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Math Help - Inverted Cone

  1. #1
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    Inverted Cone

    We want to use an inverted cone with a radius of 9cm to dispense 100mL of a chemical. To be certain that we dispense the correct quantity we want markings in 100mL increments on the outside of the cone. Where should we place the markings?(accurate to .1 of a mm).
    The volume of the inverted cone can be found with V=(32a^3)/27 where a is the radius of the cone

    So far all I have done is sub in the radius to work out the volume

    V=(32*9^3)/27
    V=864

    Any help would be greatly appreciated
    Last edited by FLOTUS; August 13th 2010 at 10:23 PM.
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  2. #2
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    What exactly is the shape of the cone? That is, what is the total height when the radius is 9 cm? The volume of a cone, of radius a and height h, is \frac{\pi}{3}a^2h. You say that the volume is given by \frac{32a^3}{27} which leads me to the conclusion that the height is \frac{a}{9\pi}. With a= 9 cm, that gives h= \frac{9}{9\pi}= \frac{1}{\pi} or about 0.318 cm. I find that hard to believe!

    If a cone has radius R and height H, then for any given height, h, above the vertex, the radius at that height is given by \frac{r}{h}= \frac{R}{H} or r= \frac{R}{H}r. A circle with that radius has area \pi r^2= \pi\frac{R^2}{H^2}h^2 and, if we think of it as a thin disk of height dh, volume \pi\frac{R^2}{H^2}h^2 dh. We can find the volume between two different height, h_1 and h_2 by integrating that:
    \frac{R^2}{H^2}\pi\int_{h_1}^{h_2} h^2dh= \frac{\piR^2}{3H^2}(h_2^3- h_1^3).

    So the distance from h_1= 0, at the vertex, to the height, h_2 where the volume is 1000 ml is \frac{\pi R^2}{3H^2}h_2^3= 1000. Once you have solved for that h, set it equal to h_1 and solve for the next h_2.
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  3. #3
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    Unfortunately I had worked out the previous question that makes you find the function for the volume of the cone. Instead of  \frac {32a^3}{27} it should actually be \frac {32\pi a^3}{81}
    Last edited by FLOTUS; August 14th 2010 at 11:09 PM.
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