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One more precalc question....

 
 
Reply Tue 3 Oct, 2006 09:31 pm
I'm obviously having some trouble with this unit in precalc - can anyone answer this question for me?Sorry I can't post the diagram that comes with the problem.

A sphere with a radius of 1 meter circumscribes a cylinder. Note that x represents half of the cylinder's height.


Express the volume, V, of the cylinder as a function of the radius, r, of the base of the cylinder. The following formula will be helpful: V(r,h) = πr2h.

-and-
Given a radius of .8 meter for the base of the cylinder, what is the cylinder's height?
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Miller
 
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Reply Tue 3 Oct, 2006 09:48 pm
Is this geometry?
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missmanda
 
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Reply Tue 3 Oct, 2006 09:53 pm
Well the class is precalc - but I guess it is more a geometry type question. I need to know how to manipulate the equation pretty much, which makes it precalc. any suggestions?
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Miller
 
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Reply Wed 4 Oct, 2006 02:55 am
I don't see any relationship to calculus.

Have you drawn a figure for the problem?
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patiodog
 
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Reply Wed 4 Oct, 2006 02:21 pm
PRE-calculus -- a condensate of geometry, algebry, and trigonometry, frequently offered or taken in lieu of trigonometry.

Unless I'm mistaken here, the problem is unnecessarily complicated by being stated in three dimensions. A cylinder is a rectangle rotated around an axis perpendicular to two of its opposite sides. The sphere circumscribing it is a circle rotated around an axis along its circumference.

So, revisualize the problem in two dimensions: you've got a rectangle drawn inside a circle, with all four corners touching the circle. The circle has a radius of 1 m (a diameter of 2 m). One dimension of the rectangle is 2*(0.8 m), or 1.6 m.

You'll also not that the distance from the center of the circle to each of the corners of the rectangle is 1 m (the radius of the circle), and that the distance between opposite corners is 2 m (the diameter of the circle).

Once you've got this, there's any number of ways to use basic geometry to figure out the unknown dimension of the rectangle. Easiest, I suppose, would be to realize that the rectangle is composed of two right triangles with a hypotenuse of 2 m and with one leg of 1.6 m. Pythagoras can tell you how to figure out the length of the other leg.



(I haven't drawn this out, so I might have screwed something up. Anyway, the train of thought is good, even if I got off at the wrong station...)
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Miller
 
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Reply Wed 4 Oct, 2006 04:31 pm
Maybe kids need to go back to algebra, geometry, trig and then colege algebra, before jumping ahead to calculus.

Why the rush?

Take time to learn the basic concepts.

Now, onto the geometry.
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