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(Easy) Calculus problem dealing with a runner

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the main broke.

    When the main broke, A runner was 6 miles east and 5000 feet north of the intersection. The runner is due west at 17 feet per second. When will the runner's feet get wet?



    3. The attempt at a solution

    Not exactly sure how to solve this problem but here's all I got:

    - Triangle, on the x-axis labelled it 31, 680 ft, on the y-axis labelled it 5,000 ft. The hypotenuse is then 32, 072.1437 ft.

    - Diffierientiated x^2 + y^2 = r^2 and got 2x dx/dt + 2y dy/dt = 2 r dr/dt

    - Looked back at the equation r = 5t, found r' to be 5, so dr/dt =5

    - Just pluged the 5 back in like any algebra problem and got

    5,000dx/dt + 31,680(-17) = 5(32072.1437)

    dx/dt = 139. 784144

    Not sure where I went wrong, I plugged the equation back into r = 5t and got 27.95 -- the book says 25.4154041 minutes so there's something wrong here
     
  2. jcsd
  3. Oct 1, 2008 #2
    Alright, well, I'm not sure if I did this the way you were supposed to, but...
    I made equations for the locations of both the runner and the water.

    For the runner, his x position: x = 31680-17t, and his y position: y = 5000.
    Since the water forms a circle, the equation of which is x^2+y^2 = r&2, the water's location can be written as: x^2 + y^2 = (5t)^2 = 25t^2.

    You want to find when the runner's position is the same as the edge of the water's position. So, you take the equations for x and y (from the runner) and put them into the equation for the water, giving you: (31680-17t)^2 + (5000)^2 = 25t^2.

    With the help of a calculator and after moving all the terms to one side, you get:
    264t^2-1077120t+1028622400=0.

    Plug those numbers into the quadratic equation and you'll get two answers for t (in seconds). Take the smaller of those two answers (when the runner FIRST gets to the edge of the water) and divide by 60 to get 25.415 minutes.
     
  4. Oct 1, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Essentially, you are asking for the intersection of two graphs. The circle giving the boundary of the water is x2+ y2= 25t2 and the runners path is given by x= 6(5280)- 17t, y= 5000. Replacing x and y in the circle formula by those gives you (6(5280)- 17t)2+ (5000)2= 25t2, a quadratic equation for t. I see no reason to differentiate. This is just an algebra problem.
     
  5. Oct 1, 2008 #4
    Thank you both for making that distinction. I am still scratching my head why this problem was assigned when the chapter is a total mismatch
     
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