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Math Help - Derivatives and application Pythagorean: A North-South highway intersects....

  1. #1
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    Question Derivatives and application Pythagorean: A North-South highway intersects....

    Question : A North-South highway intersects an East-West highway at point P. An
    automobile crosses P at 10:00 am, traveling east at a constant speed of 20 km/h.
    A car north of P, twenty minutes later passes that same point, traveling south at 60
    km/h. Find the approximate the minimum distance between the automobiles.
    Answer: 6.3 km

    I keep getting 7.8 km as my answer- don't know if I'm setting up the equations wrong or if the answer is wrong. Please help~ and thanks in advance
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  2. #2
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    Re: Derivatives and application Pythagorean: A North-South highway intersects....

    Hello, Errisa!

    The problem has a strange answer . . .


    A north-south highway intersects an east-west highway at point P.
    A car crosses P at 10:00 am, traveling east at a constant speed of 20 km/hr.
    A car north of P, twenty minutes later passes that same point, traveling south at 60 km/hr.
    Find the approximate the minimum distance between the automobiles.
    Answer: 6.3 km

    Code:
          |
          | 20/3  Q   20t   R
        P o - - - o - - - - o
          |              *
          |           *
      60t |        * x
          |     *
          |  *
        S o
    At 10:00, car A drives east from P at 20 km/hr.
    By 10:20, it has moved to point Q:\;PQ = \tfrac{20}{3} km.
    In the next t hours, it has moved 20t km to point R.

    In the same t hours, car B moves from P to S at 60 km/hr.
    . . PS\,=\,60t

    Let x = distance between the cars, RS.

    Pythagorus: . x \;=\;\sqrt{(20t + \tfrac{20}{3})^2 + (60t)^2} \;=\;\left(4000t^2 + \tfrac{800}{3}t + \tfrac{400}{9}\right)^{\frac{1}{2}}

    To minimize x, set \tfrac{dx}{dt} = 0 and solve.


    \frac{dx}{dt} \;=\;\tfrac{1}{2}\left(4000t^2 + \tfrac{800}{3}t + \tfrac{400}{9}\right)^{-\frac{1}{2}}\left(8000t + \tfrac{800}{3}\right) \;=\;0

    We have: . 8000t + \tfrac{800}{3} \:=\:0 \quad\Rightarrow\quad t \:=\:-\tfrac{1}{30}


    Negative time?
    . . . The minimum distance occured two minutes before 10:20.

    At 10:18, car A was 6 miles east of point P.
    At 10:18, car B was 2 miles north of point P.

    Their distance was: . \sqrt{6^2 + 2^2} \:=\:\sqrt{40} \:=\:6.32455532 \;\approx\;6.3\text{ km.}
    Thanks from Errisa
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    Re: Derivatives and application Pythagorean: A North-South highway intersects....

    We need to be careful with our units. I'm going to use kilometers for distance and hours for time, with t=0 representing 10:00 am. The horizontal position of the first car relative to the intersection at time t is 20t and the vertical position of the second car is 60\left(\frac13-t\right) (because 20 minutes = 1/3 hours). Draw a picture:



    Using the Pythagorean Theorem, the distance between A and B turns out to be

    s = 20\sqrt{10t^2-6t+1}

    To make the differentiation easier, let's minimize the square of the distance.

    s^2 = 400\left(10t^2-6t+1\right)

    \frac d{dt}\left[s^2\right] = 400(20t-6)

    \frac d{dt}\left[s^2\right] = 0\Rightarrow t = \frac3{10}

    So the cars are closest at 10:18 am. At this time \left(t = \frac3{10}\right), the distance between the cars is

    s = 20\sqrt{10\left(\frac3{10}\right)^2-6\left(\frac3{10}\right)+1}

    = 20\sqrt{\frac1{10}} = 2\sqrt{10} \approx 6.325\ \mathrm{km}



    Edit: Beaten by two minutes! Dang it Soroban, nice work.
    Attached Thumbnails Attached Thumbnails Derivatives and application Pythagorean:  A North-South highway intersects....-mhf_20120618b.png  
    Last edited by Reckoner; June 18th 2012 at 07:42 PM.
    Thanks from Errisa
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    Thumbs up Re: Derivatives and application Pythagorean: A North-South highway intersects....

    thank you! Both of you~ I see which step I messed up on I didn't add 20/3 to 20t!

    Sorry about late reply I was finishing up my other questions~

    Thanks once again
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