Intermediate-value theorem (approximation) ?

[njama]

intermediate-value theorem (approximation) ?

Homework Statement

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.

Homework Equations

$$V_{cylinder}=\pi r^2 h$$

$$V_{cone}=1/3 \pi r^2 h$$

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:

$$V_{cylinder}=\pi r_{1}^2 h , r_1 < r$$

$$V_{cone}=1/3 \pi r^2 h$$

Now, because we need to find r₁ so that $V_{cylinder}=V_{cone}$

$$\pi r_{1}^2 h = 1/3 \pi r^2 h$$

So we need to find:

$$r_{1}^2 = 1/3 (r^2)$$

or

$$r_{1}^2 - (1\sqrt{3})^2r^2=0$$

$$(r_1 - 1\sqrt{3}r)(r_1 + 1\sqrt{3}r)=0$$

We need to find $r_1 - 1\sqrt{3}r=0$ or $r_1 + 1\sqrt{3}r=0$

Now I've found that $r_1 = \pm 1\sqrt{3}r$

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 $1/\sqrt{3}r$ ??

 

[union68]

Let the volume of the cylinder be denoted

$$V \left(r_1\right) = \pi r_1^2h.$$

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 $$c$$ such that

$$V \left(c\right) = \frac{1}{3} \pi r^2 h,$$

with $$c<r$$. If you can show that there exists $$a$$ and $$b$$ such that

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

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

EDIT: Also, this is NOT an approximation. The $$V\left(c\right)$$ value that the IVT will give you is EXACTLY equal to $$\frac{1}{3} \pi r^2 h$$.

 

[njama]

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

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

?

 

[union68]

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, $$V\left(r_1\right)$$. 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 $$V$$ continuous? What closed interval is it continuous on?

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

Once you find c, you must show that $$c<r$$ because that is a condition stated in the problem.

 

[njama]

Thanks. For I moment I missed that part.

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

$$\lim_{r_1 \rightarrow k} V(r_1) = V(k)$$

$$\lim_{r_1 \rightarrow k} V(r_1)=\lim_{r_1 \rightarrow k} \pi r_{1}^2 h = \pi k^2 h =V(k)$$

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 $\frac{1}{3}\pi r^2 h$ between 0 and $\pi r^2 h$.

Is this correct?

 

[njama]

Is this better?