This is a related rates problem involving horizontal and vertical motion.
We know that the ship travels horizontally at 10 km/hr, and the submarine dives at an angle of 20° below the horizontal.
We want to find how quickly the submarine is descending vertically.
Step 1: Identify the Given Information
Horizontal speed = 10 km/hr
Diving angle = 20°
Vertical speed = Unknown
Since the submarine stays directly below the ship, its horizontal speed must equal the ship's horizontal speed.
Therefore:
dx/dt = 10 km/hr
Step 2: Use Trigonometry
The submarine's movement forms a right triangle.
The horizontal distance is the adjacent side, and the vertical distance is the opposite side.
Using tangent:
tan(20°) = Vertical Distance / Horizontal Distance
Let x represent horizontal distance and y represent downward vertical distance.
Then:
y = x tan(20°)
Step 3: Differentiate With Respect to Time
Since both x and y change with time, differentiate:
dy/dt = tan(20°) × dx/dt
Substitute the known values:
dy/dt = tan(20°) × 10
Since:
tan(20°) ≈ 0.36397
We get:
dy/dt = 10 × 0.36397
dy/dt ≈ 3.64 km/hr
Final Answer:
The submarine is descending at approximately 3.64 km/hr.
Easy Way to Remember:
When an object moves at an angle below the horizontal, its vertical speed can be calculated using:
Vertical Speed = Horizontal Speed × tan(Angle)
In this problem:
Vertical Speed = 10 × tan(20°)
Vertical Speed = 3.64 km/hr
The submarine is moving horizontally at the same speed as the ship while simultaneously descending at 3.64 km/hr.
