Solution to The February 1, 2006 PoW

Consider the Cartesean plane (the xy-plane).
(a) If a line is drawn from (t, 0) to (1, t) as t runs from 0 to 1, find the area of the figure formed.
{So when t = 0.4, a segment is drawn from (0.4, 0) to (1, 0.4).}

area figure 1

The difficult part of these problems is finding the equation of the curve that describes the upper boundary of the region.
First, we find the maximum y-value reached for each fixed x-value, when we draw all lines from (t, 0) to (1, t). For each fixed x, we have:
area formula 1

(b) If instead of each line segment being drawn from (t, 0) to (1, t), the segment is drawn vertically, find the area of the figure formed (as t runs from 0 to 1).
{So when t = 0.4, a segment whose length is the distance from (0.4, 0) to (1, 0.4) = sqrt[0.62 + 0.42] = sqrt(0.52) is drawn vertically.}
area figure 2

The length of the line segment from (x, 0) to (1, x) is (x2 + (1 - x)2)1/2. If each of these segments is made vertical, the area of the figure is now the area from x = 0 to x = 1 under y = (x2 + (1 - x)2)1/2 = (2(x - 1/2)2 +1/2)1/2, which is:
area formula 2


(c) Finally, consider a segment of fixed length 1 that is initially lying on the x-axis from (0, 0) to (1, 0). If the segment has its right endpoint at (1, 0) lifted straight up to (1, 1), while the left endpoint is dragged horizontally along the x-axis until the entire segment is vertical, going from (1, 0) to (1, 1), find the area of the region that is swept out.
{So when t = 0.3, a segment is drawn from (0.3, 0) to (1, sqrt(0.51)) because the distance from (0.3, 0) to (1, sqrt(0.51)) = sqrt[0.72 + 0.51] = 1.}
area figure 3

We again must fix x. Now for each t, (t, 0) is where the segment starts (left hand endpoint), and
(1, (1 - (1 - t)2)1/2) is where the line segment ends (right hand endpoint).
Fix x. For each t from 0 to x, we have:
area formula 3

Note that we should expect the final area (using the unit segment) to be greater than the first area (using the segment that's as far up at x = 1 as it is to the left of the orgin). This is because the y-value at x = 1 is (1 - (1 - t)2)1/2 > t if 0 < t < 1, so the line through (t, 0) will be steeper for the third curve than for the first curve. Also, I leave the obvious facts that all y-values from 0 to the maximum y-value are taken for each fixed x, and that what I claimed as a maximum y-value (where dy/dt = 0) is indeed a max.