How do you find the area of a parabolic segment?
If a parabolic segment is defined by the intersection of a parabola γ of equation y=ax2 (a>0) with the straight line r, oblique with respect to the axis of γ, its area can be obtained by the difference between the area of the trapezoid ABB’A’ and the areas of the mixtilineal triangles S1 and S2.
What is the area of the parabolic curve?
| \Y1=H-(H/R2)X2 | height Y |
|---|---|
| \Y2=fnInt(Y1,X,-R,R) | area under the parabola |
What is a parabolic segment?
Main theorem A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle.
What is formula for parabolic?
The general equation of a parabola is: y = a(x-h)2 + k or x = a(y-k)2 +h, where (h,k) denotes the vertex. The standard equation of a regular parabola is y2 = 4ax. Some of the important terms below are helpful to understand the features and parts of a parabola. Focus: The point (a, 0) is the focus of the parabola.
What is an Archimedes triangle?
In his proof, Archimedes first constructed a triangle whose sides consisted of two tangents of a parabola and the chord connecting the points of tangency. He then showed that the area under the parabola (shown in white and light green in the painting) is two thirds of the area of the triangle that circumscribes it.
How do you find the area of the curve?
The area under a curve between two points is found out by doing a definite integral between the two points. To find the area under the curve y = f(x) between x = a & x = b, integrate y = f(x) between the limits of a and b. This area can be calculated using integration with given limits.
What is the area of hyperbola?
The vertex of the hyperbola is (±a, 0) when the major axis is the x-axis. The hyperbola is reflected about the x-axis so the area below equals the area above. Thus, the total area is double of the area of the region from x = 2 to x = 3.
What is the equation of circle?
We know that the general equation for a circle is ( x – h )^2 + ( y – k )^2 = r^2, where ( h, k ) is the center and r is the radius.
Who developed pi?
Archimedes of Syracuse
The first calculation of π was done by Archimedes of Syracuse (287–212 BC), one of the greatest mathematicians of the ancient world.
The area of a parabolic segment. A parabolic segment is a region bounded by a parabola and a line, as indicated by the light blue region below: [See Parabola for some background on this interesting shape.] I have used the simple parabola y = x 2 and chosen the end points of the line as A (−1, 1) and B (2, 4).
The concept came via Antiphon (5th century BCE), and Eudoxus of Cnidus (4th century BCE). A parabolic segment is a region bounded by a parabola and a line, as indicated by the light blue region below: [See Parabola for some background on this interesting shape.]
What is the area of the light blue parabolic segment?
So the total area of the triangles (which gives us the area of our light blue parabolic segment) is 4X/3, which is 4/3 the area of the pink triangle, as Archimedes claimed. The thinking behind this solution is very similar to the ideas behind the development of calculus.
Does the Archimedes theorem work for every parabola?
The theorem will work for any parabola and any line passing through that parabola, intersecting in 2 points. [Of course, Archimedes did not use the x-y co-ordinate system, since it was not invented by Descartes until the 17th century.]