optimization: At what time were the ships the closest?

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dirtbagbaseball08

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At 5 P.M, an boat traveling west at 15 knots (nautical miles per hour) pases the same point as a ship that arrived at the same spot at 4 P.M. while traveling north at 25 knots. At what time were the ships closest together. Hint: The primary equation can be set up as a function of one variable so no secondary is needed.
 
Okay; so what picture did you draw? How did you label it? What equation(s) have you tried?

Please reply with all of your work and reasoning thus far. Thank you.

Eliz.
 
i drew an arrow going north and labeled it 50 nautical miles and another arrow going west and labeled it 15 nautical miles

i assume i am going to use the distance formula but that is as far as i have gotten
 
dirtbagbaseball08 said:
i drew an arrow going north and labeled it 50 nautical miles and another arrow going west and labeled it 15 nautical miles
Click to expand...
Why did you use these labels? Why did you not use any variables? Where did the lines cross? And so forth.

For instance, I would have started by noting that the first boat travelled fifty knots in the hour before the other boat crossed his path. I would have drawn an horizontal line going left ("west") from the crossing point, and a vertical line going up ("north") from the same point. I would have noted that, at time "t" hours after the second boat crossed the first boat's path (that is, "t" hours after we started counting), the distance of the first boat from the crossing point was 25 + 25t.

What is the "distance" expression for the second boat?

What do you get when you use the Distance Formula?

What do you get when you differentiate?

We won't be there to help you on the tests, so you really do need to learn how to at least start these on your own. Please reply showing your progress. Thank you.

Eliz.
 
i just know that for ever 25 miles the boat goes north, the other ship goes west 15 miles, i do not know the distance at which they are closest, could you help me set a picture? because my picture does not seem quite right
 
It can be quite discouraging when students show little or no effort, to the extent of not even appearing to read the replies they receive. I certainly hope that isn't the case here....

I have already described the picture, including my reasoning for why I drew it that way. I also gave one of the "distance" expressions. You should have been able to finish (or at least make considerable progress) from that point, just from what you learned back in algebra.

Please reply showing your progress. Thank you.

Eliz.
 
so i have the vertical line labeled as 25+25t, and i have the horizontal line labeled as 15t, does this sound right so far? is there someway to plug this into the distance formula or what now?
 
Please do not start any more threads on this exercise. If you are needing guaranteed on-demand replies, you need to contract with a paid service which offers this. Otherwise, please be considerate of the fact that the volunteers here do have "outside lives" that they tend to from time to time; you should expect to wait hours, if not days, for an online reply.

Meanwhile, have you considered the Distance Formula? Or the Pythagorean Theorem?

Did you not take algebra before taking calculus...?

Eliz.
 
i tried the distance formula but it had two variables, and i was not sure how the numbers fit into that formula, could u please help me find the right formula?
 
dirtbagbaseball08 said:
i tried the distance formula but it had two variables...
Click to expand...
What two variables?

dirtbagbaseball08 said:
could u please help me find the right formula?
Click to expand...
I named two formula options. Have you never encountered the Distance Formula or the Pythagorean Theorem before?

Try looking up the Pythagorean Theorem. (You should be able to find information on this in any search engine.) Then try drawing the picture I described. Then try using the variable I gave, along with the two "distance from the crossing point" expressions for the boats. Then see if you can figure out any way that the two "distance from the crossing point" lengths might possibly relate to the hypotenuse (being the distance between the two boats). Then see if you can figure out some way to plug that into the Formula.

If you are still stymied, then it would seem that you need to have a serious talk with your academic advisor about course placement, since you appear to be missing too much of the foundational algebra and geometry background material to be ready for calculus.

Eliz.
 
i have studied the pythagorean therom and the distance formula. i tried using the distance formula but i was unsure of what numbers to plug in and to where, it had an x and y variable, which made me usure. i used the pythagorean therom but it just gave me the hypotenuse of the 50 by 15 mile triangle and i need the distance where it was the shortest distance between the ships, you are not giving me any hint on th next step, you are telling me the same things i have been telling you, you tell me to use the distance and pythagoreon therom formulas, and i have already told you that i have and was unsuccessful
 
If I undertsand correctly, let's start the clock ticking as ship B passes point P.

Distance = rate * time.

In that time, ship A will have travelled 15-15t knots.

Ship B will have travelled 25t knots.

Use Pythagoras:

\(\displaystyle D^{2}=(15-15t)^{2}+(25t)^{2}\)

That's what you have to minimize.

boatsuz6.gif
 
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