distance minimization

sbart

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Feb 4, 2012
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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!
 
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?
 
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?
 
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.

Code:
      Q  100-12t  A    12t    P
      o - - - - - o - - - - - o
      |         *
      |       *
  10t |     * d
      |   *
      | *
    B o
Ship A starts at point \(\displaystyle P\), 100 km east of point \(\displaystyle Q.\)
In \(\displaystyle t\) hours, it travels \(\displaystyle 12t\) km west to point \(\displaystyle A.\)
. . Hence: .\(\displaystyle QA \,=\,100 - 12t\)

Ship B starts at point \(\displaystyle Q.\)
In \(\displaystyle t\) hours, it travels \(\displaystyle 10t\) km south to point \(\displaystyle B.\)
. . Hence: .\(\displaystyle QB\, =\, 10t\)

Their distance at any time is \(\displaystyle d.\) .Let \(\displaystyle D = d^2.\)

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

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

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

The time will be 4:55 pm.



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

\(\displaystyle d \:=\:\sqrt{4098.3616} \:=\:64.01844734\)

The distance is about 64 km.
 
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 \(\displaystyle \sqrt{(100-12t- 0)^2+(-10t-0)^2}= \sqrt{10000- 24t+ 244t^2}\).

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