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What the formula A = πr2 says is that the area of the circle is larger than the area of the square with side r (as you can clearly see by your drawing). More precisely, the area of the circle is the same as the area of the square with side √π ⋅ r because (√π ⋅ r)2 = πr2. So what you can say is that the π scales the length of the ...
Area of Circle = ∫r 0 2πx dx = πr2 Area of Circle = ∫ 0 r 2 π x d x = π r 2. In reality, we are approximating the circumference of each circle by an annulus of small width, and then letting that width tend to zero. This method works with finding the area of the square too.
From calculus, recall that rate of change of r2 = 2r × rate of change of r. So we have rate of growth of area = π × rate of change of r2. So the area grows at the same rate at which πr2 grows. That makes them always equal if they're equal when r = 0. And it's easy to see that they're equal when r = 0.
As r increases, we have [Math Processing Error] From calculus, recall that rate of change of r2 = 2r × rate of change of r. So we have rate of growth of area = π × rate of change of r2. So the area grows at the same rate at which πr2 grows. That makes them always equal if they're equal when r = 0.
A ′ (r) = 2πr(∗) Intuitively, the rate of change of the area of the circle is the circumference. Formally. A ′ (r) = lim Δr → 0A(r + Δr) − A(r) Δr. Now, geometrically it is pretty clear (but not really easy to prove mathematically) that the area of a corona between circles satisfies.
To solve for r, since. A 2π = r2 + rh. is a quadratic we need to make a very old trick, which is called completing the square: Note in the last steps we take the square root. We then have to think about both the positive and negative root. So you final solution is. r = − √ A 2π + (h 2)2 − h 2 or r = √ A 2π + (h 2)2 − h 2.
One geometric explanation is that $4\pi r^2$ is the derivative of $\frac{4}{3}\pi r^3$, the volume of the ...
$\begingroup$ An elementary proof does not exist. I seem to recall a piece in one of MAA's publications (from the mid 1990s---this may be it) which points out this issue, and I have colleague who has written a pretty detailed paper which tries to patch things together using the techniques available to Archimedes.
$\begingroup$ This understates the subtlety of Archimedes' proof. It not only involves using limits to calculate areas (even Euclid did that!), but also requires a convexity argument which — if fully justified for curved figures — will end up using some notion of tangent line.
0. Draw a circle with area A A and radius r r, then draw around it a larger circle. The distance from your first circle to the second is dr d r, and the difference of the areas is dA d A. As dr → 0 d r → 0, dA d A goes to the circumference of the interior circle, hence dA → 2πrdr d A → 2 π r d r. Share. Cite. Follow. edited May 26 ...