Math Problem Statement

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

Solution

To solve the problem, we can use the principles of related rates in calculus.

  1. Position of the Ships at Noon:

    • Ship A is 170 km west of Ship B.
    • Let the position of Ship B be the origin (0, 0).
    • Position of Ship A: (170,0)(-170, 0).
  2. Speeds:

    • Ship A is sailing east at 35 km/h, so its x-coordinate increases.
    • Ship B is sailing north at 20 km/h, so its y-coordinate increases.
  3. Positions at 4:00 PM:

    • Time from noon to 4 PM is 4 hours.
    • Position of Ship A at 4 PM: xA=170+35×4=170+140=30 kmx_A = -170 + 35 \times 4 = -170 + 140 = -30 \text{ km}
    • Position of Ship B at 4 PM: yB=0+20×4=80 kmy_B = 0 + 20 \times 4 = 80 \text{ km}
  4. Distance Between the Ships: Let dd be the distance between the two ships: d=(xAxB)2+(yAyB)2d = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} In our case: d=(300)2+(080)2=(30)2+(80)2=900+6400=730085.44 kmd = \sqrt{(-30 - 0)^2 + (0 - 80)^2} = \sqrt{(-30)^2 + (-80)^2} = \sqrt{900 + 6400} = \sqrt{7300} \approx 85.44 \text{ km}

  5. Using Related Rates: Let x=xA+170x = x_A + 170 (position of Ship A from Ship B) and y=yBy = y_B. Then: x=30+170=140 km(sincexAiswestofB)x = -30 + 170 = 140 \text{ km} \quad (since \, x_A \, is \, west \, of \, B) The rate of change of xx with respect to time tt is dxdt=35km/h\frac{dx}{dt} = 35 \, \text{km/h} (east) and the rate of change of yy is dydt=20km/h\frac{dy}{dt} = 20 \, \text{km/h} (north).

  6. Differentiating the Distance Formula: Using the formula for distance: d2=x2+y2d^2 = x^2 + y^2 Differentiating with respect to time tt: 2ddddt=2xdxdt+2ydydt2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} Rearranging gives: dddt=xdxdt+ydydtd\frac{dd}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{d}

  7. Substituting Values: At 4 PM:

    • x=140 kmx = 140 \text{ km}
    • y=80 kmy = 80 \text{ km}
    • d85.44 kmd \approx 85.44 \text{ km}
    • dxdt=35km/h\frac{dx}{dt} = 35 \, \text{km/h}
    • dydt=20km/h\frac{dy}{dt} = 20 \, \text{km/h}

    Now calculate: dddt=(140)(35)+(80)(20)85.44\frac{dd}{dt} = \frac{(140)(35) + (80)(20)}{85.44} =4900+160085.44=650085.4476.06 km/h= \frac{4900 + 1600}{85.44} = \frac{6500}{85.44} \approx 76.06 \text{ km/h}

Therefore, the distance between the ships is increasing at approximately 76.06 km/h at 4:00 PM.

Would you like more details on any part of this problem, or do you have other questions? Here are some related questions you might find interesting:

  1. How would the distance change if Ship A was sailing west instead?
  2. What would be the distance between the ships at 5 PM?
  3. How would the problem change if Ship B was sailing south instead of north?
  4. What is the total distance covered by each ship by 4 PM?
  5. How can we visualize the movement of the ships over time?

Tip: Always diagram the situation for related rates problems; it can clarify how the distances and directions relate to each other.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Geometry
Calculus

Formulas

d = √((x_A - x_B)² + (y_A - y_B)²)
2d (dd/dt) = 2x (dx/dt) + 2y (dy/dt)

Theorems

Pythagorean Theorem
Related Rates Theorem

Suitable Grade Level

Grades 11-12