# Math Help - Related Rates 2

1. ## Related Rates 2

At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35km/h and ship B is sailing at 25km/h.
How fast is the distance between the ships changing at 4PM.

2. Originally Posted by camherokid
At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35km/h and ship B is sailing at 25km/h.
How fast is the distance between the ships changing at 4PM.
you forgot to state what direction ship B is going in

3. B is sailing north

4. Originally Posted by camherokid
At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35km/h and ship B is sailing north at 25km/h.
How fast is the distance between the ships changing at 4PM.
see diagram below

let x be the distance ship B has travelled
let y be the distance ship A has travelled
let z be the distance between the ships at any time

after 4 hours:

ship A has travelled 35(4) = 140 km
ship B has travelled 25(4) = 100 km

by pythagoras, after 4 hours
z^2 = 100^2 + (x + y)^2 = 100^2 + (140 + 100)^2 = 67600
=> z = 260 km

Now, by pythagoras again:

z^2 = 100^2 + (x + y)^2
=> 2z dz/dt = 2(x + y)(dx/dt + dy/dt)
=> dz/dt = [2(x + y)(dx/dt + dy/dt)]/2z = [(x + y)(dx/dt + dy/dt)]/z
now plug in the values:
=> dz/dt = [(100 + 140)(25 + 35)]/260 = 240(60)/260
=> dz/dt = 55.38 km/h .........the rate at which the distance between the ships is changing

5. Hello, camherokid!

I have a slightly different approach to motion problems.
I think it's neater ... see what you think.

At noon, Ship A is 100km west of ship B.
Ship A is sailing south at 35km/h
and ship B is sailing north at 25km/h.
How fast is the distance between the ships changing at 4 PM?
Code:
                          * B
/ ↑
/   ↑ 25t
/     ↑
P * - - - - - / - - - * Q
↓       x /         :
↓       /           :
35t ↓     /             : 35t
↓   /               :
↓ /                 :
A * - - - - - - - - - * R
100

Ship A starts at point P and sails south at 35 km/hr.
. . In t hours, it has sailed 35t km south to point A.

Ship B starts at point Q and sails north at 25 km/hr.
. . In t hours, it has sailed 25t km north to point B.

Let x = AB.

In right triangle BRA, we have: . .= .(60t)² + 100² .= .3600t² + 10,000

Differentiate with respect to time: .2x·(dx/dt) .= .7200t

. . and we have: .dx/dt .= .3600t/x .[1]

At 4 PM (t = 4), the triangle has: AR = 100, BR = 240. .Hence: x = 260.

Substitute into [1]: .dx/dt .= .3600(4)/260 . .55.38 km/hr
. . Their distance is increasing.