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Another proof that uses triangles considers the area enclosed by a circle to be made up of an infinite number of triangles (i.e. the triangles each have an angle of d𝜃 at the centre of the circle), each with an area of 1 / 2 · r 2 · d𝜃 (derived from the expression for the area of a triangle: 1 / 2 · a · b · sin𝜃 ...
It was later reinvented in China by Liu Hui in the 3rd century AD in order to find the area of a circle. [2] The first use of the term was in 1647 by Gregory of Saint Vincent in Opus geometricum quadraturae circuli et sectionum. The method of exhaustion is seen as a precursor to the methods of calculus.
In calculus, the method of normals was a technique invented by Descartes for finding normal and tangent lines to curves. It represented one of the earliest methods for constructing tangents to curves. The method hinges on the observation that the radius of a circle is always normal to the circle itself. With this in mind Descartes would ...
1.3 Area of Sector of a Circle; 1.4 Application of Circular Measure; 2) Differentiation 2.1 Limit and its Relation to Differentiation; 2.2 The First Derivative; 2.3 The Second Derivative; 2.4 Application of Differentiation; 3) Integration 3.1 Integration as the Inverse of Differentiation; 3.2 Indefinite Integral; 3.3 Definite Integral; 3.4 ...
For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016 [7]) is 2π ⋅ 6371 2 | sin 90° − sin 66.56° | = 21.04 million km 2 (8.12 million sq mi), or 0.5 ⋅ | sin 90° − sin 66.56° | = 4.125% of the total surface area of the Earth.
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2 π) of the whole circle, so its area is θ/2. We assume here that θ < π /2. = = = = The area of triangle OAD is AB/2, or sin(θ)/2.
The area of a cycloid can be calculated by considering the area between it and an enclosing rectangle. These tangents can all be clustered to form a circle. If the circle generating the cycloid has radius r then this circle also has radius r and area πr 2. The area of the rectangle is 2r × 2πr = 4πr 2. Therefore, the area of the cycloid is ...
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