Transcript Document

Archimedes of Syracuse (Circa 287 BC - 212 BC), was a Greek
mathematician, astronomer, philosopher, physicist and engineer. He was
killed by a Roman soldier during the sack of the city, despite orders from
the Roman general, Marcellus, that he was not to be harmed. (The Greeks
said that he was killed while drawing an equation in the sand, and told this
story to contrast their high-mindedness with Roman ham-handedness;
however, it should be noted that Archimedes designed the siege engines
that devastated a substantial Roman invasion force, so his death may have
been out of retribution.) Some math historians consider Archimedes to be
one of history's greatest mathematicians, along with possibly Newton,
Gauss, and Euler.
Part 1
1.a
1.b
Light is known to behave in a very predictable manner. If a ray
of light could be observed approaching and reflecting off of a
flat mirror, then the behavior of the light as it reflects would
follow a predictable law known as the law of reflection.
Part 1
1.c
Archimedes constructed a kind of hexagonal mirror, and at an
interval proportionate to the size of the mirror he set similar
small mirrors with four edges, moved by links and by a form of
hinge, and made it the centre of the sun's beams--its noon-tide
beam, whether in summer or in mid-winter. Afterwards, when the
beams were reflected in the mirror, a fearful kindling of fire was
raised in the ships, and at the distance of a bow-shot he turned
them into ashes.
Part 1
2.a In my opinion plane mirror is known to behave
in a very predictable manner. If a ray of light could
be observed approaching and reflecting off of a flat
mirror, then the behavior of the light as it reflects
would follow a predictable law known as the law of
reflection. The diagram illustrates the law of
reflection. And defiantly would not affect as
Hexagonal mirror.
2.b No, It would have the same properties,
because it has different shape and has
different way of use.
www.glenbrook.k12.il.us
For example, today we use plane mirrors as a
reflection of ourselves and Hexagonal mirrors
used in surgery rooms and as a microscopic
devices.
2.c I don’t think so, because we do
not get burned from plain mirrors.
www.saburchill.com
Part 2
Question 1a. Find the Vertex of fallowing function :
Answer 1a.
We can see that the formula we should use to solve this quadratic
function is .
Now we solve
After we got our Vertex we determine if the it is a Maximum or
Minimum Point
Since a = -1, and -1 is less than 0, this parabola would open down
So our vertex (- 4, -2) is the maximum point.
.
Part 2
Question 1b. Find the Vertex of fallowing function :
Answer 1b.
We can notice that the fallowing function written as
So the formula we should use to solve this quadratic function is .
Now we Identify the values as:
And plot them
Now we plug the x to find the y
The vertex would be (1, 1).
After we got our Vertex we determine if the it is a Maximum or
Minimum Point
Since a = 1, and 1 is greater than 0, this parabola would open down
So our vertex (1, 1) is the minimum point.
.
Part 2
Question 2a. Find the Vertex of fallowing function :
Answer 2a. We can see that the formula we should use to solve this
quadratic function is
We solve
After we got our Vertex we determine if it is
a Maximum or Minimum Point. Since a = -2, and 1 < 0,
this parabola would open up .
So our vertex is (-2, 1) and in the minimum point.
Now we will find the x, y intercepts by replacing them with 0
X=0
y-intercept is (0, 5).
Y=0
Since the square
root is negative
we can not find
the x-intercept
Part 2
Question 2b. Find the Vertex of fallowing function :
Answer 2b. We can see that the formula we should use to solve this
quadratic function written as:
We use the formula
We solve
.
After we got our Vertex we determine if the it is a Maximum or
Minimum Point. Since a = 0, and 1 is less than 0, this parabola would
open down .
So our vertex (0, 1) is the minimum point.
Now we will find the x, y intercepts by replacing them with 0
X=0
Y=0
y-intercept is (0, 1).
Since the x-intercepts are (-1, 0) and (1, 0).
Part 3
1.a The name of plane curve that emerges the functions y=ax^2,
a = 0 is Parabolic curve.
1.b In my opinion it related next way:
1. Both share the same shape, which
changes only if the values of a were
different in both equations.
Click here to see better
example in
Maple Illustration.
Part 3
2.a We use parabolas in everything that relates to curve and second power such
as a cup, a bowl , and car lights as shown downward.
Invariant under certain Dilation
Parabola have the property that when scaled (stretching/shrinking) along a direction
parallel or perpendicular to its axis, the curve remain unchanged. (For example, line
also have this property, but circle do not. A stretched line is still a line, but a stretched
circle is no longer a circle) When a parabola is stretched along the direct x a units and
along the axis by b units, the resulting curve is the original parabola scaled in both
direction by a^2/b.
2.b Given a parameterization of a parabola {x f(t), y f(t)} with vertex at Origin and
focus along the y-axis, it's focus is {0, xf(t)^2/[4y f(t)] }.
Optical Property
A radiant point at the focus will reflect or refract off the parabola into parallel lines. The
figure shows three parabolas, two of which share a common focus.
http://xahlee.org
http://xahlee.org
Part 3
3.a
First of all, when we see functions as: y=f(x) or f(x)=0 we know
that the relation is: Discriminate=0
We use the standard form
Where a= 0
When a quadratic equation is in standard form,
the
expression
that is found under the square root part of the
quadratic formula is called the discriminate.
The discriminate can tell you how many solutions there are going to
be and if the solutions are real numbers or complex imaginary
numbers.
Kinds of solution
Discriminate
1) If
for
Two distinct real solutions
Note that the value of the discriminate is found
under the square root and there is a + or - in front
of it. So, if that value is positive, then there would
be two distinct real number answers
Part 3
3.a
2) If
One real solution
Note that the value of the discriminate is found
under the square root and there is a + or - in front
of it. So, if that value is zero, + or - zero is the
same number, so there would be only one
real number solution.
Two distinct complex imaginary solution
3) If
Note that the value of the discriminate is found
under the square root and there is a + or - in front of it.
So, if that value is negative, then there
would be two distinct complex imaginary
number answers.
Part 3
4 Real coefficients in a second-degree polynomial are
crucial keys to determine and draw the graph of the function.
To find out vertex, and whether the graph curve up or down,
real coefficients give us the image of the graph of function.
For example:
f (x) =ax^2 (a) is a positive real coefficient in a second-degree
polynomial, this gives us information, which is the direction of the
curve. In the general case, f (x) =ax^2 + bx + c , we find the
coordinates of the vertex of the parabola by using the formula:
[-b/2a, f (-b/2a)], a, and b, are keys to find the vertex, and c, is the yintercept.
Another function is: f (x) = -ax^2
In this case, the property of (a) is same as a in the first function.
As a result, it is a reflection over the x- axis. The real coefficient a
is an integral key to form the shape of the graph of the function.
Moreover, it depends on it, and the parabola varies wider or
narrower.
Part 4
Yes, I think that all of the quadratic equations has focal property, because conic functions
sections can be defined as curves such that the ratio of the distances from a point on the
curve to a straight line (the directrix) and a point (the focus) is a constant, called the
eccentricity e.
This is called the focal definition. If the directrix is taken as the y-axis, and the focus is at
(p,0), then this definition gives [(x - p)2 + y2]1/2/x = e. Multiplying out,
x2(1 - e2) + y2 - 2px + p2 = 0. It's easy to see that this quadric is an ellipse if e < 1. a
hyperbola if e > 1, and a parabola if e = 1. The ellipse and hyperbola are called central
conics, for which the origin is usually taken at the centre. If this is done by making the
substitution x = x' + h in the equation, then h = p/(1 - e2). For an ellipse, this puts the
directrixes at x' = ±(a/e); that is, on the two sides. As these directrixes recede to infinity, the
ellipse becomes a circle. We can show that e = c/a, which is the usual definition of the
eccentricity of an ellipse. Here, a is the semi major axis and c is the focal distance. The
directrixes of a hyperbola are between the branches, again at distances ±(a/e), but here e > 1.
The focal definitions of the ellipse and hyperbola are not of much use in applications, in
contrast with the focal definition of the parabola.
Paraboloids are found very commonly as reflectors of waves, in satellite
antennas ("dishes"), searchlights and listening devices. They bring any
kind of waves to a focus, whether light, radio waves, sound, or even water
waves. A bay with a parabolic beach will concentrate waves at the focus. I
do not know of any examples of this, but similar effects of ocean wave
refraction are well known. All of these uses depend on the focal property.
A spherical concave mirror also has this property, but it is only
approximate. To do better than a spherical mirror, a paraboloid must be
rather accurate. In small sizes, it is easier to make an accurate spherical
mirror than an accurate paraboloid, and it is also better than a bad
paraboloid. The paraboloid must be accurately pointed to profit from its
advantages. Otherwise, a spherical mirror is actually superior.
http://www.du.edu/
Student: Elyahu Pinhasov
Course : Math 200
Professor : Orlando Alonso
> a:=3*x^2;h:=3*(x+2)^2;j:=3*(x+2)^2-5;
a := 3 x
2
h := 3 (x + 2)2
2
h: = 3(x +4x+4)
2
h: = 3x +12x+12
h: = (-2,0)
2
j := 3 (x + 2) - 5
2
j: = 3(x +4x+4) -5
2
j: = 3x +12x+12-5
2
j: = 3x +12x+7
> plot([a,h,j],x=-5..3,y=-5..27);
As we can see, the functions in front of us have
congruent geometrical representations. They
represent vertical and horizontal shift of each
other.
> a:=3*x^2;
a := 3 x
2
> plot([a],x=-3..3,y=-5..27);
> b:=-3*x^2;
b := -3 x
2
> a:=3*x^2;plot([a,b],x=-3..3,y=-27..27);
a := 3 x
2
> a:=3*x^2;c:=3*x^2+12*x+12;d:=3*x^2+12*x+7;
a := 3 x
2
2
c := 3 x + 12 x + 12
2
d := 3 x + 12 x + 7
> plot([a,c,d],x=-5..3,y=-5..27);