Seems : The thkim1011 Problem (to be cont)

Quote:

A tank in the shape of an inverted circular cone with height $\ 12\text{m}$ and base radius $\ 3\text{m}$ is filled with water to a height of $\ 9\text{m}$ . Find the work required to pump all the water up and out, over the top.
(Note: Density of water is $\ 1000 \dfrac{\text{kg}}{\text{m}^3}$ )

thkim1011 loves math . thkim1011 loves physics. thkim1011 is a boss at math. thkim1011 is a boss at physics. $\implies$ thkim1011 is a boss.
Well all of those statements are true. I've known thkim ever since I first joined AoPS (hm those middle school nights on FTW). We have each other's emails and we chat a lot (well kinda used to; now we're busy). He's a boss at all kinds of math but I know him for his calculus skills. I've asked him for help on a lot of calculus and physics problems in the past. He also helps me with CSS (oh yeah he's a boss at computer science but that's irrelevant right now).
When I first saw this problem I immediately thought "thkim1011! hm I should blog about this. . ." I think I saw this on Thursday; well here we are.

The problem struck me because: 1) in calculus class -hey a "complicated" calculus problem. thkim1011 could probably do this. 2) work? hm so physics too . . . oh hey thkim1011 is good at that too lol.
To me this problem represents thkim. A rich problem of math and physics (well not so much physics, but it's an application thing). This right here is the essence of thkim1011. (lol to me at least)
In class I was like "oh hey I love thkim1011. I will do this for him." So I sat up and paid attention and followed along . . . hm well I kinda started working ahead at times but hey this is a "complicated" problem so I had to wait for the professor occasionally to know what to do.

So this is part of a section about calculus applications in physics, more specifically integral applications to force, work, power, etc.
In one simple equation it's: $\ \text{F} = \text{m} \cdot a(t)$ where F is force, m is mass, and a is the acceleration function for time t. We know that acceleration is the derivative of the velocity function v(t), which is the derivative of the position function s(t). Or:

Quote:

$a(t) = \dfrac{d}{dt} v(t)$
$v(t) = \dfrac{d}{dt} s(t)$
$\implies a(t) = \dfrac{d}{dt} v(t) = \dfrac{d^2}{dt^2} s(t)$
So our expression for force $\ F=ma$ we can actually write as:
$\text{F} = \text{m} \cdot \dfrac{d^2}{dt^2} s(t) = \text{m}\dfrac{d^2s}{dt^2}$

Now what is work? Work in physics is defined as the force applied (F) times the distance traveled (d). $\ W=Fd$ .
We know what "force" is in the context of this problem. But what about "distance?" Well distance is the change in . . . oh hey it's $\Delta x$ or $\Delta y$ . . . or delta whatever. The change in position (at least in one direction(!)) can be written like that.
But . . . if it's in one direction(!), why do we care about calculus? Why not just solve problems using force times distance? It's rather straightforward; no calculus is needed.
Well. . . what if it's not in one direction..? What if the movement of the object (or whatever) is not straight, but changes direction? Then the force isn't always pointing in the same direction, and hence is not the same at all times. However, we can look at individuals sections of "times" when force is being applied in a certain direction. If someone is pushing a box around on a curvy line, he isn't always applying the same force in the same direction. However, we can look at moments in time, when he is pushing with a certain force in a certain direction. Of course, if the path is truly a curve, only infinitely small moments in time will give specific forces and specific directions, otherwise everything's still changing. WOah but in calculus we often talk about "infinitely" small times and stuff like that.
If we look at tiny tiny time spans, we can use calculus to find the total amount of the tiny pieces of force times distance.
Instead of times, let's look at certain points on the curve, like an arbitrary point $\ x_i$ . The work done at this point is: $\ W_i=F_i\cdot d_i$ .
We know our force, it's just mass times acceleration. For our purposes we will denote F(x) as $\ f(x)$ , our function. (Keep in mind that most problems won't give an explicit function for force, we'll have to find the mass times accel. ourselves)
Our expression for work becomes: $\ W_i = f\left(x_i\right) \Delta x$
Now what if we look at every single point $\ x_i$ along the whole curve or whatever. We'll take the sum of the work done at every single point x.
$W= \lim_{n\to \infty} \sum_{i=1}^{n} f\left(x_i\right) \Delta x$
Well whaddaya know this is an expression for an integral!
$W= \lim_{n\to \infty} \sum_{i=1}^{n} f\left(x_i\right) \Delta x = \int_{a}^{b} f(x) dx$
This is the expression we will use for work.

We will use our expression for work, solving for all the variables. We will use $\ 9.81$ meters per second squared for acceleration due to gravity.
Btw "inverted" means the cone is upside down. So it's like an ice-cream cone.
Let's put axes on this. Let the center of the cone and center of the surface of the water (3 meters from the top of the inverted cone) be the origin. Right/left is obviously the x axis, but we'll let the y axis point down. (Now don't get this confused) Just remember that positive y does down/negative y goes up.
We're trying to find work required. $\ W=Fd$ . Force is mass times acceleration: the force here is gravity pulling on the water. Our work is overcoming this force of (mass times acceleration due to gravity).
Well the work required to pump all the water out is not uniform for the whole "piece" of water. We'll look at teeny sections of the water, more specifically "disks" of water. For each $\ y_i$ we pick, a "disk" of water goes with it. We'll let $\ y_i$ be a tiny width (or rather height) of water.
Remember we're just looking for the mass of the water. If we have the mass, we can find the Force (mass times accel, which is gravity) and distance is pretty straightforward.

So what's $\ V_i$ ? After we find the volume, we multiply by density to get the mass, and then acceleration to get the total Force. Well taking little sections of the cone won't give us cylinders, but they won't give us cones either. We'll be getting extremely thin frustrums. (Frustrums are usually described as being slices of a cone, with the top (sharp part) cut off, so it's flat on both ends)
However, if our intervals $\ y_i$ are extremely tiny, we can approximate these sections as cylinders. (Volume of frustrum is too hard to deal wit)
Well let's do some math!

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The volume of each cylinder is:
$V_i = \pi r_i^2h_i$
The height is just the $\delta y$ .
$= \pi r_i^2 \Delta y$
Now what's the radius? Well if we consider the cone in 2 dimensions (the x/y axes we put there), we can draw a right triangle. The leg on top is 3 meters. The other leg (which is the y axis going down the middle of the cone) can be divided into two parts for a ratio. We let a random radius in the cone (perpendicular to the y axis to the hypotenuse) be $\ r_i$ . It splits up/down leg into two parts, we'll say $\ y_i$ and $\ 12-y_i$ . (Remember that the y axis goes down, so $\ y_i$ will be positive)
From all this we can set up a ratio:
$\dfrac{3}{r_i}=\dfrac{12}{12-y_i}$
$r_i = \dfrac{36-3y_i}{12} = \dfrac{1}{4}\left(12-y_i\right)$

We can plug this into our expression for volume:
$V_i = \pi \left(\dfrac{1}{4}\left(12-y_i\right)^2\right) \Delta y$
$= \dfrac{\pi}{16} \left(12-y_i\right)^2 \Delta y$
Now mass is volume times density, so we multiply this by 1000 because water is 1000 kg per cubic meter.
$m_i = \dfrac{1000\pi}{16}\left(12-y_i\right)^2 \Delta y$
$a_i = 9.81 \dfrac{m}{s^2}$
$F_i=m_ia_i = \dfrac{9810\pi}{16}\left(12-y_i\right)^2\Delta y$
We now have an expression for the force required to move the infinitely many "slices" of water. The sub i indicates that this is only for each individual "slice" of water, each being infinitely thin.
Now we know that $\ W=Fd$ , and to figure out what $\ W_i$ would be we need $\ d_i$ .
However remember back when we were solving for the radius $\ r_i$ we let the leg be divided into $\ y_i$ and $\ 12-y_i$ ? Well if you look at a diagram (or think really hard about it) you'll notice that for each $\ r_i$ we pick, the distance to the top of the cone is just the $\ y_i$ . (darn I really need to learn asy)
And how far are we moving anyway? To get each "slice" of water out the top of the cone, we need to pump each i(th) slice a distance of $\ y_i$ .
So $\ d_i = y_i$ .

We can now set up a full equation for work:
$W_i = F_i \cdot d_i = \dfrac{9810\pi}{16} \left(12-y_i\right)^2 \Delta y \cdot y_i$
(Remember this is just the work for each i(th) slice.
hm we need to take every single $\ W_i$ . There are an infinite number of them. . .
$W= \lim_{n\to \infty} \sum_{i=1}^{n} W_i = \lim_{n\to\infty}^{n} \dfrac{9810\pi}{16}\left(12-y_i\right)^2 \Delta \cdot y_i$
This is an integral! We write it as:
$W= \int \dfrac{9810\pi}{16} (12-y)^2 y \, dy$
...hm but we should have a definite integral no? Wat are our limits of integration . . .?
The work we've found is the work required to move each piece up the cone and to the top. The water at the bottom of the cone is at $\ y= 12$ and the top of the water is at $\ y= 3$ . (Remember the y axis points down) To go through all that water we go from $\ y= 3$ to $\ y= 12$ .
$W=\int_{3}^{12} \dfrac{9810\pi}{16} (12-y)^2 y \, dy$
Now we solve this integral:
$W = \dfrac{9810\pi}{16} \int_{3}^{12} (12-y)^2 y \, dy$
$W= \dfrac{4905\pi}{8} \int_{3}^{12} \left(144-24y+y^2\right) y \, dy$
$W = \dfrac{4905\pi}{8} \int_{3}^{12} \left[ 144y - 24y^2 + y^3 \right] dy$
$W = \dfrac{4905\pi}{8} \left(72y^2 - 8 y^3 + \dfrac{1}{4}y^4 \right)|_{3}^{12}$
$W = \dfrac{4905\pi}{8}\left( \left(72(12)^2 - 8(12)^3 + \dfrac{1}{4}(12)^4 \right) - \left(72(3)^2-8(3)^3 + \dfrac{1}{4}(3)^4\right) \right)$
$W = \dfrac{4905\pi}{8} \left((10368-13824+5184) - (648-216+20.25)\right)$
$W = \dfrac{4905\pi}{8} (1728 - 452.25)$
$W = \dfrac{4905\pi}{8} (1275.75) = 2457335.611... \approx \boxed{2.4573 \cdot 10^6} \text{J}$
The final answer will be in Joules. (Work out the units if you want)

Seems

The thkim1011 Problem (to be cont)

by EpicSkills32, Feb 15, 2015, 6:30 AM

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