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:

  • AA be the position of Ship A.
  • BB be the position of Ship B.
  • xx be the horizontal distance (east-west direction) between the ships.
  • yy be the vertical distance (north-south direction) between the ships.

We are given:

  • At noon, x=40x = 40 nautical miles (Ship A is 40 nautical miles due west of Ship B).
  • Ship A is sailing west at 17 knots, meaning dxdt=17\frac{dx}{dt} = -17.
  • Ship B is sailing north at 16 knots, meaning dydt=16\frac{dy}{dt} = 16.

Step 2: Determine xx and yy at 7 PM

Since the ships start moving at noon, 7 PM is 7 hours later.

For Ship A:

  • x=40+(17 knots×7 hours)x = 40 + (17 \text{ knots} \times 7 \text{ hours})
  • x=40+119=159x = 40 + 119 = 159 nautical miles.

For Ship B:

  • y=16 knots×7 hoursy = 16 \text{ knots} \times 7 \text{ hours}
  • y=112y = 112 nautical miles.

Step 3: Use the Distance Formula

Let zz be the distance between the ships. Using the Pythagorean theorem:

z=x2+y2z = \sqrt{x^2 + y^2}

Step 4: Differentiate zz with Respect to tt

To find dzdt\frac{dz}{dt}, the rate of change of the distance between the ships, we differentiate both sides of the equation z=x2+y2z = \sqrt{x^2 + y^2} with respect to tt:

dzdt=12x2+y22xdxdt+2ydydt\frac{dz}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x\frac{dx}{dt} + 2y\frac{dy}{dt}

Simplifying:

dzdt=xdxdt+ydydtx2+y2\frac{dz}{dt} = \frac{x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}}{\sqrt{x^2 + y^2}}

Step 5: Substitute Known Values

Now we substitute x=159x = 159, y=112y = 112, dxdt=17\frac{dx}{dt} = -17, and dydt=16\frac{dy}{dt} = 16:

  1. First, calculate zz: z=1592+1122z = \sqrt{159^2 + 112^2} z=25281+12544=37825194.5 nautical milesz = \sqrt{25281 + 12544} = \sqrt{37825} \approx 194.5 \text{ nautical miles}

  2. Then, substitute into dzdt\frac{dz}{dt}: dzdt=159(17)+112(16)194.5\frac{dz}{dt} = \frac{159(-17) + 112(16)}{194.5} =2703+1792194.5= \frac{-2703 + 1792}{194.5} =911194.5= \frac{-911}{194.5} 4.68 knots\approx -4.68 \text{ knots}

Answer:

The distance between the two ships is decreasing at a rate of approximately 4.684.68 knots at 7 PM.

Would you like further details or have any questions?


Here are some related questions that may interest you:

  1. How would the solution change if one of the ships changed course?
  2. What if the ships were initially positioned differently, such as one being north or south instead of due west?
  3. How would you calculate the distance if only the time period is different (e.g., 5 PM)?
  4. Can we solve this problem using parametric equations instead?
  5. 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