiate the equation that relates the position of the submarine with respect to time.
Let's assume that the submarine's altitude is represented by the variable y and time by t.
From the given information, we can draw a right triangle to represent the situation. The side opposite the 20-degree angle represents the horizontal distance traveled by the submarine, and the side adjacent to the angle represents the vertical distance or altitude.
Since the ship is moving horizontally, the horizontal distance traveled by the submarine can be expressed as x = 8t, where t is the time in hours and 8 km/hr is the speed of the ship.
The equation relating the position of the submarine can be defined using trigonometry as follows:
tan(20°) = y / x
Substituting the value of x in terms of t:
tan(20°) = y / (8t)
To find the rate of change of altitude (dy/dt), we need to differentiate the equation with respect to time:
sec^2(20°) * (dθ/dt) = (dy/dt) / (8t)
where sec^2(20°) is the derivative of tan(20°), and dθ/dt represents the rate at which the angle between the horizontal and the submarine's path changes. Since the angle is not changing in this scenario, dθ/dt is equal to 0.
Therefore, the equation simplifies to:
0 = (dy/dt) / (8t)
Solving for (dy/dt), we get:
(dy/dt) = 0 * 8t
(dy/dt) = 0
Hence, the submarine's altitude is not changing or decreasing with respect to time in this scenario.