Differentiating and Integrating the Formula for Area of CircleDate: 05/11/98 at 23:22:53 From: Joey Bohanon Subject: Derivatives with circles I have recently been reading up on calculus in a book my dad got me. I noticed that if f = pi*r^2 (the area of a circle), then f' = 2*pi*r (the circumference of a circle). I also integrated pi*r^2 and came up with (pi*r^3)/3. Is the result of this integral a formula for some sort of solid? If so, could this be used in figuring out volumes (I'm not sure what they're called) of 4-dimensional objects, and so on? What is the significance of the circumference being the derivative of the area? Thanks a bunch, Joey Date: 05/14/98 at 11:39:27 From: Doctor Schwa Subject: Re: Derivatives with circles Joey, I'll take your third question first. It is such a good question, I thought for sure someone would have asked it before, but when I searched the Dr. Math archives for "circumeference," "area," and "derivative," I didn't find anything. So it's up to me to meet the challenge. The derivative is the rate of change. If I let A = pi*r^2, then what does dA/dr mean? It's the limit of: change in area ---------------- change in radius So your question becomes: what happens to the area when the radius is increased a tiny amount? The change in area is the area of a thin washer, of thickness dr in the radius direction. If you imagine cutting that washer and unbending it to make a straight line, you'd have (approximately) a rectangle, with a thickness of dr and a length equal to the circumference of the circle. Thus dA = (circumference) * dr. This is similar in some ways to the 2pi*r*h formula for the area of the side (lateral area) of a cylinder, with the cutting and unrolling idea. You also ask about the integral of pi*r^2, which is equal to (pi*r^3)/3. First, this integral is a formula for some sort of solid, but not quite in the same way as the relationship between area and circumference. This one I would think of as adding up the areas of a pile of circles, increasing in radius -- that is, a cone. If you want more explanation of why this is a good way to visualize it, let me know. But indeed, (pi*r^3)/3 is the volume of a cone of radius r and height r. The derivative of the volume of a sphere is its surface area, and yes, the same sort of thing holds in higher dimensions. So if you wanted the "volume" of a four-dimensional "cone" made by stacking up spheres in the fourth dimension (in the same way as a 3D cone is made by stacking up circles), you could integrate (4/3)*pi*r^3*dr. And if you knew the formula for the "hypervolume" of a 4D sphere, you could find it by integrating the "hyper surface area" formula, if only you had that. Or you could find the "surface area" if you only knew the volume. But actually, you can find the "volume" of a 4D "hypersphere." Think about how to use integrals to find the area of a circle, or the volume of a sphere, and then try to extend the pattern. Let me know how it goes! -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/