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Following Archimedes' argument in The Measurement of a Circle (c. 260 BCE), compare the area enclosed by a circle to a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less.
The unit vector ^ has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle θ the same as the angle of (). If the particle displacement rotates through an angle dθ in time dt , so does u ^ R ( t ) {\displaystyle {\hat {\mathbf {u} }}_{R}(t)} , describing an arc on the unit circle ...
The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product = For example, if a force of 10 newtons ( F = 10 N ) acts along a point that travels 2 metres ( s = 2 m ), then W = Fs = (10 N) (2 m) = 20 J .
The curve is a cycloid, and the time is equal to π times the square root of the radius (of the circle which generates the cycloid) over the acceleration of gravity. The tautochrone curve is related to the brachistochrone curve, which is also a cycloid.
A page from Archimedes' Measurement of a Circle. Measurement of a Circle or Dimension of the Circle (Greek: Κύκλου μέτρησις, Kuklou metrēsis) [1] is a treatise that consists of three propositions, probably made by Archimedes, ca. 250 BCE. [2] [3] The treatise is only a fraction of what was a longer work. [4] [5]
Liu Hui's method of calculating the area of a circle. Liu Hui's π algorithm was invented by Liu Hui (fl. 3rd century), a mathematician of the state of Cao Wei.Before his time, the ratio of the circumference of a circle to its diameter was often taken experimentally as three in China, while Zhang Heng (78–139) rendered it as 3.1724 (from the proportion of the celestial circle to the diameter ...
Since the angular velocity ω = v/r is constant, the area swept out in a time Δt equals ω r 2 Δt; hence, equal areas are swept out in equal times Δt. In uniform linear motion (i.e., motion in the absence of a force, by Newton's first law of motion), the particle moves with constant velocity, that is, with constant speed v along a line.
Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the time cancels out, and only displacement remains.)