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Calculus word problem

  1. Apr 17, 2006 #1
    This is what am trying to solve

    A balloon is rising at the rate of 5ft/s. A boy is cycling along a straight road at a speed of 15ft/s.when he passes under the balloon it is 45 ft above him.
    How fast is the distance between the balloon and the boy increasing 3 seconds later

    Please let me know if am on the right path

    dy/dt = 5
    dx/dt = 15
    y= 45


    2s ds/dt = 2x dx/dt + 2y dy/dt

    is this correct but also we can conclude that ds/dt = 15.81,
    but i think we have to find ds/dt.
    In the above equation am getting two unkowns thats were am getting stuck. Please help, thanks.
  2. jcsd
  3. Apr 17, 2006 #2


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    The "y = 45" you should write as y(0) = 45 since y is really a function. In the equation you have last, 2s ds/dt = 2x dx/dt + 2y dy/dt, you have all the information you need. You can find s, x, and y by simple algebra and geometry, and you already know dx/dt and dy/dt. Then you can solve easily.
    Last edited: Apr 17, 2006
  4. Apr 17, 2006 #3
    Thanks for replying. In this problem since its given to us that the cyclist is travelling at 15 ft/s, so in 3 seconds he will cover 45 ft and this will be the value of x. Now i have x, and y so i can find s and that will be s = 63.64 ft.

    Now i can plug everything in the equation

    ds/dt = 1/s (x dx/dt + y dy/dt)

    so am getting ds/dt = 14.14 ft/s

    please let me know if this is correct, sorry bout 15.81.
    Last edited: Apr 17, 2006
  5. Apr 17, 2006 #4


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    Not quite, you don't have the right value for s or y. Remember the balloon is moving too.

    My attitude is, it's far better to make errors at first but be right in the end than say nothing.
    Last edited: Apr 17, 2006
  6. Apr 18, 2006 #5
    I used pythagorean theorem to get 15.81. I was thinking 15 to be the base and 5 to be the adjacent side.

    Y= 45 ft high only when the cyclist is right under it. As the cyclist is moving away the balloon keeps on rising so the y value is changing. Balloon is rising at the rate of 5 ft/s, so this is dy/dt and 15 ft/s is dx/dt. My value of x is correct right. The cyclist is moving at 15 ft/s and in 3 seconds he would cover 45 ft. When the cyclist is 45 ft away from the balloon the height of the balloon will be 4 Y = 45 + y true. Thanks for looking into this problem.
  7. Apr 18, 2006 #6


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    The height of the balloon will be 60 feet. It is at 45 feet at time 0 and it moves with a speed of 5 ft/s over a period of 3 seconds.
  8. Apr 18, 2006 #7
    Thanks orthodontist for ur help, am getting the answer
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