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Math Help - Using differentials in an applied problem

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    Using differentials in an applied problem

    Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 cm thick to a hemispherical dome with a diameter 50m.

    Okay, so I've determined that the dome has a surface area of 3927, but after this point I'm not sure what to do, or if knowing the surface area is even helpful. I don't know how to put this into a workable function...
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    Here's another one that I also find difficult in terms of setting it up:

    Sand is poured onto the top of a conical pile at a rate of 100m^3/minute. The coefficient of friction of the sand is such that the height and radius are always the same. At what rate is the height increasing when the pile is 12 meters high? The formula for the volume of a cone is V = 1/3pir^2h, but since the height and radius are the same this can be rewritten as V = 1/pih^3.

    Again, I'm pretty much lost. I vaguely understand that the dV/dt is related to the dh/dt, but I'm not sure how to make it work.
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    Quote Originally Posted by bobsanchez View Post
    Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 cm thick to a hemispherical dome with a diameter 50m.

    Okay, so I've determined that the dome has a surface area of 3927, but after this point I'm not sure what to do, or if knowing the surface area is even helpful. I don't know how to put this into a workable function...
    \displaystyle V = \frac{2\pi}{3} r^3

    \displaystyle \frac{dV}{dr} = 2\pi r^2

    \displaystyle dV = 2\pi r^2 \, dr = 2\pi (25^2)(0.0005) = 0.625 \pi \, m^3
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    Hello, bobsanchez!

    Use differentials to estimate the amount of paint needed to apply a coat of paint
    0.05 cm thick to a hemispherical dome with a diameter of 50 m.
    The volume of a hemisphere is: . \frac{2}{3}\pi r^3


    The radius of the dome is 2500 cm.

    The volume of the dome is: . V_1 \;=\;\tfrac{2}{3}\pi(2500^2)\text{ cm}^3


    With the coat of paint, the radius of the dome is: 2500.05\text{ cm.}

    The new volume of the dome is: . V_2 \:=\:\frac{2}{3}\pi(2500.5^3) \text{ cm}^3


    The volume of paint is the change in Volume:

    . . \Delta V \;=\;V_2-V_1 \;=\;\tfrac{2}{3}\pi(2500.5^3) - \tfrac{2}{3}\pi(2500^3)

    which we could crank out on our calculator:

    . . \Delta V \;=\;1,\!963,\!534.678\text{ cm}^3


    Therefore: . \Delta V \;\approx1.963535\text{ m}^3 .
    (actual volume of paint)



    But if we need only an estimate of the volume of the paint,
    . . we can use the differential.

    The volume is: . V \;=\;\tfrac{2}{3}\pi r^3

    The differential is: . dV \:=\:2\pi r^2\,dr

    We have: . . r = 2500\text{ cm}
    . . . .and: . dr \:=\:\text{change in radius} \:=\:\text{thickness of paint} \:=\:0.05

    Hence: . dV \;=\;\tfrac{2}{3}\pi(2500^2)(0.05) \;=\;1,\!963,\!495.408 \text{ cm}^3

    Therefore: . dV \;\approx\;1.963495\text{ m}^3 .
    (approximate volume of paint)


    See? . . . Pretty close!
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    Quote Originally Posted by bobsanchez View Post
    Here's another one that I also find difficult in terms of setting it up:

    Sand is poured onto the top of a conical pile at a rate of 100m^3/minute. The coefficient of friction of the sand is such that the height and radius are always the same. At what rate is the height increasing when the pile is 12 meters high? The formula for the volume of a cone is V = 1/3pir^2h, but since the height and radius are the same this can be rewritten as V = 1/pih^3.

    Again, I'm pretty much lost. I vaguely understand that the dV/dt is related to the dh/dt, but I'm not sure how to make it work.

    \displaystyle V = \frac{\pi}{3} h^3

    take the derivative of the above equation implicitly w/r to time, t ... sub in your given values, then determine \displaystyle \frac{dh}{dt}
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    Quote Originally Posted by bobsanchez View Post
    Here's another one that I also find difficult in terms of setting it up:

    Sand is poured onto the top of a conical pile at a rate of 100m^3/minute. The coefficient of friction of the sand is such that the height and radius are always the same. At what rate is the height increasing when the pile is 12 meters high? The formula for the volume of a cone is V = 1/3pir^2h, but since the height and radius are the same this can be rewritten as V = 1/pih^3.

    Again, I'm pretty much lost. I vaguely understand that the dV/dt is related to the dh/dt, but I'm not sure how to make it work.
    Since you have gotten three answers to your first question, for this one, since V= \frac{1}{3}\pi h^3, by the chain rule, [tex]\frac{dV}{dt}= \frac{1}{3}\pi\left(3h^2\frac{dh}{dt}\right)= \pi h^2\frac{dh}dt}[/quote].

    When the pile is 12 m high, h= 12, and dh/dt= 100. \frac{dV}{dt}= \pi(12)^2(1000)
    Last edited by HallsofIvy; August 7th 2010 at 04:52 AM.
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    Quote Originally Posted by HallsofIvy View Post
    Since you have gotten three answers to your first question, ...
    ?
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    Okay, only two - yourself and Soroban. I miscounted- I never was good at higher math!
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