My math teacher told us that car lights used to no not have a parabolic shape so the light was spread all around making the light reaching the road in front of the car really dim. Car manifactureers made a parabolic mirror behind the lights which made them much more effecient.
Volume of paraboloid via Cavalieri's principle: (I discovered this proof) Type *Without integration, why is the volume of a paraboloid half of its inscribing cylinder?*
As both the paraboloid and the cylinder are volumes of revolution, let's consider a half-parabola y=x^2 from 0 to 2 and a rectangle from (0,0) to (2,4). Rotating both through 360° creates the solids of interest, but the areas of the 2D shapes are easier to work with. For both 2D shapes: the base lengths are 2; the heights are 4. A formula for the area within a half-parabola is here: www.structx.com/Shape_Formulas_001.html The area within the parabola = 2/3 * 2 * 4. The area of the rectangle = 2 * 4. The ratio of the area within the parabola to the area within the rectangle is 2/3.
You are correct. en.wikipedia.org/wiki/Pappus%27s_centroid_theorem "the volume V of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d *_traveled by the geometric centroid_* of F." From my previous link for a half-parabola, the distance from the vertical axis to the centroid of the half-area is 3*b/8, so the volume swept by the half-parabola in a full rotation is (2 * pi)*(3 * b/8)*(2/3 * b * h) = pi / 2 * b^2 * h. For the rectangle, the distance from the vertical axis to its centroid is b/2, so the volume swept by the rectangle in a full rotation is (2 * pi)*(b/2)*(b * h) = pi * b^2 * h. So the volume of the paraboloid *_is_* half that of the cylinder, as you said.
Having trouble visually grasping the straight line example as it appears straight but over a curved surface. And in that case, wouldn't a straight line be able to be superimposed across any surface ad infinitum? Sorry, I am stupid.
I want to see another video on conic sections...
PLEASE!
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I was studying conic sections for finals , which is nearby , and this helped me . Thank you
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Nice! Good that you mentioned the Shukhov tower.
Thank you.
My math teacher told us that car lights used to no not have a parabolic shape so the light was spread all around making the light reaching the road in front of the car really dim. Car manifactureers made a parabolic mirror behind the lights which made them much more effecient.
Beautiful animation, perfect explanation!!
Great video! Would love to see more
Great video😉👍
Nice!
Superb💥💥
Can I ask why do parabola and hyperbola appear in graphs of polynomials and polynomial fractions (not sure how they're called..)
Could you do a video about mathematics apply to medicine? Please.
Volume of paraboloid via Cavalieri's principle: (I discovered this proof)
Type *Without integration, why is the volume of a paraboloid half of its inscribing cylinder?*
Shouldn't that be 2/3?
@@artsmith1347 That is for a sphere inscribed in the cylinder.
As both the paraboloid and the cylinder are volumes of revolution, let's consider a half-parabola y=x^2 from 0 to 2 and a rectangle from (0,0) to (2,4). Rotating both through 360° creates the solids of interest, but the areas of the 2D shapes are easier to work with.
For both 2D shapes: the base lengths are 2; the heights are 4.
A formula for the area within a half-parabola is here:
www.structx.com/Shape_Formulas_001.html
The area within the parabola = 2/3 * 2 * 4.
The area of the rectangle = 2 * 4.
The ratio of the area within the parabola to the area within the rectangle is 2/3.
You are correct.
en.wikipedia.org/wiki/Pappus%27s_centroid_theorem
"the volume V of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d *_traveled by the geometric centroid_* of F."
From my previous link for a half-parabola, the distance from the vertical axis to the centroid of the half-area is 3*b/8, so the volume swept by the half-parabola in a full rotation is
(2 * pi)*(3 * b/8)*(2/3 * b * h) = pi / 2 * b^2 * h.
For the rectangle, the distance from the vertical axis to its centroid is b/2, so the volume swept by the rectangle in a full rotation is
(2 * pi)*(b/2)*(b * h) = pi * b^2 * h.
So the volume of the paraboloid *_is_* half that of the cylinder, as you said.
Having trouble visually grasping the straight line example as it appears straight but over a curved surface. And in that case, wouldn't a straight line be able to be superimposed across any surface ad infinitum? Sorry, I am stupid.
Try explaining it again. Sorry, I don’ t get your concern
Conic sections rule :)