Elysian Shadows

Posted: **Thu Aug 27, 2009 8:59 am**

So every week I get something called 'Enrichments' in my math class. Just for Clarification of my question here's the 'enrichment'

The rate at which people enter a High School on the first day is modeled by the function Q given by Q(t) = 1680t^2 - 776t^3 for
0 <= t <= 2; Q(t) is measured in people per hour. No one is in the school at time t = 0, when the gates open. The gates close and school begins at time t = 2.

a.) Graph the function on the time interval for 0 <= t <= 2.
b.) Find the time when the rate at which people enter school is a maximum. Justify your answer.
c.) How would you find out how many people attend school on the first day of school, assuming no other students attended when the gates closed?

Okay so a.) and b.) are easy, got those done in no time at all. But c.) is what I don't really know. My teacher said it was calculus because t could be an infinite amount of numbers in between 0 and 2. Really, the only 'explanation' I can come up with is simple; Solve for t an infinitely amount of times, increasing t by an infinitely small number that is in between 0 and 2, and find the total of all that.

Something tells me there's a better way to explain it, so I thought you guys might be able to help =D.
I've used google, didn't find anything much at all relevant, so I came here. Also, I need the answer so I can help everyone else, my teacher told everyone to stop asking him questions and ask me... Basically most people are epically retarded at my school, so I'm like that nerd guy that everyone asks for homework help from >.>.

In another note... dang I didn't think the difference from AP and CP was that bad. I couldn't stand my AP english teacher, she gave me 2 referrals before she was even my teacher. They also put me in there without my knowledge of some summer reading we had to do, so I was like waaaay behind in that class before it even started. Had like 5-6 books to read over the summer + analytical essays and all that stuff.... Anyways, that CP class was filled with so many stupid people... Half the people hadn't the slightest clue of what a thesaurus might be. The other half thought was a dictionary. No one in the room had a clue of what the word cryptic could mean. Combined IQ of that class = IQ of a desk

Posted: **Thu Aug 27, 2009 10:08 am**

c) is a "definite integral." Where you are integrating the function from t = 0 to t = 2.

Q(t) = 1680t^2 - 776t^3

Integrated is:

q(t) = 560t^3 - 194t^4 + c (where C is an arbitrary constant)

Now this function must be evaluated on the interval [0, 2].
q(2) - q(0) (which is 0) so just:

q(2) = 560(2)^3 - 194(2)^4 = 4480 - 3104 = 1376 students

-----------------------------------------------
What I just did is integral calculus. I am not so sure that I could explain it very well, but I will say that

XianForce wrote:Really, the only 'explanation' I can come up with is simple; Solve for t an infinitely amount of times, increasing t by an infinitely small number that is in between 0 and 2, and find the total of all that.

You are a very smart kid. You just pulled the entire concept of a "Reiman Sum" out of your ass without even knowing integral Calculus. That is exactly what you are doing:

The image is a Reiman sum with a FINITE number of rectangles used to approximate the area under the curve. An integral is using infinitely many rectangles for the approximation at an infinitely small resolution.

Actually doing it is extremely easy. I divided the coefficient by the exponent+1 and added one more to the exponent to get the "integral."

Geometrically, think about it this way. I just graphed the function from Q(0) to Q(2) and took the area under the curve.

Here for further reading. It was a simple definite integral: http://en.wikipedia.org/wiki/Integral

Posted: **Thu Aug 27, 2009 10:27 am**

this is what falco means in a picture form.
the integral of the original equation evaluated between any two given points will give you the area underneath the unintegrated equation between those two points.

if you wanna test this to make sure its actually true, (which i did back when i started) was i would take equations that were straight lines. ie y=x or y = 2x etc.
that would be a triangle between 2 points, preferably [0,+n] and you can get the area of the triangle with 1/2bh or via the integral. they always end up being the same.

anyways. nice explanation falco.

Posted: **Thu Aug 27, 2009 10:46 am**

Holy shit, avansc. What program is that?!

Posted: **Thu Aug 27, 2009 10:52 am**

http://en.wikipedia.org/wiki/Grapher

come free with osx.

i have a maple license if you would like that. i don't ever use it anymore.
http://www.maplesoft.com/

http://www.maplesoft.com/products/Maple/demo/index.aspx

look at the clickable math video.

Posted: **Thu Aug 27, 2009 1:40 pm**

c.) How would you find out how many people attend school on the first day of school, assuming no other students attended when the gates closed?

I know Falco explained this, but just to give you a leg up in your calculus class, they're going to make you do Riemann sums first. This is hard. To get the right answer easily* (or perhaps, check your answer since they'll make you show your work), you can do two things:
1) Get a TI-86, put the equation in, set graph boundaries to 0 and 2.1, go to graph mode, and on one of the "2nd" functions is some graph operations, one of these is integral. Do 0 to 2.
2) To take the integral of (find the area under the graph of, ie, infinite sum of, hence why the symbol in avansc's picture looks like a big, stylized 's') 1680t^2 - 776t^3, you increase the powers by one and divide. I'm denoting "between 0 and 2" as [0, 2]. Sorry, I can't do proper notation easily here.

Code: Select all

Integral(1680t^2 - 776t^3) between 0 and 2
= Integral(1680t^2) [0, 2] - Integral(776t^3) [0, 2]
= ((1/3) 1680t^3 - (1/4) 776t^4) [0, 2] // This has been integrated, that's why "Integral" has disappeared
= ((1/3) 1680(2)^3 - (1/4) 776(2)^4) - ((1/3) 1680(0)^3 - (1/4) 776(0)^4) // This is maximum minus minimum
= ((1/3) 1680(2)^3 - (1/4) 776(2)^4) //This is your answer

This is the usual case, but they'll have some where the graph goes below 0, in which case you have to break up your integral, iirc. Also, Isaac Newton is one of the two inventors of calculus :)

*At first, they'll have you doing finite Riemann sums, Trapezoidal method, etc, so you'll only be able to tell if you're in the ballpark.

Posted: **Thu Aug 27, 2009 7:13 pm**

GyroVorbis wrote:
XianForce wrote:Really, the only 'explanation' I can come up with is simple; Solve for t an infinitely amount of times, increasing t by an infinitely small number that is in between 0 and 2, and find the total of all that.
You are a very smart kid. You just pulled the entire concept of a "Reiman Sum" out of your ass without even knowing integral Calculus. That is exactly what you are doing:

Lol, thanks =D.

Posted: **Thu Aug 27, 2009 7:27 pm**

MarauderIIC wrote:
c.) How would you find out how many people attend school on the first day of school, assuming no other students attended when the gates closed?
I know Falco explained this, but just to give you a leg up in your calculus class, they're going to make you do Riemann sums first. This is hard. To get the right answer easily* (or perhaps, check your answer since they'll make you show your work), you can do two things:
1) Get a TI-86, put the equation in, set graph boundaries to 0 and 2.1, go to graph mode, and on one of the "2nd" functions is some graph operations, one of these is integral. Do 0 to 2.
2) To take the integral of (find the area under the graph of, ie, infinite sum of, hence why the symbol in avansc's picture looks like a big, stylized 's') 1680t^2 - 776t^3, you increase the powers by one and divide. I'm denoting "between 0 and 2" as [0, 2]. Sorry, I can't do proper notation easily here.
Code: Select all
Integral(1680t^2 - 776t^3) between 0 and 2
= Integral(1680t^2) [0, 2] - Integral(776t^3) [0, 2]
= ((1/3) 1680t^3 - (1/4) 776t^4) [0, 2] // This has been integrated, that's why "Integral" has disappeared
= ((1/3) 1680(2)^3 - (1/4) 776(2)^4) - ((1/3) 1680(0)^3 - (1/4) 776(0)^4) // This is maximum minus minimum
= ((1/3) 1680(2)^3 - (1/4) 776(2)^4) //This is your answer
This is the usual case, but they'll have some where the graph goes below 0, in which case you have to break up your integral, iirc. Also, Isaac Newton is one of the two inventors of calculus

*At first, they'll have you doing finite Riemann sums, Trapezoidal method, etc, so you'll only be able to tell if you're in the ballpark.

Yeah I graphed it, that portion doesn't go below 0. But I guess I'll take a look at all this and actually solve the damn problem instead of explaining it. Kiss ass + Extra credit =p.

BTW, I forgot my manners -.-

Thanks for all that help guys =D

I wish I could get a closer approximation xD, oh well.

Elysian Shadows

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