anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
Okay, why don't you assign a set of coordinates to the ships, and write equations showing their position at time \(t\)?|dw:1371401707581:dw| Then use the distance formula to write an equation showing the distance between them at time \(t\). That's a going to be a parabola, and you just need to find the vertex.
anonymous
  • anonymous
can't we just use the Pythagorean therom?
whpalmer4
  • whpalmer4
What is the position equation for the ship sailing north? Well, the \(x\) value remains constant at 0, and the \(y\) value is given by \(y = 0+v_1 t\) or \(y = 12t\). So the position of ship 1 at time \(t\) is \((0, 12t)\) — does that make sense to you?

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whpalmer4
  • whpalmer4
We will use the Pythagorean theorem (in the form of the distance formula between two points). Of course, if you know already where each ship is when they are closest, we could skip right to that. Do you? :-)
whpalmer4
  • whpalmer4
I've got a medal ready for you, if so :-)
anonymous
  • anonymous
Very interesting problem. Tracing this out on my desk with my fingers, I believe the closest approach would be the point where A intersects the path of B, which will be directly west of B. At the point of observation, this intersect is 15km ahead of A, and at 12km/h, A will arrive there in 15/12 hours. In that same time, B will have traveled 15/12*9km, and that will be their closest approach.
whpalmer4
  • whpalmer4
Right now, they are 15 km apart. After 1 hour, the first ship will be at (0,12) and the second ship will be at (15,9). Distance at that point will be \[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{(15-0)^2+(9-12)^2} = \sqrt{225+9} \approx 15.3\] After 2 hours, the first ship will be at (0,24) and the second ship at (15,18). Distance then will be \[d=\sqrt{(15-0)^2+(18-24)^2} \approx 16.16\]
whpalmer4
  • whpalmer4
@qweqwe123123123123111 what will their separation be at that point?
anonymous
  • anonymous
I'm getting 11.25 km
whpalmer4
  • whpalmer4
What if I told you they get closer than that? :-)
anonymous
  • anonymous
Then I'd like to see how. :-)
whpalmer4
  • whpalmer4
Think about it — the hypotenuse of a right triangle is always less than the sum of the other two sides, right? So there's a point before A gets to B's wake where it is shorter to go across...
whpalmer4
  • whpalmer4
If we have the first ship start sailing north at 12km/h from (0,0) and the second ship sailing due east at 9km/h from (0,15), the position of ship 1 as a function of \(t\) will be \((0,12t)\) and ship 2 will be \((9t,15)\) right?
anonymous
  • anonymous
B is sighted 15km ahead. A is traveling 12km/h, so it will take 15/12 or 1.25 hours to arrive at that same point. FROM that point, B will be traveling at 9km/h for that same 1.25 hours, or 11.25 km. |dw:1371403043264:dw|
whpalmer4
  • whpalmer4
Now, apply the distance formula to get the distance as a function of \(t\): \[d = \sqrt{(9t-0)^2+(15-12t)^2}\]If we square both sides we get \[d^2 = 81t^2+225-360t+144t^2\] which is a parabola.
whpalmer4
  • whpalmer4
We can find the vertex of that parabola, and that will be the point of closest approach.
whpalmer4
  • whpalmer4
Here's a plot of distance vs t
1 Attachment
whpalmer4
  • whpalmer4
Actually I guess my formula is a hyperbola not a parabola...no matter :-) Find the solution, and you know d and t for the point of closest approach.
whpalmer4
  • whpalmer4
It happens before A gets to B's wake.
anonymous
  • anonymous
Drat. I thought this problem would only last a couple of minutes, which is why I decided to comment. I apologize, but I have to go get ready for dinner, so I've run out of time to study your graph or your rationale, both of which are quite intriguing to me and which I really do want to examine. :-) But are you saying closest approach occurs after only about .8 hours after observation? I suppose for now, we could ask @0202 what his answer is! :-)
anonymous
  • anonymous
@qweqwe123123123123111 lol i'm a girl. sorry guys i haven't been replying, my computer was messing up but now i'm back....but for the answer, the textbook states that 9km is the closest distance
whpalmer4
  • whpalmer4
That's correct. It takes place at t = 4/5 hour. Do you know calculus?
whpalmer4
  • whpalmer4
(I could tell from your icon that you're a cute girl, btw ;-)
anonymous
  • anonymous
You aren't the only one with computer problems... :-( I just tried responding to whpalmer4's PM, and the damned system just sits there staring at me. Anyway, @whpalmer4, you're correct. I just graphed your figures on a plot (rather than using my fingers on my desk), and I concede: Their closest approach is 9km after .8 hours. Good job!! :-)
whpalmer4
  • whpalmer4
Here's how I got the figures: the distance formula gave the distance as a function of \(t\) as \(225-360t+225t^2\) I want to find the minimum, so I take the first derivative, which gives me \(-360+450t\)At a minimum, the first derivative will be 0 (remember, it's the slope of the tangent to the curve at that point), so \[-360+450t=0 \rightarrow t=4/5\]\[d=\sqrt{225-360(4/5)+225(4/5)^2} = \sqrt{225-72*4+16*11} = \sqrt{81}=9\]
whpalmer4
  • whpalmer4
It's worth drawing the triangle representing the positions of the ships at t = 4/5 to convince yourself...
anonymous
  • anonymous
Yes, that's exactly what I did. Which is why I conceded... :-) I'm going to bookmark this thread so I can examine your math later until I understand exactly what you did. But for now, I'm already late for dinner...I have GOT to go! :-) Take care all! :-)
whpalmer4
  • whpalmer4
So, I have no idea if I solved this the way the instructor wanted it solved, but hopefully you've grasped the approach I used. If not, ask some more questions :-)

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