Harper Vance AP Calculus BC Final Project

OK. Hi, I'm Harper and this is my A P calculus final calculus in the real world.

And to start off this project for the first half of it,

we will be looking at differential equations

and how they can be applied to insulation and creating insulating

protectors.

And to first do this, we ought to establish what a differential equation is.

And a differential equation or AD

Q is an equation that has a rate in it.

Thus relating one or more functions in their respective derivatives.

For example, Dydx equals X squared over Y.

And in order to solve ba

basic first order differential equations, so we'll be dealing with today,

one must follow the steps separate and great itself.

And we can see how these can be applied to our problem here.

So if we're being asked to solve the DVQDYDX equals X squared over IY, first,

we can separate the variables making it Ydy equals X squared DX.

And then we'll integrate both sides and add constant to the right hand

side Y squared over Y equals X cubed over three plus C.

And then from here we just solve for Y to get this general equation.

So it's Y

plus a minus square root to execute over three plus two C.

And as I said,

that's the general equation as we still have this constant C component.

And if we're wanting to make this a more partic,

if we're wanting to turn this into a particular solution,

we must be given another coordinate point.

And in this example, we're given the 0.03.

And so in order

her to work with this, we have to plug that into the equation.

And then from that, we are able to solve for same, you see that in steps one and two here

and then you're able to substitute that same value back

into the general solution to get you your particular solution,

which

in virtually all occasions is a lot more useful.

Perfect.

OK.

So now that we at least have a pretty introductory

or like working knowledge of what a differential equation is,

we could see how it's applied.

And differential equations can be found in the real world scenarios.

That's key through the production of insulating containers such as surfaces.

For to see the engineers rely on the fact that

at

the rate at which the substance placed in

an insulated container cools down or warms up

is proportional to the difference between the substance

current temperature and the ambient external temperature.

The rate of change

in temperature over time can be represented as D capital T over D lowercase D.

Thus DT is proportional to delta T,

which is a change in temperature or DT TT equals

K delta T when K is the proportionality context.

And we can further this saying that delta T can additionally be

written as T minus B where B is the ambient external temperature,

thus making DTDT

equal to K times the quantity of T minus B.

And when applying our separate integrate solve steps to this, we can find

that um this initial differential equation

ends up being T equals ce to the KT plus B I have that um derivation there.

That's what that is.

Um And so we get T equals ce to the KT plus B or C is

a constant K is a constant of proportionality

T is time B is the ambient external temperature

and C plus B is equal to the objects this temperature.

And let's see how we can apply this in our example here.

So given the scenario that water has just been boiled for tea and is thus at 100

°C

and has been poured into a mug in a room that is 22

°C

at T equals 10 minutes. Its temperature is measured to be 90

°C.

How long from the time it's poured will take the water to have cooled down to 60

°C.

So first, we're gonna

assume for this whole problem, we're gonna be utilizing the

equation T equals ce to the KT plus B.

And so we'll know that B equals 22

C will be 100 minus 22 or 78.

And from there, we can plug them into the equation.

So we get T equals 78 E to the KT plus 22.

And as we know when T equals 10 minutes, its temperature is measured to be 90

°C, we can then plug those values in

yielding us 90 equals 78 E to the

10-K plus 22.

And working that through that will eventually get us as K is equal to negative 0.014.

And then

moving up.

Thus, we know that T equals 78 E raised to the negative 0.014

T plus 22 as we just subbed in that K value that we just found in the previous slide.

And

um

as we're being asked how much time has passed,

how much time will pass when the water reaches 60

°C

to figure this out, we just sub the 60

for tea as that's the end, ultimate temperature that we're trying to reach.

And then we sell for TN for tea.

Um That's gonna be 51.37 minutes. So that's what it ends up being.

So thus, we can include that 51.37 minutes after the water was poured,

it will be 60 degrees septic grade.

Perfect. And that's all you have to do

for these types of problems. OK, you just

work out the diff eq like that.

It's really not that bad.

OK. Perfect.

And just a little note on this scenario, it's very much parenthetical.

However, I do feel as if it's worth noting

even if it is intrinsically a little nerdy.

Um

but just on the physics of it, all Newton's law of heating and cooling,

which is what we were just using purely

serves to model the convection of heat energy.

However,

only the surface of water will exhibit such

behavior as it possesses a water air interface.

Meanwhile, on the sides and bottom of the mine,

the water will experience thermal conduction should be

subject to fourier's law of thermal conduction.

However, for this discussion,

the thermal conduction through the container would be affected,

affected by

characteristics such as surface area,

thickness and thermal conductivity of the container.

However,

here all these factors can be assumed to be rolled together to influence the K

variable as there are constants that contributed to the value of K that was reached

A K. This note is just making sure that like the physics

within this problem can be assumed to be correct,

but it does have some basis to it.

OK.

And so now that we've looked at some real world applications of it,

looked at the math side

even taken a little glimpse into the physics side, let's see how

people can make a career out of this.

And so if the aforementioned assumptions regarding

fourier's law about a material sickness,

thermal conductivity

and surface area being subsumed into this K

variable of the differential equation is accepted.

Then these factors would be pertinent to the decision making

of engineers designing such thermoses or other insulating containers.

Furthermore,

a second design consideration focusing on Newton's law of heating

and cooling would focus on minimizing the amount of convection

that takes place at the air surface interface.

Although air excels at heat transfer through convection,

it is relatively poor at heat conduction.

And for that reason,

a design feature as simple as a cap would help limit the rate of heating and cooling.

So here we're just kind of exhibiting how

or um some of our prior like our differential equations can me

with physics and be very much so pertinent to engineers and their careers if they're

tasked to make these type of materials.

And then here are my sources for the section.

OK. And so for the second half of this

um presentation and

honestly, the half that I'm most excited about,

we are going to be looking at equestrian training and arc length.

And once again, let's establish what this topic is that we're talking about.

So for arc length,

arc length is often represented by the variable L

and can be defined as the difference between distance

between two points A and B if they are connected by a curved line

and then within the rest of the slide,

it's just kind of working through this

um derivation of these arc length formulas with

these bottom three integral equations being which

will what will be most salient to us

today.

And we can see an example problem here

if we're asked to find the length of the curve X equal, no,

find the length of the curve Y equals X to the

three halves from X equals zero to X equals four.

We're first gonna be asked to find Dydx

and then,

which will be three halves X to the one

half and then plug that into this arc length equation

and evaluate it from X one to X two. We have this on the prior slide.

It's just one of our standard arc length equations.

And then we're gonna evaluate this

integral once we've plugged in Dydx at X equals

zero to X equals four as our given parameters.

And

although it looks a little hairy, it works out well

and it will eventually lead to the answer being 9.073.

So it's just a baseline example.

Like I said, some of the math looks really hairy, but overall,

it is a pretty straightforward process.

OK?

And so now that we have a working understanding of that,

let's see how this can be applied.

And arc length can be understood in real

world context as it relates to equestrian activities.

As when a horse jumps over an obstacle,

its path will be naturally curved line over the jump.

Thus, the path of the horse provides a curve and space where both the X

coordinate, horizontal distance

and Y coordinate vertical distance vary as a function of time.

Thus establishing a set of parametric equations.

Furthermore, by establishing the point where the horse leaves the ground at point A

and the point where the horse lands after the jump as point B one can

appropriately set up an equation to find the arc length of a horse's shop.

And let's do just that, let's see how this works.

Um And for all these subsequent discussion, consider the

G shown in the section video one,

this is a video once we're just going to be considering

this and this natural parabola that this horse's movement is creating.

And this gift depicts a horse jumping a 33 obstacle,

the training session that the depicted event occurred and was

videotaped and processed through the open source program tracker.

This program allows one to trace the objects movements through serial frames.

The yellow marks on the video indicate the position

of the horse's shoulder in each frame of the video

tracker is able to assess the X and Y

coordinates as a function of time and then plot them

a screen print of the analysis Y of T is depicted in the section

figure one, here's figure one.

Um And this graph Y of T measuring in feet

resembles a parabola that is superimposed in the picture.

And based on the data in the curve tracker calculated the following equation

Y of T equals negative 16.4 T squared plus 12.1 T minus 0.21.

And just,

it's gratifying to note that our um acceleration

coefficient is very similar to that of gravity.

So math is working out how we should

uh

continued.

And so further going on this similar analysis of X

of T can be seen in the section figure two

perfect.

Um and can be used to assess the horse's movement along the X

axis which yields the equation X of T equals 6.2 minus 0.06.

Furthermore, the total horizontal

distance covered by the horse in the video is 11.44 ft.

The total distance along the Y axis, the vertex of the parabola is around 2.072 ft.

It can also be noted that the horse achieved the

maximum vertical distance of around 1.5 ft before the actual jump

tracker itself does not calculate the length of the jump.

But this can be easily derived from the,

using the

arc

length equation.

And then it's just gonna be this one that we're plugging in our values too.

And you will see how

we do this here.

So you're given the set of parametric equations Y of T and X of T.

And from here,

you can find Dydt and DXCT to be the following with Dydt ending up to be negative

32.8 plus 12.1 and DXCT being 16.2.

And then it is also very much important to note that the time in between

the horse taken off the jump where the front feet lift off the ground.

And from the, when the fro front feet first land back on the ground is 0.733 seconds.

There,

the relevant length of the arc is equal to this integral here,

evaluated from zero to 0.733.

And here we just plugged

nr to derivative equations and to the formula that was shown on

the slide prior and evaluate it from the relevant time span.

And overall this yield, see information

that the overall arc length is 12.889 ft.

Therefore, the horse travels 12.889 ft over the jump.

Ok.

So now that we've made this request,

this connection between arc length and equestrian activities,

let's see how this can be a career topic.

And the models discussed in the section of the presentation can be of use to horse

trainers and equestrian competitors as they work to

get optimal performance out of the horse,

equestrian discipline such as hunters, which is what I do.

I promise, I know what I'm talking about,

work to ensure that both the rider and the horse

appear graceful and elegant over the jump within the discipline.

A horse's stride is expected to be around 12 ft in length,

but then taking off 6 ft before the jump and landing 6 ft after it.

Thus, since the horse must clear a vertical jump in that distance,

the arc length of the jump should be a value that efficiently exceeds 12 ft.

And it is the job of hunter trainers to ensure that their students are

equipped to meet such standards as competitions

will use these criteria when scoring competitors

for. It's this expected arc length that is um

symbolic that is

portrayed as very elegant, as very graceful.

It's what you want to achieve as a smooth arc over the jump.

It was just a little greater than 12

fit.

Here were my sources with my printed review

textbook and um also the credits for tracker,

which is very cool

and that

should be it.

Thank you so much

for listening to that spiel, sorry that it was rushed.

Um My program would not allow me to videotape more than 15 minutes.

Harper Vance AP Calculus BC Final Project

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