Calculas h/w help

Question

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 24 knots and ship B is sailing north at 21 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

Solution:

Set up the coordinate plane so that ship A is moving along the negative x-axis with the coordinate x and ship B is moving along the positive y-axis with the coordinate y. Let z be the distance between A and B. Then

z^2 = x^2 +y^2

Differentiate both sides with respect to time t:

2z dz/dt = 2x dx/dt +2y dy/dt

dz/dt = (x dx/dt + y dy/dt) / z (1)

At 4 PM,

x = -(10 + 24*4)= -106

y = 21*4 = 84

z = sqrt[(-106)^2 + 84^2] = sqrt(18292)

We also know that at any moment,

dx/dt = -24

dy/dt = 21

Substitute all the values of x, y, z, and dx/dt, dy/dt above into equation (1):

dz/dt = [-106*(-24) + 84*21] / sqrt(18292)

≈ 31.85

Answer: The distance between the ships is increasing at a rate of 31.85 knots at 4 PM.