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Math Help - Multi-Calc Help

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    Multi-Calc Help

    Suppose there are all these bubbles, where each bubble is shaped as a different type of surface. One of the bubbles is shaped as a circular cylinder, and it will keep that general shape over time, although the height increases at a rate of 3 cm/sec while the radius will decrease at a rate of 1 cm/sec (and when the bubble gets thin enough it will eventually pop).

    1.) What rate is the volume, V, of this bubble changing when it is 10 cm high and 6 cm across at the bottom?

    2.) What rate is the surface area of the bubble changing at the same instant?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    Suppose there are all these bubbles, where each bubble is shaped as a different type of surface. One of the bubbles is shaped as a circular cylinder, and it will keep that general shape over time, although the height increases at a rate of 3 cm/sec while the radius will decrease at a rate of 1 cm/sec (and when the bubble gets thin enough it will eventually pop).

    1.) What rate is the volume, V, of this bubble changing when it is 10 cm high and 6 cm across at the bottom?

    2.) What rate is the surface area of the bubble changing at the same instant?
    1. V = (pi)r^2*h

    dV/dt = (pi)*2rh*dr/dt + (pi)r^2*dh/dt

    Just plug in the numbers.

    2. SA = 2(pi)r^2 + 2(pi)r*h

    d(SA)/dt = 2(pi)*2r*dr/dt + 2(pi)r*dh/dt

    -Dan
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    Quote Originally Posted by topsquark View Post
    1. V = (pi)r^2*h

    dV/dt = (pi)*2rh*dr/dt + (pi)r^2*dh/dt

    Just plug in the numbers.

    2. SA = 2(pi)r^2 + 2(pi)r*h

    d(SA)/dt = 2(pi)*2r*dr/dt + 2(pi)r*dh/dt

    -Dan
    Thanks, topsquark.

    So using the info. given, it would be:

    Pi*2*3*10*1 + Pi*3^2*3 = 60*Pi + 27*Pi = 87*Pi and so it is changing at a rate of 87*Pi cm/sec ?

    I have the 3 in bold since when the prob. says "6 cm across" I assume it means the diameter, and therefore the radius is 3.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    Thanks, topsquark.

    So using the info. given, it would be:

    Pi*2*3*10*1 + Pi*3^2*3 = 60*Pi + 27*Pi = 87*Pi and so it is changing at a rate of 87*Pi cm/sec ?

    I have the 3 in bold since when the prob. says "6 cm across" I assume it means the diameter, and therefore the radius is 3.
    The radius is DECREASING at a rate of 1 cm/s, so dr/dt = -1 cm/s. The correct answer is:

    Pi*2*3*10*(-1) + Pi*3^2*3 = -60*Pi + 27*Pi = -23*Pi (cm^3/s)
    so the volume is decreasing at a rate of 23(pi) cm^3/s. (Beware the units!)

    -Dan
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