Post a New Question

Calculus

posted by .

At 4 PM ship A is 40 miles due South Ship B. Ship A is sailing due South at a rate of 20 knots while ship B is sailing due East at a rate of 25 knots. Find the rate of change of the distance between the ships at 3 PM and at 5 PM. At what time were the ships closest together ?

[ Hint let t = 0 correspond to 4 PM , so t = –1 corresponds to 3 PM and t = 1 corresponds to 5 PM
Find dL/dt at these times where L is the distance between the ships at time t ]

  • Calculus -

    The distance L between the ships, at t hours after 4pm, is

    L^2 = (40 + 20t)^2 + (25t)^2
    2L dL/dt = 2(40+20t)*20 + 2*25t*25
    L dL/dt = 800 + 40t + 625t
    plug in L at t=-1 and +1 to get dL/dt at 3pm and 5pm

    dL/dt = 0 when t = --32/41, or at 3:13 pm, when L was 31.23mi

  • Calculus -

    make that L dL/dt = 800 + 400t + 625t

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question