I understand the equation is in the form of y=mx+b but how did he derive it for this specific line? Did he plot all the points on the y axis and then came up with 2piR by trial and error?
3 Answers
$\begingroup$I had the same question. This is how I made it click for me. From the equation, $y = mx + b$, we can see right away that $b = 0$ and $x = r$. To calculate $m$, we can use rise over run. The easiest way to get that is putting the height of the triangle over the width of the triangle (note that this is the circumference of the circle over the radius of the circle). That gives $2 \pi(r)/r$, or $2 \pi$. Thus the slope, or $m$, is $2 \pi$. $y = 2 \pi (r) + 0$
$\endgroup$ 1 $\begingroup$The function y = 2(pi)r corresponds to the value of each point at a certain value of r. For example, at a radius of 3, the area of the disk with thickness dr is 6pi. He uses this relation to plot the area of the disks as a function of r. Hope that helps!
$\endgroup$ $\begingroup$Let's work from the inside out, since that wasn't specifically addressed in the video. Every ring got "unwrapped" into a trapezoidal shape, with thickness $dr$ and left-side height $2\pi r,$ which we can approximate with a rectangle of thickness $dr$ and height $2\pi r.$ In each case, we're sort of "cutting off" a triangular region from the top of each unwrapped ring, so that our rectangles are always slight underestimations.
But what about the tiny circle in the center? Well, that one instead unwraps into a triangular shape with base $dr$ and (right-side) height $2\pi r,$ which is precisely the area we were ignoring from the unwrapped rings, and we'll do the same here, for consistency's sake. That is, we'll approximate this one by a "rectangle" of thickness $dr$ and height $0=2\pi\cdot 0.$
Now, we begin laying out our approximating rectangles along the positive $r$-axis, starting from the origin. We lay our zero-height "rectangle" on the interval from $0$ to $dr$ first, then we lay out the other rectangles side by side in order of increasing height. So we lay our next largest rectangle on the interval from $dr$ to $2dr,$ then the next from $2dr$ to $3dr,$ and so on. Recall now that the height of each rectangle is the same as the inner circumference of the ring whose area it approximates.
The inner radius of our smallest ring is the same as the outer radius of the tiny circle at the center, which is just $dr,$ and so the height of our smallest (actual) rectangle is $2\pi dr.$ Note, though, that the left side of this rectangle is at the value $dr$ on the $r$-axis, so the height of this rectangle is equal to $2\pi$ times its left end's $r$-coordinate. Put another way, the top-left corner of this rectangle lies on the line $y=2\pi r,$ which is also true for the "rectangle" we started with.
The inner radius of our second smallest ring is the outer radius of the smallest ring, which is $dr+dr=2dr,$ and so the height of its approximating rectangle is $2\pi\cdot 2dr.$ On the other hand, the left end of this rectangle is laid out at the right end of the previous rectangle--since the left end of the previous rectangle is at $dr,$ and since the rectangle has thickness $dr,$ then the left end of this rectangle is at $dr+dr=2dr.$ So, once again, the top-left corner of this rectangle lies on $y=2\pi r.$
More generally, let's say we know that the ring of radius $r_0$ is being approximated by a rectangle of height $2\pi r_0,$ whose top-left corner lies on $y=2\pi r,$ at the $r$-value $r_0.$ The next ring has inner radius $r_0+dr,$ so has height $2\pi(r_0+dr).$ On the other hand, since the left end of the prior rectangle was at $r_0,$ and since the prior rectangle had thickness $dr,$ then the $r$-coordinate of the previous rectangle's right end is $r_0+dr,$ and so the $r$-coordinate of this rectangle's left end is likewise $r_0+dr.$ Thus, this rectangle's top-left point lies on $y=2\pi r,$ above the $r$-value $r_0+dr.$
This is what's called a proof by induction. Roughly speaking, we started with the smallest rectangle, and showed how knowing the truth for one rectangle allowed us to prove that it was true for the next one, and thus, by iterating the argument, that it was true for all of them. We had to handle the zero-height "rectangle" separately, of course.
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