# related rates

• Oct 17th 2009, 03:48 PM
samtheman17
related rates
At noon, ship A passes 9 km west of a harbour, heading due south at 8 km/h. At 11 a.m., ship B had left the harbour, sailing due south at 6 km/h. Give an expression for d, the distance between the two ships, in terms of t, the number of hours after noon.

i really dont even know how to start on this problem,.
• Oct 18th 2009, 06:28 AM
skeeter
Quote:

Originally Posted by samtheman17
At noon, ship A passes 9 km west of a harbour, heading due south at 8 km/h. At 11 a.m., ship B had left the harbour, sailing due south at 6 km/h. Give an expression for d, the distance between the two ships, in terms of t, the number of hours after noon.
i really dont even know how to start on this problem,.

start by making a sketch of the situation.

let the harbor be at the origin and let t = 0 be noon

at t = 0 , ship A is at position (-9,0) and is heading south at 8 km/hr ...

its position at any time t is (-9, -8t)

at t = 0, ship B is at position (0, -6) and is heading south at 6 km/hr ...

its position at any time t is (0, -6 - 6t)

use the distance formula to find the distance between the two ships at any time t.
• Oct 18th 2009, 07:57 AM
Soroban
Hello, samtheman17!

Quote:

At 11 a.m., ship $b$ leaves the harbor, sailing due south at 6 km/hr,
At noon, ship $a$ passes 9 km west of the harbor, heading due south at 8 km/h.

Give an expression for $d$, the distance between the two ships,
in terms of $t$, the number of hours after noon.

First, make a sketch!
Code:

              9     Q * - - - - - - - - * H       |                |   8t |                |       |                | 6     A *                |       :  *              |       :    * d        * P  6-2t :        *        |       :          *    | 6t       :              *  |     C * - - - - - - - - * B               9

At 11 a.m. ship $b$ starts at $H$, sails south at 6 km/hr for one hour.
At noon, ship $b$ is at $P\!:\;HP = 6$
In the next $t$ hours, it sails $6t$ km to point $B\!:\;\;PB = 6t$

At noon, ship $a$ starts at $Q\!:\;HQ = 9$ km,
. . and sails south at 8 km/hr,
In the next $t$ hours, it sails $8t$ km to point $A\!:\;QA = 8t$

Note that: . $AC \:=\:(6 + 6t) - 8t \:=\:6 - 2t$

In right triangle $ACB\!:\;d^2 \;=\;(6-2t)^2 + 9^2$

Therefore: . $d \;=\;\sqrt{4t^2 - 24t + 117}$

• Oct 18th 2009, 05:05 PM
samtheman17
thank you so much, that really helped! :)