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Word problem two ships departing -- One north the other west...

  1. Sep 21, 2015 #1
    Question:
    Two ships leave port at the same time. One travels north at 80 knots (that is, 80 nautical miles per hour), and the other west at 80 knots. The distance between the ships increases at a constant rate.

    At what rate is the distance between the two ships increasing?


    3. The attempt at a solution
    The distance traveled by the north bound ship = 80t [because distance traveled = speed x time taken]

    The distance traveled by the west bound ship = 80t

    Both leave port at the same time ... so they are making an ever increasing right triangle

    Let the distance between the boats at time t be d

    so d = √[(80t)² + (80t)²]

    d = √12800t²

    so dd/dt = 80sqrt(2)

    Why is this incorrect?
     
  2. jcsd
  3. Sep 21, 2015 #2

    Mark44

    Staff: Mentor

    To minimize confusion, call the distance D, not d.
    Edit: Incorrect response edited out...
    BTW, questions about derivatives should be posted in the Calculus section. I have moved your post to that section.
     
    Last edited: Sep 22, 2015
  4. Sep 21, 2015 #3

    Ray Vickson

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    Why do you think it is incorrect?
     
  5. Sep 21, 2015 #4
    The program I keep using says it is incorrect
     
  6. Sep 22, 2015 #5

    HallsofIvy

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    Exactly what form does the program expect? (That is the problem with such formats- it is not enough to get the correct answer, you must enter it in the right way.)

    However, here, if I were a teacher and a student were to enter "[itex]80\sqrt{2}[/itex]" as the answer, I would not mark it as completely correct. The correct answer is "[itex]80\sqrt{2}[/itex] knots" since it is a speed.

    (And I disagree that it was necessary to post this in "Calculus". Since everything is linear here, it can be done as an algebra problem without taking the derivative.)
     
  7. Sep 22, 2015 #6

    Ray Vickson

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    Well ##80 \sqrt{2}## is numerically correct, but perhaps the program you are using wants a decimal approximation ##80 \sqrt{2} \doteq 125.865007 \doteq 125.86 \doteq 125.9 \doteq 126##, depending on accuracy requested. And, of course, as HallsofIvy has indicated, you (may) need to supply units as well.
     
  8. Sep 22, 2015 #7
    Wait a sec...
    D = sqrt(12800)t^2
    = sqrt(6400) * sqrt(2) * t^2
    = 80 * sqrt(2) * t^2

    dD/dt = 160 * sqrt(2) * t

    Isn't it just an error with your arithmetic?
     
  9. Sep 22, 2015 #8

    Ray Vickson

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    No, he got it right: ##D = \sqrt{12800 \, t^2} = \sqrt{12800} \: t##.
     
  10. Sep 22, 2015 #9
    oops didnt see the brackets my bad :S
     
  11. Sep 23, 2015 #10
    I will see if my teacher can give me an extension on this assignment. Because this does not make any sense. The program I am using is called WebAssign and it is nortious for it's bad programming.
     
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