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Intermediate-value theorem (approximation) ?

  1. Nov 2, 2009 #1
    intermediate-value theorem (approximation) !?!

    1. The problem statement, all variables and given/known data

    Use the Intermediate-Value Theorem to show that there is a right circular cylinder of height h and radius less than r, whose volume is equal to that of a right circular cone of height h and radius r.

    2. Relevant equations

    [tex]V_{cylinder}=\pi r^2 h [/tex]

    [tex]V_{cone}=1/3 \pi r^2 h[/tex]

    3. The attempt at a solution

    Here is my assumption:

    Since the radius r of the basis of the cone is greater than the radius r1 of the cylinder, we can write the formulas as follows:

    [tex]V_{cylinder}=\pi r_{1}^2 h , r_1 < r[/tex]

    [tex]V_{cone}=1/3 \pi r^2 h[/tex]

    Now, because we need to find r1 so that [itex]V_{cylinder}=V_{cone}[/itex]

    [tex]\pi r_{1}^2 h = 1/3 \pi r^2 h [/tex]

    So we need to find:

    [tex]r_{1}^2 = 1/3 (r^2)[/tex]

    or

    [tex]r_{1}^2 - (1\sqrt{3})^2r^2=0[/tex]

    [tex](r_1 - 1\sqrt{3}r)(r_1 + 1\sqrt{3}r)=0[/tex]

    We need to find [itex]r_1 - 1\sqrt{3}r=0[/itex] or [itex]r_1 + 1\sqrt{3}r=0[/itex]

    Now I've found that [itex]r_1 = \pm 1\sqrt{3}r[/itex]

    Now is my goal to prove using the intermediate value theorem that the statement above is true?

    Now the length of the radius r1 is between (0,r) because (r1 < r )

    (0+r)/2 = r/2

    Now the solution is between (r/2,r).

    (r/2 + r)/2 = 3r/4

    Now the solution is between (r/2,3r/4).

    Should I continue doing this until I get close number to [itex]1/\sqrt{3}r[/itex] ??
     
    Last edited: Nov 2, 2009
  2. jcsd
  3. Nov 2, 2009 #2
    Re: intermediate-value theorem (approximation) !?!

    Let the volume of the cylinder be denoted

    [tex]V \left(r_1\right) = \pi r_1^2h.[/tex]

    It's clear that this function is continuous on any closed interval, hence the IVT may be applied. Your goal is to show that there exists some [tex]c[/tex] such that

    [tex] V \left(c\right) = \frac{1}{3} \pi r^2 h, [/tex]

    with [tex]c<r[/tex]. If you can show that there exists [tex]a [/tex] and [tex]b[/tex] such that

    [tex] V \left(a\right) < \frac{1}{3} \pi r^2 h[/tex] and [tex] V \left(b\right) > \frac{1}{3} \pi r^2 h, [/tex]

    then the IVT will guarantee the existence of your [tex]c[/tex] value. Then, you must show that [tex]c<r[/tex]. You've already done most of the work already...can you see it? How would you find a suitable [tex]a[/tex] and [tex]b[/tex] value?

    EDIT: Also, this is NOT an approximation. The [tex]V\left(c\right)[/tex] value that the IVT will give you is EXACTLY equal to [tex]\frac{1}{3} \pi r^2 h[/tex].
     
    Last edited: Nov 2, 2009
  4. Nov 2, 2009 #3
    Re: intermediate-value theorem (approximation) !?!

    But why c < r, is it because r1 < r?

    Why I need to show
    [tex]
    V \left(a\right) < \frac{1}{3} \pi r^2 h
    [/tex]
    and [tex]V \left(b\right) > \frac{1}{3} \pi r^2 h[/tex]

    ?
     
  5. Nov 2, 2009 #4
    Re: intermediate-value theorem (approximation) !?!

    I suggest you go back and look at the precise statement of the IVT. Here it is from MathWorld (I've modified their variables to fit my hint):

    If f is continuous on [a,b], and x is any number between f(a) and f(b) inclusive, then there is at least one number c in the closed interval such that f(c)=x.

    Our function is the volume of the cylinder, [tex]V\left(r_1\right)[/tex]. To apply the IVT to this function, we need to make sure the theorem's hypotheses are satisfied. This is a step that many students skip, and to their detriment. Is [tex]V[/tex] continuous? What closed interval is it continuous on?

    In terms of how the theorem is stated, we're dealing with [tex]x= \frac{1}{3}\pi r^2 h[/tex]. If you can find a,b such that [tex] V\left(a\right) < x[/tex] and [tex]V\left(b\right) >x[/tex], then the IVT guarantees the existence of c such that [tex]V\left(c\right) = x[/tex]. Can you see that? How do you find a and b?

    Once you find c, you must show that [tex]c<r[/tex] because that is a condition stated in the problem.
     
  6. Nov 3, 2009 #5
    Re: intermediate-value theorem (approximation) !?!

    Thanks. For I moment I missed that part.

    V is continuous everywhere because for every number k in the domain the limit:

    [tex]\lim_{r_1 \rightarrow k} V(r_1) = V(k)[/tex]

    [tex]\lim_{r_1 \rightarrow k} V(r_1)=\lim_{r_1 \rightarrow k} \pi r_{1}^2 h = \pi k^2 h =V(k)[/tex]

    Because it is continuous everywhere, we can choose any numbers [a,b].

    Because the volume of the cylinder is 3 * V(of cone), I would choose b=r so that V(b)= пb^2h and a=0 so that V(a)=0.

    Now I trapped [itex]\frac{1}{3}\pi r^2 h [/itex] between 0 and [itex]\pi r^2 h[/itex].

    Is this correct?
     
  7. Nov 4, 2009 #6
    Re: intermediate-value theorem (approximation) !?!

    Is this better?
     
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