Related Rates

[Goldy]

hey i am very much new to this and in desperate need of help. I have been assigned to answer thinking and inquire questions on related rates. The question i am stuck on is:

A ship leaves a port at 12 noon and travels due west at 20 knots. At 12 noon the next day, a second ship leaves the same port and travels northwest at 15 knots. how fast are the two ships separating when the second ship has traveled 90 nautical miles?

if someone can give me some guidelines to help me answer these question it would be great!

 

[soroban]

Hello, Goldy!

I'll try to explain the set-up . . .

A ship leaves a port at 12 noon and travels due west at 20 knots.
At 12 noon the next day, a second ship leaves the same port and travels northwest at 15 knots.
How fast are the two ships separating when the second ship has traveled 90 nautical miles?

First, I'll change <u>everything</u> to nautical miles.
Then "20 knots" = 20 mph, and "15 knots" = 15 mph (nautical miles per hour).

                          B
                          *            The port is at P.
                d     *     \
                  *           \15t     Ship A has had a 24-hour headstart
              *                 \      and has moved 480 miles to point X.
          *                   45° \ 
      * - - - - - * - - - - - - - -*   Then ship B starts moving northwest.
      A    20t    X      480     P

In the next \(\displaystyle t\) hours, ship A will move \(\displaystyle 20t\) more miles to point A
. . and ship B will move \(\displaystyle 15t\) miles to point B.

For the distance between them, \(\displaystyle D\), use the Law of Cosines:

. . . \(\displaystyle D^2\;=\;(20t+480)^2\,+\,(15t)^2\,-\,2(20t+480)(15t)\cos45^o\)

Differentiate with respect to time to get \(\displaystyle \frac{dD}{dt}\)
. . and evaluate when \(\displaystyle t = 6.\) *

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* . That is when the second ship has traveled 90 miles, you see.