Parabola Applications Maximizing Minimizing Reflectors, etc


HELLO! Mr. Tarrou. Now we just got done talking about parabola. We found.. um.. their vertexes, their x intercepts,
their y intercepts. We talked about their axis of symmetry which
if the x term is the one that is squared, it is a vertical line going through the x
component of the vertex. Now let’s talk about applications of parabolas
for a minute. Why do we care about these? Well, anytime you though something like a
football or baseball, or just anything that is not under it’s own power, the path of that
item will generally follow the shape of a parabola. That is good for modeling objects. The shape of a light reflectors and satellites. When you… When we do a topic later and actually do conic
sections as an entire chapter and we go through the conic sections like parabolas, ellipses,
circles, and hyperbolas you are going to learn that there is something in a parabola called
a focus point. Well, if you take… If this were the shape of a light reflector
or a satellite, we are going to do a light though, and I let some light emit from the
light bulb which if done properly is in the place of the focus point all of that light
is going to hit the parabola and reflect out in a series of parallel lines. That is why when you take a flashlight, you
can look around that there is just that little bright spot of white light because the reflector
is in the shape of a parabola. We can also use them to maximize revenue and
maximize area. If you go back and you look at my video about
modeling data with functions, we a lot of times ended up with quadratic equations. That meant parabolas… which meant that we
could find the vertex, which we are going to do today, of some parabolas to maximize
some variables. Ok. So what is all of that going to look like? Well, it is going to look like this. Let’s take a look at modeling the flight of
a non-powered object. Let’s say that you throw a baseball or a football,
or something ball shaped, and you get a function that models the flight of that object which
is going to be f(x) equals -.01x^2+1.18x+2. And this is modeling some kind of flight. And you know, this 2 means that it was initially
tossed at 2 meters off the ground. So this a parabola that opens down. So the graph would look something like this. And we are going to find the vertex of this
parabola. That vertex is going to represent the highest
that object gets and horizontally how far over will the… How far is the object going to travel before
it reaches its maximum height. Well this parabola is already in General form,
and general form is ax^2+bx+c. And if you want the x of the vertex without
using a graphing calculator to find the relative maximum, then you would have to find the x
of the vertex by going through this formula -b/2a. Well a, b, and c means that the x of the vertex
is found by negative… or the opposite of b which is 1.18… over 2a. Well ‘a’ is -.01, so 2 times -.01. And when you put this into your calculator
you are going to get a value that equals… let’s see… I have it somewhere… 59 units. So if this word problem were about you throwing
something and it was initially released 2 feet off the ground, then the maximum height
reached would be in 59 feet. If this is meters, then the maximum height
is going to be reached after 59 meters of flight… yards.. and so on. For this parabola, if you travel over from
the initial place where it was thrown 59 units, then above that x somewhere is going to be
the maximum height the ball travels… or flies up. So the y of the vertex is going to be found
by taking this x value and plugging it back into the function. So f(59) is equal to -.01(59)^2+1.18(59)+2. And I am going to cheat and pull my number
off of the side board here. That is 36.81 units. So the maximum height
of 36.8 units is reached in 59 units of horizontal travel. There is the vertex right there. So any problem that talks about objects being
thrown, tossed off a building, just on the football field. Even shooting a bullet. Once that leaves the gun it is following a
parabolic that because that is not underneath its own power. Those are all going to be modeled with parabolas
and quite often we find the vertex. And you might even find out how far that ball
travelled by finding the x intercept by doing the Quadratic Formula or Factoring. So let’s take a look at another example. Let’s talk about an example where you are
going to maximize the area. Again if you watched my videos or if you have
studied recently a bunch of word problems involving functions, a very common one is
about areas. And relating the area of a rectangle to its
perimeter. So we have a river. A really common textbook question. Against that river you want to fence in a
certain amount of area. Well to make this fence we are going to use
500 yards of fencing material. Now we don’t have to fence against the river
bed because… well, the river is acting as its own barrier. So we are going to just run three lines of
fencing. Now how much fencing do we have? Well I just said 500 yards. So we are going to label this as x, and this
side x if it is going to be a rectangle. Then what is left? Normally we just say x and y for the shape
of a rectangle. But we know that these three sides have to
add up to 500 yards. We have already used two of those sides. So what is left? Well, let’s say this were a 100 yards. 100 and 100 makes 200, and there would be
300 yards left. So as an expression, because I don’t really
know that this 100 yards, but what did I just do with the numbers? 500 minus two of whatever these lengths are. So now we have got three sides of a rectangle
set to a perimeter of 500 yards. How much area can we inclose with this rectangle? Well, area is length times width. So the area of a rectangle is length times
width. But, we don’t have any random rectangle, or
normal rectangle, we have one with a perimeter of 500. That means that we have some set values. We know something about the length and the
width. So we are not going to use this generic L
and W, we are going to write area of terms of x. Well that is length times the width of 500-2x. And this forms a parabola 500x minus 2 x squared. We can use that parabola to find the area. This is what it is going to look like. We have a downward opening parabola. So it is going to look something like this. We will find out exactly what it looks like
in a minute. And the length of the side, how far we are
away from the river x is going to be of course the x axis. And the formula is giving us the area. We plug in an x and we get an area. So the area is going to be on the y axis. We want to maximize, we want to find the greatest
area. We want to find again the vertex of this parabola. And it is pretty easy to do that with a graphing
calculator. You throw the graph up there, you hit second
calculate, and ask it to find a relative maximum. I will do a calculator lab for that for you
if you don’t know how to do that soon. But we are going to find the vertex of this
by… And I just realized that x times 500 is 500x. So I forgot that x. Thank you for pointing that out for me [no
one in classroom] How do you find the x of the vertex is again -b/2a. Well my b value, I did not write this in general
form so be aware of that. The squared term, in front of that is ‘a’. The single degree term, in front of that is
‘b’. Then if there is a constant, that is ‘c’. Well be is 500 so take that ‘b’ out. ‘a’ is negative two. And the x of my vertex is negative 500 divided
by negative 4. That means 500/4 is 125. So over here at 125 yards we are going to
find our maximum area. Now if I want to actually find that area. Then that would be A of 125 equals 500 times
125 minus 2 times 125 squared. And when you work that out with your calculator
you are going to find out whatever that maximum area is. So we took a length and a width as it related
to a fixed perimeter of 500. We did the distributive property. We got a higher order of two and the parabola
opens down with the leading coefficient that is negative. We worked out -b/2a and found that the maximum
x, or the x of the vertex… the x that is going to maximize the area was 125. So if this is 125 feet on each side and this
would be 500-2(125), and that would be 250, we would get the maximum area that we could
possibly enclose with this fence that is against the river. And we will do calculator labs soon about
finding relative maximums with a graphing calculator. If you go back and you watch the modeling
with functions videos, there are three of them, a lot of those real life applications
ended in a parabola. You can use this type of idea with -b/2a to
find the vertex of those parabolas. And maximize revenue of a business, maximizing
area, maximize I think the volume of a box. So now you can do those. But… There you go. Finding the maximum value of some functions,
some real life applications using parabolas and vertexes. BAM! Have a great day:D

9 thoughts on “Parabola Applications Maximizing Minimizing Reflectors, etc

  1. I think the solution in this parabola/area problem may be incorrect. I believe the maximum area of a rectangle is when all sides are equal, i.e. a square, which would be inconsistent with the solution, 250 x 125. Thanks for your great videos and I look forward to your comment!

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