Math @ MindSay

Well, it's kind of a big day in the life of a little guy.

Cartoon Ranger goes to a "gen-ed" class today for the first time.  As I type, the "math class" time is over.  He only went for math, which is cool. It's a good introduction to a more mainstreamed educational environment, in my opinion. 

I'm terribly proud of my little guy.  I have prayed he'll have a positive experience.  His personal math level is above grade level, so I'm mostly concerned about the social aspect of this new part of his academic day.

Here, it has been Date Day.  :)  My Spousal Unit sacrificed exceedingly for his wife and...

Ready?

...watched Mansfield Park with me on the IFC channel.  After watching The Importance of Being Earnest.  The latter makes us both laugh frequently, the former was a first-time-ever for Spousal Unit and he was kind enough not to fall asleep. :) 

To top it all off, we made a return visit to The Arizona Pizza Company!  It has, indeed, been a beautiful day. 

I am now waiting with anticipation for Ms. Brown's daily note for me to find out how CR did in math today. :)

Hey all

Its been a while since i sat down and actually wrote an actual blog that was more than a few sentences (since like, july :|) Ive wanted to for a while to write an update blog, so, here it goes...

 

Grade 11 started a few weeks ago. Its been pretty hectic. This semester goes: Chemistry, Physics, Biology, Grade 10 math. Easiest classes right now are grade 10 math (were basically doing grade 9 review right now(and in case anyones wondering why im taking grade 10 math when im in grade 11, its because i slacked off severly last year) and physics. Hardest is Biology (chapter we are finishing up is about different kinds of sugars (glucose, glycogen, etc) and their structures..its just not  sinking in with me....however, my teacher assured us its going to get easier) Chemistry is somewhere in the middle. We are basically doing mostly what we did in the grade 10 chem unit right now, so its fairly easy.

 

Besides classes and everything though, my attitude is alot better towards homework, and I'm doing most of it:) Basically i just make sure i understand whats going on, and if i dont do all tyhe homework beyond that point, it doesnt matter to me. So im really happy about that:)

 

In more personal news, Mikko and I are doing great. Our 9 month was on the 17th:) My longest relationship before him was 6 weeks :|

 

Also, I am still on prozac, and it has been about 9 months of that to. It helps alot with mood swings and the such, but i still have my days. The one thing VERY noticible its doing is that everynight for the last, 4 months or so, i have weird and random dreams. Examples: One night i had a dream one of my guy friends was in my math class and we were putting on lime-green mascara:P Then the teacher took my glasses away because i was lying on my desk on my stomack :|:P:P LOL. Another time i had a dream where i was in a concentration camp and I was pregnant so they tried cutting open my stomach (i was awake, not medicated) to get the baby out. Outside of the dream i forced my eyes open (not with my hands, just with well...my eyelids) because i didnt want myself to dream that because it was terrifying. This isnt lying either. It was probably the most horrifying dream I ever had :|:|:|

 

And in just in case your wondering, i DO check mindsay pretty much everyday, its just i havent blogged in forever.

 

anyways, thats all for now. till later,

 

-:PKristal:|

 

 

Ike continues to have its sights set on Chicago. We've recieved 6-8" of rain from other storms before Ike, and now Ike is feeding the front that is sloooowly making its way across the state. The sump pump is holding, but I think that this water should be shared with everyone, not just me. But that's me being greedy/generous.

 

I wish I could deal with the stress that this is bringing. I am holding out ok mentally, but this is an issue that seems to have been getting worse as the years go on. My dad suffers greatly from the same affliction, but his is more anxiety on what could happen; mine is small pockets of anxiety coupled with actual stress now.

 

I was tutoring a young boy the other day who did not want to be at Huntington at all, and it was evident that his parents probably coddle him because of his attitude for a 5th grader was more along the lines of my friend's oldest son who just turned 7. Actually, my friend's son is more mature than this kid. I think I was able to at least persuade him to do some work in math. What amazes me is the attitude that some kids take - of all ages! This young boy said flat out, that not only did he hate math, he didn't need it, and he learned everything that he needed to know about it. He also hated reading. I asked him, has he ever read Harry Potter and he replied that he never reads 'chapter books'. Whoo. Dick and Jane then I guess are still his bedtime stories, if he has any at all. 

 

Ironic/funny thing was, he loves science and history, especially astronomy and WWII. LMAO! Will he be in for a rude awakening. He had quite the vivid imagination and perception on history too - point blank stating that the Nazis had won all the wars they were involved in.

 

Yeah.

 

Don't know what version of history he's subscribing to, but the book is a little off. So then, after I showed him an astronomy problem in an Algebra II book that we have, (all it was was 3D distance formula application) in order to equate his love for the sciences with a lot more math that he'll have to take, so he better get used to the fact of at least tolerating math, he asked me this:

 

"Do you know what's harder than that math?"

 

I quickly commented, "Calculus?" *fwoosh* Right over his head.

 

He then said that there's a problem that he sees all the time:

 

A + d =

 

There's no typos there. That's all he wrote. He asked me if I could solve it? And I said, it's impossible. It has no answer, and there's two unknowns but only one equation. You need a second equation relating these two variables again to solve for them. And he said yes! But I think he said yes, more because he thought that he knew something I didn't.  

 

After him was an older girl who for some reason, I notice, always wears ballerina flats - not actual ones, but the ones that are in style now - saying essentially the same thing - She hates doing long division. Now she's probably in high school, says she hasn't done this since 5th grade (meanwhile, thinking back to the boy who hated math who was in 5th grade, do you know what he was doing? Addition and subtraction) and why should she have to do it the long way when there are calculators that can do the work for you now?

 

*sigh*

 

And of course, kids like this think they have it all figured out already, and that they're the first ones to do so.

No matter what logical reason I tossed at her, she had a ready answer:

 

me: What if you don't have a calculator?

her: Why wouldn't I have a calculator?

me: What if you forgot it?

her: I'll go ask to get it out of my locker.

me:  What if it's an ACT or SAT where they don't allow you to go get it?

her: Then I'll fail it and retake it later on.

me: *mental smack of forehead* What if you have it and it runs out of battery power?

her: I'll make sure and put in fresh ones before the test, or get a solar-powered one.

me: *mental smack of forehead*

 

Yep. Got it all figured out. Congrats, you are a genius and noone's thought of this before.

 

I saved the good ammunition rationales for another time. Again, what gets me is that kids forget it right after they learn it. They don't think that the previous knowledge will build onto knew knowledge. Oh well.

 

This is just a good ole summary section.  Nothing new.  Now you can just get all your main equations from the chapter in one nice easy section!

Fundamental Theorem of Calculus:

∫[a,b] (F'(x)dx) = F(b) - F(a)

Produces a distance between two values, aka a straight line between values at a and b.  Ultimate relates an integral to function values.

Fundamental Theorem for Line Integrals

∫[C] (Ñf ∙ dr) = f(r(b)) - f(r(a))

Produces an arc length between two values, aka a curve C between values at r(a) and r(b).  Ultimately relates a line integral to values of vector-valued functions.

Green's Theorem

∫∫[D] ((∂Q/∂x - ∂P/∂y)dA) = ∫[C] (Pdx + Qdy)

Produces the area of a region D bounded by a curve C, taking counterclockwise as positive orientation.  Ultimately relates a double integral to a line integral.

Stokes's Theorem

∫∫[S] (curl F ∙ dS) = ∫[C] (F ∙ dr)

Considers only a surface (positive oriented with n outward) and a boundary curve on some area of this surface and produces the area of this section.  Ultimately relates a surface integral to a line integral.

Divergence Theorem

∫∫∫[E] (div F dV) = ∫∫[S] (F ∙ dS)

Produces a volume of a surface given by an enclosed region E, positive oriented with n outward.  Ultimately relates a triple integral to a surface integral.

Earlier, we said Green's Theorem could be written in a vector form as

∫[C] (F ∙ n ds) = ∫∫[D] (div F(x,y) dA)

For giggles, is it possible that we could create a similar set up for a surface integral?  Why yes!  Yes we can!  It works for a special type of region, called a simple solid region.  In Chapter 16, in triple integration, we had three types of regions.  Well, if you take something like, say, a rectangular box, that can be ANY of the three types of regions, all of your choosing, and all at once.  Such a region is a simple solid region if it can be all three types of regions at once.  Also, we chose positive orientation for surfaces to be outward, that is, n, the unit normal vector, points outward, not inward.  With those two in mind, we can now create the Divergence Theorem:

Let E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation.  Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E.  Then

∫∫[S] (F ∙ dS) = ∫∫∫[E] (div F dV)

And we'll prove it at the end.

And that is IT.  We'll do an example.

We wanna evaluate ∫∫[S] (F ∙ dS), where F(x,y,z) = xyi + (y² + e^{xz^2})j + sin(xy)k, and S is the surface of the region E bounded by the parabolic cylinder z = 1 - x² and the planes z = 0, y = 0, and y + z = 2.

Normal methods of surface integrals would be horrendous.  Not only would we have to evaluate FOUR surface integrals (because this solid region has four surface boundaries), but look at F, and you see it's sort of a nightmare.

However, we'll take div F:  ∂[xy]/∂x + ∂[y² + e^{xz^2}]/∂y + ∂[sin(xy)]/∂z
= y + 2y = 3y

So, instead of some "mess" for F ∙ dS, we have div F dV, which would be 3ydxdydz, though not necessarily the "d_" in that order.  Let's look at our surface.

If we say E is a type 3 region, then we have E = {(x,y,z) | -1 < x < 1, 0 < z < 1 - x², 0 < y < 2 - z}, and so we finally wind up with

∫∫[S] (F ∙ dS) = ∫∫∫[E] (div F dV) = ∫∫∫[E] (3ydV)
= 3*∫[-1,1]∫[0,1-x²]∫[0,2-z] (ydydzdx)

Integrating to y, we get y²/2 from 0 (trivial) to 2 - z, which is just (2 - z)²/2.  To integrate that, we let u = 2 - z, du = -dz, so now we have (-3/2)*stuff (we pull out the 1/2), so we have u² in there and that integrates to u³/3, or (2 - z)³/3 from 0 to 1 - x², which is just (1 + x²)³/3 - 2³/3 = (1 + x²)³/3 - 8/3, or (1/3)[(1 + x²)³ - 8], which simplifies to x⁶ + 3x⁴ + 3x² + 1 - 8, or

(-1/2)*∫[-1,1] ((x⁶ + 3x⁴ + 3x² - 7)dx)

Integrating to x gives x⁷/7 + 3x⁵/5 + x³ - 7x from -1 to 1.  That's (1/7 + 3/5 + 1 - 7) - (-1/7 - 3/5 - 1 + 7) = (2/7 + 6/5 + 2 - 14) = -368/35.  Multiply that by -1/2 from earlier, and you get 184/35.  The end.

At the end, this theorem is proved for simple solid regions, but it can also be proved for finite unions of simple solid regions (like, say, an egg within a ball, for example).

Suppose E lies between the closed surfaces S₁ and S₂, where S₁ lies inside of S₂.  Let n₁ and n₂ be outward normals of S₁ and S₂.  Then the boundary surface of E is S = S₁ unioned with S₂, and its normal n is given by n = -n₁ on S₁ and n = n₂ on S₂.  Applying the Divergence Theorem,

∫∫∫[E] (div F dV) = ∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS)
= ∫∫[S₁] (F ∙ (-n₁) dS) + ∫∫[S₂] (F ∙ n₂ dS)
= -∫∫[S₁] (F ∙ dS) + ∫∫[S₂] (F ∙ dS)

To continue, let's pull an application.  Suppose we have an electric field E(x) = εQx/|x|³, where S₁ is a small sphere with radius a and center at the origin.  div E = 0, and so we have

∫∫[S₂] (E ∙ dS) = ∫∫[S₁] (E ∙ dS) + ∫∫∫[E] (E ∙ n dS) from a rearrangement of the above mess.  The normal vector at x is x/|x|, so

E ∙ n = εQx/|x|³ . (x/|x|) = εQx/|x|⁴  ∙ x
= εQ/|x|² = εQ/a² because the equation of S₁ is simply |x| = a.  And so, we have

∫∫[S₂] (E ∙ dS) = ∫∫[S₁] (E ∙ n dS)
= (εQ/a²)*∫∫[S₁] (dS) = (εQ/a²)*A(S₁)
= (εQ/a²)4πa² = 4πεQ

So, the flux of E is 4πεQ through ANY closed surface S₂ that contains the origin.

Another application comes from fluid flow.  Let v(x,y,z) be the velocity field of a fluid with constant density ρ.  Then F = ρv is the rate of flow per unit area.  If P₀(x₀,y₀,z₀) is a point in the fluid and B_a is a ball with center P₀ and a VERY small radius a, then div F(P) ~ div F(P₀) for all points in B_a since div F is continuous.  We approximate the flux over the boundary sphere S_a as

∫∫[S_a] (F ∙ dS) = ∫∫∫[B_a] (div F dV)
~ ∫∫∫[B_a] (div F(P₀) dV)
= div F(P₀) *V(B_a), where V(B_a) is the volume of the ball.

This approximate becomes better as a -> 0, and so we can suggest that

div F(P₀) = lim (1/V(B_a))*∫∫[S_a] (F ∙ dS) as a -> 0

div F(P₀) is the net rate of outward flux per unit volume at P₀, and if div F(P) > 0, the net flow is outward near P and P is called a source.  If div F(P) < 0, the net flow is inward near P and P is called a sink.

Anyway, that's really it.  Let's do some problems!

-----

1)  Use the divergence theorem to prove that ∫∫[S] (D_nf dS) = ∫∫∫[E] (Ѳf dV), assuming that S and E satisfy the conditions of the divergence theorem and the scalar functions and components of the vector fields have continuous second order partial derivatives.

Not hard.  D_nf = Ñf ∙ n, so

∫∫[S] (D_nf dS) = ∫∫[S] (Ñf ∙ n dS) = ∫∫∫[E] (div (Ñf)dV) = ∫∫∫[E] (Ѳf dV)

2)  Use the divergence theorem to evaluate ∫∫[S] (F ∙ dS) where F(x,y,z) = x⁴i - x³z²j + 4xy²zk, where S is the surface of the solid bounded by the cylinder x² + y² = 1 and the planes z = x + 2 and z = 0.

Well, we need div F = 4x³ + 0 + 4xy², so we have div F = 4x(x² + y²).  If we look at the plane z = x + 2, we see an ellipse of x² + y² = 1, and we can convert this to cylindrical coordinates by saying that E = {(r,θ,z) | 0 < r < 1, 0 < θ < 2π, 0 < z < rcos(θ) + 2} by simply saying x = rcos(θ), y = rsin(θ), z = z, and so z = rcos(θ) + 2, x² + y² = 1 = r², r = 1.

That gives us ∫[0,2π]∫[0,1]∫[0,rcos(θ)] (4cos(θ)r²rdzdrdθ) when we convert 4x(x² + y²) into 4cos(θ)r².

So, we have to integrate 4cos(θ)r³dzdrdtheta with respect to z first.  That's just 4cos(θ)r³z from 0 (trivial) to rcos(θ), giving us 4cos²(θ)r⁴.  Integrating to r, we get (4/5)cos²(θ)r⁵ from 0 (trivial) to 1 gives us (4/5)cos²(θ).  cos²(θ) = 1/2 + cos(2θ)/2, so integrating THAT gives us θ/2 + sin(2θ)/4 from 0 (trivial) to 2π (trivial in the sine), giving us 2π/2 = π.  So, we have (4/5)*π = 4π/5.

3)  Use the divergence theorem to evaluate ∫∫[S] ((2x + 2y + z²)dS) where S is the sphere x² + y² + z² = 1.

We need to find some F ∙ n = 2x + 2y + z².  If we take the partial derivatives on S, we get 2x + 2y + 2z = 0, or x + y + z = 0, and we can turn that into our n via n = (the partial)/sqrt(sum of squares of partial), which ultimately amounts to (xi + yj + zk)/sqrt(x² + y² + z²), ahh, but we said the sphere has a radius of 1, so we can replace that bottom bit with just sqrt(1), giving us 1, so n = xi + yj + zk.

Now we have F ∙ (xi + yj + zk) = 2x + 2y + z².  Simple analysis will tell us that if we multiply the first two terms in xi + yj + zk by 2, we'd get our 2x + 2y.  Then if we multiply the last term by z, we'd get our z², so F = 2i + 2j + zk.

div F = 1, then, and if B = {(x,y,z) | x² + y² + z² < 1}, then we can easily say ∫∫[S] ((2x + 2y + z²)dS) = ∫∫∫[B] (1dV) = V(B) = (4/3)πr³ = 4π/3.

-----

Why do you care?  Well, the initial example gave us a MUCH easier way of evaluating surface integrals, for one.  The divergence theorem also allows us to analyze vector fields with more relative ease than just defining control surfaces/boundaries/volumes and integrating/evaluating from there.  You saw two applications in fluid mechanics and electricity right there.

-----

Proof of Divergence Theorem for simple solid regions.

Let F = Pi + Qj + Rk, then div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z

∫∫∫[E] (div F dV) = ∫∫∫[E] ((∂P/∂x)dV) + ∫∫∫[E] ((∂Q/∂y)dV) + ∫∫∫[E] ((∂R/∂z)dV)

If n is the unit outward normal of S, then the surface integral ∫∫[S] (F ∙ dS) is

∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS) = ∫∫[S] ((Pi + Qj + Rk) ∙ n dS)
= ∫∫[S] (Pi ∙ n dS) + ∫∫[S] (Qj ∙ n dS) + ∫∫[S] (Rk ∙ n dS)

To prove the divergence theorem, we need to prove the following three equations:
∫∫[S] (Pi ∙ n dS) = ∫∫∫[E] ((∂P/∂x)dV)
∫∫[S] (Qj ∙ n dS) = ∫∫∫[E] ((∂Q/∂y)dV)
∫∫[S] (Rk ∙ n dS) = ∫∫∫[E] ((∂R/∂z)dV)

We'll do the last one and make a note that they can all be done similarly.

We'll let E be a type 1 region and so E = {(x,y,z) | (x,y) in D, u₁(x,y) < z < u₂(x,y)}, where D is the projection of E onto the XY plane.  So

∫∫∫[E] ((∂R/∂z)dV) = ∫∫[D] (Int[u₁,u₂] ((∂R/∂z)(x,y,z)dz)  dA), and by the Fundamental Theorem of Calculus:

∫∫∫[E] ((∂R/∂z)dV) = ∫∫[D] ((R(x,y,u₂) - R(x,y,u₁))dA)

Hold onto that.  Now let's look at the other part of the equation.  The boundary surface S consists of three pieces:  bottom S₁, the top S₂, and a possibly vertical S₃ that lies above the boundary curve of D.  If, say, S were a sphere, S₃ wouldn't even really exist.

On S₃, we have k ∙ n = 0, because k is vertical and n is horizontal, and so

∫∫[S₃] (Rk ∙ n dS) = ∫∫[S₃] (0dS) = 0

And so, regardless of what surface we have, we can say, then, that

∫∫[S] (Rk ∙ n dS) = ∫∫[S₁] (Rk ∙ n dS) + ∫∫[S₂] (Rk ∙ n dS)

S₂ is z = u₂(x,y), (x,y) in D, ad the outward normal n points upward, so

∫∫[S₂] (Rk ∙ n dS) = ∫∫[D] (R(x,y,u₂)dA).

On S₁, z = u₁(x,y), (x,y) in D, but here the outward normal n points downward, so we have

∫∫[S₁] (Rk ∙ n dS) = -∫∫[D] (R(x,y,u₁)dA)

Thus,

∫∫[S] (Rk ∙ n dS) = ∫∫[D] ((R(x,y,u₂) - R(x,y,u₁))dA), which is what we had written before, and so the initial 3 equations we have sought out to prove, well, we proved one of them, and we could do this exact same sequence to prove the others by changing their region types.