1. ## 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,.

2. 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.

3. Hello, samtheman17!

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

Give an expression for $\displaystyle d$, the distance between the two ships,
in terms of $\displaystyle t$, 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 $\displaystyle b$ starts at $\displaystyle H$, sails south at 6 km/hr for one hour.
At noon, ship $\displaystyle b$ is at $\displaystyle P\!:\;HP = 6$
In the next $\displaystyle t$ hours, it sails $\displaystyle 6t$ km to point $\displaystyle B\!:\;\;PB = 6t$

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

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

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

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

4. thank you so much, that really helped!