View Full Version : distance minimization

sbart

02-25-2012, 04:55 PM

At noon, ship A is 100 kilometers due east of ship B. Ship A is sailing west at 12 kilometers per hour, and ship B is sailing south at 10 kilometers per hour. At what time will the ships be nearest to each other, and what will this distance be? I know that the answer is: t is approximately 4.92, making the time approximately 4:55 PM, and distance is approximately 64km. My teacher has told me to use two formulas, one being the distance formula, but other than that I do not know how to go about the problem. Please help. Thanks!

HallsofIvy

02-25-2012, 05:27 PM

I suggest this- set up a coordinate system so that ship B is at the origin and the positive x-axis runs through ship A. Ship A starts at x= 100, y= 0 and moves along the negative x axis at -12 km/hr. What will be the x and y coordinates of ship A after t hours? Ship B starts at x= 0, y= 0 and moves along the negative y axis at -10 km/hr. What will be the x and y coordinates of ship B after t hours?

sbart

02-25-2012, 05:58 PM

Ship A (100-12t, t), Ship B (-10t, t). In the distance formula, would I use these points giving me d=[(100-2t)^2]^(1/2). How do I proceed?

soroban

02-25-2012, 08:09 PM

Hello, sbart!

At noon, ship A is 100 kilometers due east of ship B.

Ship A is sailing west at 12 km/hr, and ship B is sailing south at 10 km/hr.

At what time will the ships be nearest to each other, and what will this distance be?

I know that the answer is: t is approximately 4.92, making the time approximately 4:55 PM,

and distance is approximately 64km.

Q 100-12t A 12t P

o - - - - - o - - - - - o

| *

| *

10t | * d

| *

| *

B o

Ship A starts at point P, 100 km east of point Q.

In t hours, it travels 12t km west to point A.

. . Hence: .QA \,=\,100 - 12t

Ship B starts at point Q.

In t hours, it travels 10t km south to point B.

. . Hence: .QB\, =\, 10t

Their distance at any time is d. .Let D = d^2.

. . D \:=\:(10t)^2 + (100-12t)^2 \:=\:244t^2 - 2400t + 10,000

. . D' \:=\:488t - 2400 \:=\:0 \quad\Rightarrow\quad t \:=\:\dfrac{2400}{488} \:=\:4.918032787

Hence: .t \:\approx\:4.92\text{ hours} \:\approx\:\text{4 hours, 55 minutes}

The time will be 4:55 pm.

d^2 \:=\:244(4.92)^2 - 2400(4.92) + 10,000 \:=\:4098.3616

d \:=\:\sqrt{4098.3616} \:=\:64.01844734

The distance is about 64 km.

HallsofIvy

02-25-2012, 10:02 PM

Ship A (100-12t, t), Ship B (-10t, t). In the distance formula, would I use these points giving me d=[(100-2t)^2]^(1/2). How do I proceed?

No. "Ship A starts at x= 100, y= 0 and moves along the negative x axis at -12 km/hr."

Since it moves along the x axis, y does not change. x= 100- 12t, y= 0.

"Ship B starts at x= 0, y= 0 and moves along the negative y axis at -10 km/hr."

Since it moves along the y axis, x does not change. x= 0, y= -10t

The distance between A and B is given by \sqrt{(100-12t- 0)^2+(-10t-0)^2}= \sqrt{10000- 24t+ 244t^2}.

Now, do you know how to find max and min values for a function?

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.