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Show using the intermediate value theorem

  1. Sep 30, 2009 #1
    Use the Intermediate Value Theorem to show that there is a cylinder of height h and radius less than r whose volume is equal to that of a cone of height h and radius r.
    IVT states that: if f is continuous on a closed interval [a,b] and k is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a,b] such that f(x)=k.
    The volume of the cylinder is [tex] \pi x^2h[/tex] and the volume of the cone is [tex]\frac{\pi}{3} r^2h[/tex], where x< r. If r > x then, the curve [tex] \frac{\pi}{3} r^2h[/tex] is streched more than the curve [tex] \pi x^2h[/tex], i.e it is closer to the y-axis than the curve [\pi x^2h[/tex], so I cannot find k. And is not a<0 and b>r? Then f(a)<0 and f(b)>f(r)?I am studing this on my own. Please help. Thanks.
     
    Last edited: Sep 30, 2009
  2. jcsd
  3. Sep 30, 2009 #2
    I dont see how y=x^2/3 is closer to the y axis than y=x^2? It's the other way around

    and if you haven't solved the problem,
    look at the interval [0,r]
     
  4. Sep 30, 2009 #3

    HallsofIvy

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    As you say, a cone of height h and radius r has volume [itex]\frac{1}{3}\pi r^2 h[/itex].

    What is the volume of a cylinder of height h and radius 0? What is the volume of a cylinder of height h and radius r? Now, what does the intermediate value theorem tell you?
     
  5. Oct 1, 2009 #4
    The volume of the cylinder is 0 and [tex] \pi r^2h [/tex] respectively. I think you mean that f(a)=0 and [tex]f(b)= \pi r^2h [/tex]. In that case surely there is an infinite number of values that k can have between f(a) and f(b). But we only want one value of k such that the volume of the cylinder equals the volume of the cone. I just don't understand yet.
     
  6. Oct 1, 2009 #5

    HallsofIvy

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    Yes, there are an infinite number of values between 0 and [itex]\pi h r^2[/itex].

    And one of them is [itex](1/3)\pi h r^2[/itex]!

    The intermediate value theorem tells you that as [itex]\rho[/itex] goes from 0 to r [itex]\pi h \rho^2[/itex] takes on all values between 0 and [itex]\pi h r^2[/itex].
     
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