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March 26, 2017

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t: 0 2 5 7 11 12
r'(t): 5.7 4.0 2.0 1.2 0.6 0.5

The volume of a spherical hot air balloon expands as the air inside the balloon is heated. The radius of the balloon in feet is modeled by a twice-differentiable function of r of time t, where t is measured in minutes. For 0<t<12, the graph of r is concave down. The table above give selected values of the rate of change, r'(t), of the radius of the balloon over the time interval 0<t<12. The radius of the balloon is 30ft when t=5.

a)est. radius of balloon when t=5.4 using tan line approximated at t=5. Why?

b)Find rate of change of the volume of the balloon w/ respect to time when t=5. what is the units of measure?

  • Calculus - ,

    sorry about that: when t=0, r't=5.7;
    t=2, r't=4; t=5, r't=2; t=7, r't=1.2; t=11, r't=.6; t=12, r't=.5

  • please help!!! - ,

    i really don't know how to start this problem

  • Calculus - ,

    I have not answered your question since I am somewhat puzzled by the wording of your question.

    by <<The radius of the balloon in feet is modeled by a twice-differentiable function of r of time t>> do you mean that r is a quadratic function of t, a cubic or what?

    As I see it, I could differentiate a function any number of times, just because I might reach a derivative of zero does not mean I couldn't do it once more.

    e.g y = x^3 + ...

    y' = 3x^2 ....
    y'' = 6x ....
    y''' = 6
    y'''' = 0
    y ''''' = 0

  • Calculus - ,

    i'm guessing that twice differentiable means y''

  • Calculus - ,

    agree, but what level is the function?

    e.g. If I knew that y = ... is quadratic, I could say y = at^2 + bt + c and go from there, but how do I know it's not a cubic?

  • Calculus - ,

    It does not matter I think.
    We are not fitting the entire data set with a function but doing linear, then quadratic, interpolation.
    For part a use radius at 5 and dr/dt at 5
    r(5.4) = r(5) + .4 r'(5)
    r(5.4) = 30 + .4 (2)
    r(5.4) = 30 + .8
    r(5.4) = 30.8

  • Calculus - ,

    V = Volume = (4/3) pi r^3 in ft^3
    DV/dt = (4/3) pi (3) r^2 dr/dt in ft^3/sec
    at t = 5
    dV/dt = 4 pi r^2 (2)
    = 8 pi (30)^2 = 240 pi ft^3/second

    NOTE - the rate of change of volume is the surface area, 4 pi r^2 times the rate of change of radius - think about it.

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