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Related Rates Help

  1. Jan 3, 2005 #1
    A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

    i know how to do related rates problems...i just don't know where to start with this one...please help!!!!!
     
    Last edited: Jan 3, 2005
  2. jcsd
  3. Jan 3, 2005 #2

    quasar987

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    Do you know what the answer is?

    Personally I would just say: "There is no telling what the rate of change of the baloon and of the car will be 1 second later"... maybe a big wind will blow the balloon sideway and maybe in 0.5 seconds later the car will crash into a tree.
     
  4. Jan 3, 2005 #3
    here is my suggestion, have to wait for some one else to check to see if its right.

    i formed a right angled triangle
    the vertical side has length 200+15t, the horizontal side has length 66t and let the hypotenuse be x.
    we want to find dx/dt
    using pythagorus we get x^2=291t^2+600t+40000
    from this calculate dx/dt
    pop in t=1 second to get dx/dt=2.9ft/s
     
  5. Jan 11, 2005 #4
    I did sort of the same thing, creating a right triangle with sides x, y, and hyp of d. the taking d^2=x^2+y^2, then taking the derivative, getting d(dd/dt)=x(dx/dt)+y(dy/dt), then subbing in the given numbers, and finding d after one second, I get (224.9ft)(dd/dt)=(66ft)(66ft/sec)+(215ft)(15ft/sec), and when solved, I get (dd/dt)=33.7ft/sec at t=1
     
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