CONSTRUCTION:
Construct line segment LN. Construct the perpendicular bisector of line segment LN
intersecting line segment LN at point K and the cycloid at point M.
Construct the circle with diameter KM. Construct the line through point P
parallel to line segment LN and intersecting the circle ( label as Q the intersection point
closest to P). Construct the tangent line through P and perpendicular to line KQ.
PROOF:
If L is the point (0,0), N is the point (2πa,0), and the humps of the cycloid are oriented upward, its parametric equation is
x = a[t - sin(t)] (1)
y = a[1 - cos(t)] (2)
Taking derivatives with respect to t we have
dx/dt = a[1 - cos(t)] (3)
dy/dt = a*sin(t) (4)
Therefore, the slope of the tangent is
dy/dx = (dy/dt)/(dx/dt)
= sin(t)/[1 - cos(t)] (5)
From this we see that the slope is positive for the first half of the hump, horizontal at point M, and negative for
the second half of the hump.
The intersection of the line y = a[1 - cos(t)] and the circle
(x - πa)2 + (y - a)2 = a2 gives (a[π ± sin(t)],a[1 - cos(t)])
for point Q. The slope of line KQ is then
a[1 - cos(t)] - 0 1 - cos(t)
-------------------- = ------------
a[π ± sin(t)] - πa ±sin(t)
Picking the right sign and knowing that the slope of the tangent to the cycloid is -1/slope(KQ) gives equation (5).
See the post by Harry for an alternate construction.
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