Math Problem Statement
At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 7 PM
Solution
To find the rate at which the distance between the two ships is changing at 7 PM, we can break down the problem into several steps.
Step 1: Define the Positions of the Ships
Let:
- be the position of Ship A.
- be the position of Ship B.
- be the horizontal distance (east-west direction) between the ships.
- be the vertical distance (north-south direction) between the ships.
We are given:
- At noon, nautical miles (Ship A is 40 nautical miles due west of Ship B).
- Ship A is sailing west at 17 knots, meaning .
- Ship B is sailing north at 16 knots, meaning .
Step 2: Determine and at 7 PM
Since the ships start moving at noon, 7 PM is 7 hours later.
For Ship A:
- nautical miles.
For Ship B:
- nautical miles.
Step 3: Use the Distance Formula
Let be the distance between the ships. Using the Pythagorean theorem:
Step 4: Differentiate with Respect to
To find , the rate of change of the distance between the ships, we differentiate both sides of the equation with respect to :
Simplifying:
Step 5: Substitute Known Values
Now we substitute , , , and :
-
First, calculate :
-
Then, substitute into :
Answer:
The distance between the two ships is decreasing at a rate of approximately knots at 7 PM.
Would you like further details or have any questions?
Here are some related questions that may interest you:
- How would the solution change if one of the ships changed course?
- What if the ships were initially positioned differently, such as one being north or south instead of due west?
- How would you calculate the distance if only the time period is different (e.g., 5 PM)?
- Can we solve this problem using parametric equations instead?
- What if both ships were moving in the same direction at different speeds?
Tip: When analyzing related rates problems, always start by defining what you know and finding an equation that links the variables you need to differentiate.
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Math Problem Analysis
Mathematical Concepts
Related Rates
Pythagorean Theorem
Differentiation
Formulas
z = √(x² + y²) (Distance formula using Pythagorean theorem)
dz/dt = (x * dx/dt + y * dy/dt) / √(x² + y²) (Derivative of distance with respect to time)
Theorems
Pythagorean Theorem
Chain Rule (for differentiation)
Suitable Grade Level
Grades 11-12