Show using the intermediate value theorem

[John O' Meara]

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 \pi x^2h and the volume of the cone is \frac{\pi}{3} r^2h, where x< r. If r > x then, the curve \frac{\pi}{3} r^2h is streched more than the curve \pi x^2h, 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.

 

[boboYO]

I don't 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]

 

[HallsofIvy]

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

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?

 

[John O' Meara]

The volume of the cylinder is 0 and \pi r^2h respectively. I think you mean that f(a)=0 and f(b)= \pi r^2h. 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.

 

[HallsofIvy]

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

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

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