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 ]
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
make that L dL/dt = 800 + 400t + 625t
To find the rate of change of the distance between the ships at 3 PM (t = -1) and 5 PM (t = 1), we need to determine the equation for the distance between the ships as a function of time.
Let's break down the problem and solve it step by step:
Step 1: Set up a coordinate system.
Assume an x-y coordinate system, where ship A is located at (0, 0) and ship B initially at (0, 40).
Step 2: Determine the position of the ships at any given time t.
Since ship A is moving due South at a rate of 20 knots, its position at time t is given by (0, -20t).
Similarly, ship B is moving due East at a rate of 25 knots, so its position is given by (25t, 40).
Step 3: Calculate the distance between the ships at any given time t.
The distance between two points (x1, y1) and (x2, y2) is given by the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Using this formula, we can calculate the distance, L, between the ships at time t:
L = sqrt((25t - 0)^2 + (40 - (-20t))^2).
Step 4: Find the derivative of L with respect to t.
To find the rate of change of the distance between the ships, we need to find dL/dt.
dL/dt = (dL/dx)*(dx/dt) + (dL/dy)*(dy/dt),
where x and y are the variables in the distance formula, and t is the independent variable.
Differentiating L with respect to t gives:
dL/dt = 25*2t + (-20)*2(-20t) = 50t + 800t.
So, dL/dt = 850t.
Step 5: Substitute t values to find the rate of change at 3 PM and 5 PM.
At t = -1 (3 PM), dL/dt = 850*(-1) = -850 knots.
At t = 1 (5 PM), dL/dt = 850*1 = 850 knots.
Therefore, the rate of change of the distance between the ships at 3 PM is -850 knots, and at 5 PM is 850 knots.
Step 6: Determine when the ships were closest together.
The time when the rate of change is 0 corresponds to the ships being closest together.
To find this time, we set dL/dt = 0 and solve for t:
850t = 0,
t = 0.
Since t = 0 corresponds to 4 PM, the ships were closest together at 4 PM.
So, the rate of change at 3 PM and 5 PM is -850 knots and 850 knots, respectively, and the ships were closest together at 4 PM.