Is it signifigant that the derivative of the volume of a sphere is equal to the area of a circle? Or that the derivative of the area of a circle is equal to the circumference of a circle?
What exactly do you mean by “significant?” Do you think it’s a coincidence?
It’s the same as that whole position/speed/acceleration/jerk thing in physics.
That isn’t obvious to me. I want to say that it’s a consequence of one of the higher-dimensional analogues of the fundamental theorem, but I can’t remember which one.
[nitpick]The derivative of the volume of a sphere is equal to four times the area of its circle.
V = 4/3 X pi X r[sup]3[/sup]
dV = 4 X pi X r[sup]2[/sup]
yes, my mistake on the volume. the area vs circumference is still quite interesting
Well, it’s not too surprising that the derivative of the area pir[sup]2[/sup] is the circumference 2pir. Think of the derivative of the area as the instantaneous rate of change in the area with respect to r. If you change r a tiny bit, say increase r to r[sup][/sup], what happens to the area? The incremental area is roughly a ring of thickness r[sup][/sup]-r. This is pretty close to the circumference when r[sup][/sup]-r is very small, which is exactly what happens when you take the limit as r[sup]*[/sup] approaches r in the definition of the derivative for the area.
Which is the surface area of a sphere, so it still fits in the non-coincidental category.
And that’s explained in the same way that nivlac showed earlier. (Well done, by the way, nivlac.)
When you increase the sphere’s radius by an infinitesimal amount, the incremental volume increase is a thin shell. That infinitely thin shell is the surface area of the sphere.
Let me put it this way. Suppose you have a sphere of radius R that you want to paint to a thickness t, and you want to figure out how much paint you need (assume t is much less than R). There are two ways you could calculate it: First, you could find the surface area of the sphere, and multiply that by the thickness. That gives you a volume of paint equal to 4piR*[sup]2[/sup]t . Second, you could calculate the volume of a sphere of radius R, and then calculate the volume of a sphere of radius R + t, and subtract the two. This would give you a volume of paint equal to (4/3)pi(R + t)[sup]3[/sup] - (4/3)piR*[sup]3[/sup] . Assuming that t << R, both of these methods should work, so those two results should be equal to each other:
4piR*[sup]2[/sup]t = (4/3)pi(R + t)[sup]3[/sup] - (4/3)piR*[sup]3[/sup]
or, if we divide both sides by t,
4piR*[sup]2[/sup] = [(4/3)pi*(R + t)[sup]3[/sup] - (4/3)piR*[sup]3[/sup]]/t
But if we’re considering t to be very small, the thing on the right turns into the definition of the derivative of the area with respect to radius. And the thing on the left is just the surface area. So that’s why the two are equal.
Odd, I was just thinking about this the other day.
If we take the derivative of the surface area, 4piR[sup]2[/sup], we get 8piR. Does this correspond to any significant feature of a circle or sphere?
Well. it’s four times the circumference of a great circle of the sphere, although it’s not clear to me what significance that has (unlike the fact that the derivative of the volume with respect to the radius gives the surface area).
Well, derivatives of area are necessarily lengths – the trick is figuring out which length they mean. I like the four-great-circles explanation, but I feel like there ought to be something else. Given that constants tend to propagate downstream in calculus, that might be all there is to it, though.
If you had a string of length 8* pi * R, and a sphere of radius R, wrapping it exactly four times around the middle is a pretty good fit. Maybe you could even do something with the string and a steradian? I don’t know.
It’s simply the Fundamental Theorem of Calculus. How do you get the volume of an n-ball of radius R? Integrate the areas of (n-1)-spheres of radius r as r runs from 0 to R.
What’s more interesting is the derivative of the volume of an n-sphere as you let n vary.