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.

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- Mar 25th 2007, 07:58 PMcamherokidRelated 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. - Mar 25th 2007, 08:29 PMJhevon
- Mar 25th 2007, 09:19 PMcamherokid
B is sailing north

- Mar 25th 2007, 10:24 PMJhevon
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 - Mar 26th 2007, 06:23 AMSoroban
Hello, camherokid!

I have a slightly different approach to motion problems.

I think it's neater ... see what you think.

Quote:

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: .x² .= .(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.*