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Lateral area and volume of a cone divided equally by a plane parallel to base

  1. Oct 1, 2012 #1
    I am working independently from the book Precalculus Mathematics in a Nutshell by George F. Simmons. Although the book is fairly small, many of the problems are quite challenging, at least for me.

    I am stuck on this problem:

    "The height of a cone is h. A plane parallel to the base intersects the axis at a certain point. How far from the vertex must this point be if the plane divides the lateral area in two equal parts? If the plane divides the volume into two equal parts?"

    I cannot provide much information on my attempt, as I do not really know where to begin with this one. The way I understand the problem so far, this plane basically runs through the cone horizontally and cuts it into a smaller cone at top and a frustrum at the bottom. The "certain point" is a point on the height plane that runs vertically and right through the middle of the cone, through the vertex. And the answer, I take it, must be written in relation to/terms of h. (I often find that figuring out what the problem actually says takes much longer than doing the problem itself.)

    Any help or hints would be greatly appreciated.
     
  2. jcsd
  3. Oct 1, 2012 #2

    vela

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    Sounds good up to here.

    This doesn't make sense to me.

    A plane that's parallel to the base will split the cone into the two pieces. The question is asking for two things:
    1. How far from the vertex does the plane have to be so that the volume of the two pieces are equal?
    2. How far from the vertex does the plane have to be so that the surface area of the two pieces are equal (not counting the flat surfaces)?
     
  4. Oct 1, 2012 #3
    Thank you for the reply.

    What I mean is that the "certain point" is where the horizontal plane (the plane parallel to the base) will intersect with the height plane (the line that runs from the vertex to the center of the circular base, creating a right triangle on both sides and dividing the cone into two equal halves). The height plane is the height, I mean, the line that designates the height. So when you say "how far from the vertex," I take that to mean how far up or down the line designating the height, as the horizontal plane will intersect more than one point of the entire cone.

    Speaking of right triangles, I thought that I was onto something earlier when I saw that the shapes can be broken down into multiple triangles, and perhaps the answer can be derived from using 45/45/90 or 30/60/90 triangle rules. The problem is that I'm not sure what kind of triangles are created minus the one 90 degree side.

    I already suspected that my answer will divide the cone into smaller cones and frustrums, that, despite their differing shapes, will have identical lateral areas and volumes. In that sense, I know what the question is asking. But I simply do not know how to proceed. What should be my next step?
     
  5. Oct 1, 2012 #4

    vela

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    Let x be the distance from the vertex to the plane. Write down expressions for the lateral area and volume of the (small) cone and the frustrum.
     
  6. Oct 1, 2012 #5

    Ray Vickson

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    Using 45/45/90 or 30/60/90 triangle rules would not help, because you are not told what is the angle at the tip (i.e., vertex) of the cone. Anyway, you don't need to do anything fancy like that; you just need to know how the lateral area and volume vary with the height of the cone when the cone's vertex angle stays constant. A bit of intuition helps a lot here; then you can worry later about proofs.

    RGV
     
  7. Oct 1, 2012 #6
    The attached image shows what I have so far.

    But as you can see, I do not know how to proceed from here. I'm assuming that I must write the answer in terms of h, but with so many variables, I'm not sure if I will ever reach an elegant expression.

    Should I set A1 and A2 equal to one another? What do I do with x and h?
     

    Attached Files:

  8. Oct 2, 2012 #7

    Ray Vickson

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    You have everything you need: just figure out how r and s vary with x. That is, if I double x, what happens to r? What happens to s?

    RGV
     
  9. Oct 2, 2012 #8

    vela

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    Your expressions for the volume aren't correct. The volume of a cone of height h and with a base of radius r is given by ##V = \frac{1}{3}\pi r^2 h##. Where you have s and s1 in your expressions, you should have h and x, respectively.
     
  10. Oct 2, 2012 #9
    Thanks for the replies. I tend to do my math in the evenings, so I will report back later tonight.
     
  11. Oct 15, 2012 #10
    I have taken a break from this problem and decided to come back to it tonight. Nearly two (!) hours later, I still do not understand how to derive the relationship between x and h. I wrote out the formulas (correctly this time) for the lateral area and the volumes of the big cone, the small cone, and the frustum. But I just don't see how to use x here.
     
  12. Oct 16, 2012 #11

    HallsofIvy

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    If you draw a "side section", you can represent the cone as an isosceles triangle and, drawing a perpendicular from the vertex to the base as two right triangle. With R the radius and h the height of the cone, they are the lengths of the two legs of the right triangle. Now, suppose we draw a line parallel to the base, at distance x from the vertex, representing the plane dividing the cone into two parts. The top part of the cone is another, smaller cone. Its height is x and we now have two similar triangles, the large, original right triangle and the smaller right triangle. Letting r be the radius of the base of the new, top, cone, we must have r/R= x/h or r= (R/h)x. The large cone has volume [itex](1/3)\pi R^2h[/itex] and the smaller has volume [itex](1/3)\pi r^2 x= (1/3)\pi (R^2/h^2)x^3[/itex]. You want to find x so that the volume of the smaller cone is 1/2 the volume of the larger cone: [itex](1/3)\pi (R^2/h^2)x^3= (1/2)(1/3)\pi R^2h[/itex]. Solve that for x.
     
  13. Oct 16, 2012 #12
    Thank you for your detailed reply, HallsofIvy.

    I understand everything until this following point: r/R=x/h or r=(R/h)x. How does r divided by R equal x divided by h? Why are we dividing one side by the other?
     
  14. Oct 18, 2012 #13
    I now see that the equation r/R=x/h is simply a way of deriving the ratio between the two triangles. I figured it out by drawing 2 equilateral triangles, one with each side equal to 3 and the other with each side equal to 9, plugging in the values for the variables, and getting a ratio as the result, 0.333... or 1/3, which is the ratio between the sides 3 and 9.

    I have two more questions now:

    1. I see how r/R=x/h is the same as r=(R/h)x. But when checking to see if they are identical, I simply multiplied by R on both sides of the first expression and got r = (xR)/h. That to me seems like the more intuitive way of writing this expression. Is there a reason why it was written the other way, r=(R/h)x? The reason I ask is because it seems like it was derived a little differently than my method of multiplying by R and I'd like to know how it was done. Maybe there is another reason for setting it up this way?

    2. I do not see how the right side was derived from the left side of this expression for the smaller cone: (1/3)πr2x=(1/3)π(R2/h2)x3. Plugging in some numbers, I see that it works, but I do not understand why. How were the R2, h2, and x3 values derived? I do see the similarity between r=(R/h)x and (1/3)π(R2/h2)x3, namely the (R2/h2)x3, but I do not understand it.
     
    Last edited: Oct 18, 2012
  15. Nov 13, 2012 #14
    Difficult problem.
     
    Last edited: Nov 13, 2012
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