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[thesaucetastic]

From wikipedia it looks like u_t + J_x = 0 is conservation of mass. If you assume v does not change with x, you get an equation like yours above.

We assume v is constant in this context. The problem I have with that equation is that I don't know how to get to that from u_t(x,t) = -Du_xx(x,t) + vu_x(x,t). I know J is supposed to be the -Du_xx(x,t) + vu_x(x,t), because then if you add it to both sides you get the u_t = alpha^2*u_xx - vu_x equation. But how do I get there?

Just for reference, here's the chapter if that helps at all (Farlow's lessons are mostly self-contained, so you don't have to skim through anything but lesson 4).

edit: I guess the real question here is why is J_x = -Du_xx(x,t) + vu_x(x,t). I don't really get that

 

[simplayer]

edit: I guess the real question here is why is J_x = -Du_xx(x,t) + vu_x(x,t). I don't really get that

J is the flux. As per your Fick's law, the flux due to diffusion is -Du_x(x,t). The flux due to the convection would simply be vu(x,t). Thus the total flux would be -Du_x(x,t) + vu(x,t). Since u_t(x,t) + J_x(x,t) = 0, we get:
u_t(x,t) + (-Du_xx(x,t) + vu_x(x,t)) = 0, or, u_t(x,t) = Du_xx(x,t) - vu_x(x,t) where we assume v is constant over x.

 

[thesaucetastic]

J is the flux. As per your Fick's law, the flux due to diffusion is -Du_x(x,t). The flux due to the convection would simply be vu(x,t). Thus the total flux would be -Du_x(x,t) + vu(x,t). Since u_t(x,t) + J_x(x,t) = 0, we get:
u_t(x,t) + (-Du_xx(x,t) + vu_x(x,t)) = 0, or, u_t(x,t) = Du_xx(x,t) - vu_x(x,t) where we assume v is constant over x.

Yeah, that seems obvious to me now that I've finally gotten some sleep. I think what was tripping me up before was the solution manual, which was saying that the derivation was similar to the heat equation derivation (a really unhelpful nonanswer if you ask me).

Thanks for answering my dumb questions all the time.

 

[triplestation]

hey guys

im doing some college algebra/trig homework and im getting a nice kick out of it

i know this is baby stuff but im going into computer science soon and after research i found it involves discrete math rather than continuous

i know that ill have to take calculus but afterwards can i expect discrete math to be easier than the higher level continuous stuff?

thanks

 

[cpp_is_king]

hey guys

im doing some college algebra/trig homework and im getting a nice kick out of it

i know this is baby stuff but im going into computer science soon and after research i found it involves discrete math rather than continuous

i know that ill have to take calculus but afterwards can i expect discrete math to be easier than the higher level continuous stuff?

thanks

Most people consider Discrete Math to be easier than Calculus / Analysis, yes. Of course nobody can guarantee that the same will be true for you, but it is pretty common.

 

[Memento_Mori15]

Stupid question but I can't remember atm. Two matrices are given, A^-1=[..] B^-1=[..].
I need to find (A*B)^-1.

First I would find A and B and then multiply it and then take the inverse? I remember (A*B)^T=B*A. There's no special condition for inverses, right? I've already completely forgotten about linear algebra. :\

 

[harriet the spy]

Stupid question but I can't remember atm. Two matrices are given, A^-1=[..] B^-1=[..].
I need to find (A*B)^-1.

First I would find A and B and then multiply it and then take the inverse? I remember (A*B)^T=B*A. There's no special condition for inverses, right? I've already completely forgotten about linear algebra. :\

Well, recall that the inverse of a matrix M is a matrix M^-1 such that M . M^(-1) = I

You want to find the inverse of (A.B). You can guess the solution X=B^-1.A^-1 (note the order matters, since matrix product is not commutative), and verify it is the solution:

(A.B).(B^-1 A^-1)=A (B B^-1) A^-1 = A I A^-1 (since B B^-1 =I) = A A^-1 = I

Therefore proving that (A.B)^-1 = B^-1 A^-1

In general, inverse of 'sequences' show as sequence of inverses in the reverse order (which makes sense if you think about it intuitively: the inverse of putting your pants, then your shoes, is to remove your shoes first, then your pants).

 

[triplestation]

Most people consider Discrete Math to be easier than Calculus / Analysis, yes. Of course nobody can guarantee that the same will be true for you, but it is pretty common.

thank you for your reply

i hated doing math before and now im enjoying it, its really weird

 

[Memento_Mori15]

Well, recall that the inverse of a matrix M is a matrix M^-1 such that M . M^(-1) = I

You want to find the inverse of (A.B). You can guess the solution X=B^-1.A^-1 (note the order matters, since matrix product is not commutative), and verify it is the solution:

(A.B).(B^-1 A^-1)=A (B B^-1) A^-1 = A I A^-1 (since B B^-1 =I) = A A^-1 = I

Therefore proving that (A.B)^-1 = B^-1 A^-1

In general, inverse of 'sequences' show as sequence of inverses in the reverse order (which makes sense if you think about it intuitively: the inverse of putting your pants, then your shoes, is to remove your shoes first, then your pants).

Thank you! I have no idea where I kept my linear algebra notes when I took it in the summer which is odd since I usually keep all math related notes.

 

[Lonely1]

Most people consider Discrete Math to be easier than Calculus / Analysis, yes. Of course nobody can guarantee that the same will be true for you, but it is pretty common.

I would say that higher level discrete is harder, though.

 

[Epsilon-delta]

thank you for your reply

i hated doing math before and now im enjoying it, its really weird

People start enjoying math once they get over the psychological barrier of 'math is too hard.'

I thought so too at one time but the anxiety went away once I started to learn math on my own. I taught myself advanced algebra, trig, and calculus through self-study and with the help of sites like Khan Academy and YouTube videos of MIT lectures on calculus.

 

[cpp_is_king]

I would say that higher level discrete is harder, though.

At that point I call it combinatorics though, and yes that can be extremely difficult. You won't need any of the hard stuff for a regular CS degree.

 

[VegiHam]

So, can anyone help me with an ordinary differential equation? I'm supposed to find a general solution to :

2x(y^2) dy/dx = (y^2)x + 3(x^2)y + 2y^3

and I'm totally stumped. I think it's supposed to be homogeneous or something, but I can't find anyway to rearrange it where I don't get stuck with like a half left over; so I can't get a substitution where I can move all the x to one side and all the substitution to the other to integrate? Am I looking at it totally wrong or something?

Sorry this is kinda simple compared to some of the stuff in here but I'm pretty bad at maths. Any help would be appreciated.

 

[luoapp]

So, can anyone help me with an ordinary differential equation? I'm supposed to find a general solution to :

2x(y^2) dy/dx = (y^2)x + 3(x^2)y + 2y^3

and I'm totally stumped. I think it's supposed to be homogeneous or something, but I can't find anyway to rearrange it where I don't get stuck with like a half left over; so I can't get a substitution where I can move all the x to one side and all the substitution to the other to integrate? Am I looking at it totally wrong or something?

Sorry this is kinda simple compared to some of the stuff in here but I'm pretty bad at maths. Any help would be appreciated.

My ODE is rusty, but doesn't it look like this?

,

solution

From wikipedia, http://en.wikipedia.org/wiki/Ordinary_differential_equation

 

[The Lamp]

So, can anyone help me with an ordinary differential equation? I'm supposed to find a general solution to :

2x(y^2) dy/dx = (y^2)x + 3(x^2)y + 2y^3

and I'm totally stumped. I think it's supposed to be homogeneous or something, but I can't find anyway to rearrange it where I don't get stuck with like a half left over; so I can't get a substitution where I can move all the x to one side and all the substitution to the other to integrate? Am I looking at it totally wrong or something?

Sorry this is kinda simple compared to some of the stuff in here but I'm pretty bad at maths. Any help would be appreciated.

one: if you're taking differential equations, you're not "bad at maths" lol.

also, I don't think this equation can be solved by just isolating the two variables on each side and integrating.

 

[VegiHam]

My ODE is rusty, but doesn't it look like this?

,

solution

From wikipedia, http://en.wikipedia.org/wiki/Ordinary_differential_equation

Yeah, I think I'm aiming for something like that, but I can't express the right-hand side as a function of (x/y), cus one of the terms cancels to leave a half.

one: if you're taking differential equations, you're not "bad at maths" lol.

also, I don't think this equation can be solved by just isolating the two variables on each side and integrating.

You're right, sorry, I'm just frustrated. And the variables I'm trying to separate are x and v; which is a substitution for (y/x); but I'm failing.

 

[luoapp]

Yeah, I think I'm aiming for something like that, but I can't express the right-hand side as a function of (x/y), cus one of the terms cancels to leave a half.

2x(y^2) dy/dx = (y^2)x + 3(x^2)y + 2y^3

dy/dx=1/2 + 3x/(2y) + y/x = 1/2 + 3/2u + u            (u=y/x)

Am I missing something here?

 

[VegiHam]

2x(y^2) dy/dx = (y^2)x + 3(x^2)y + 2y^3

dy/dx=1/2 + 3x/(2y) + y/x = 1/2 + 3/2u + u            (u=y/x)

Am I missing something here?

Nope, that's what I have too. I just can't do what I want cus of that half...

Edit: WAIT can I just divide the whole of the other side by that function?

 

[luoapp]

Nope, that's what I have too. I just can't do what I want cus of that half...

Edit: WAIT can I just divide the whole of the other side by that function?

where F(u) = 1/2 + 3/2u + u

 

[VegiHam]

where F(u) = 1/2 + 3/2u + u

OK, I can see where ln(cx) is coming from but I don't get the other side. Since I was never taught that formula I'm assuming I'm not supposed to use it.

 

[luoapp]

OK, I can see where ln(cx) is coming from but I don't get the other side. Since I was never taught that formula I'm assuming I'm not supposed to use it.

it's in the wiki, "Set y = ux, then solve by separation of variables in u and x."

d(ux)/dx=u+du/dx=F(u)

dx/du=1/(F-u)

...

 

[VegiHam]

it's in the wiki, "Set y = ux, then solve by separation of variables in u and x."

d(ux)/dx=u+du/dx=F(u)

dx/du=1/(F-u)

...

OK, thanks! Except now that I've done that I have xln Cx = 2y + (y^2)/3x; so I can't solve it for y. Have I gone wrong somewhere?

 

[simplayer]

OK, thanks! Except now that I've done that I have xln Cx = 2y + (y^2)/3x; so I can't solve it for y. Have I gone wrong somewhere?

Both sides should have a ln. You should have something like ln(Cx) = (2/3)*ln(3/2u+1/2)

Ln's should cancel out (barring some constant).

 

[VegiHam]

Both sides should have a ln. You should have something like ln(Cx) = (2/3)*ln(3/2u+1/2)

Ln's should cancel out (barring some constant).

Hey, you're right! I forgot that 1/some stuff added isn't the same as 1/stuff + 1/over stuff hahaha being sleepy makes me foolish I guess.

Edit: think I have it solved: y=((2(cx)^5/3) -x)/3. I'll check it again when I wake up. Thanks for the help everyone, you guys are awesome!

 

[luoapp]

Hey, you're right! I forgot that 1/some stuff added isn't the same as 1/stuff + 1/over stuff hahaha being sleepy makes me foolish I guess.

No offense, I actually pictured you as your avatar.

 

[VegiHam]

No offense, I actually pictured you as your avatar.

Haha, none taken, that's basically the face I was making.

 

[Mabef]

I have a weird theoretical question. I came across it while programming something, and though it was easy to find a way around by adding some extra rules, I wondered if there's a word for it in math-land.

Basically, I want to assign numbers to a looping pattern so I may easily compare proximities.

So as an example, there ^^ are 6 colors I care about. I want a rule that can solve which colors are closest to color x using the simplest math possible. Also, I can assign whatever number I want to each color (E.g. Yellow is 0, Orange is 2, Red is 5...).


Is there a word for this kind of circular math gunk idea?

 

[Mabef]

Basically, I want to assign numbers to a looping pattern so I may easily compare proximities.
[...]
Is there a word for this kind of circular math gunk idea?

Is the answer... circles? Ugh my weird example may have solved my question.

For the color-wheel example I gave, I knew I could do some geometry to find the distance between points on a circle. But that solution requires there being a circle; the problem as I first came upon it was just a looping pattern (like "a b c d e a b c d e a "). So is the solution to just create a circle out of the pattern, so I may compare points? I guess so. Maybe, I don't know what I'm asking really. Blechchhch

 

[luoapp]

Is the answer... circles? Ugh my weird example may have solved my question.

For the color-wheel example I gave, I knew I could do some geometry to find the distance between points on a circle. But that solution requires there being a circle; the problem as I first came upon it was just a looping pattern (like "a b c d e a b c d e a "). So is the solution to just create a circle out of the pattern, so I may compare points? I guess so. Maybe, I don't know what I'm asking really. Blechchhch

modular?

 

[simplayer]

Is the answer... circles? Ugh my weird example may have solved my question.

For the color-wheel example I gave, I knew I could do some geometry to find the distance between points on a circle. But that solution requires there being a circle; the problem as I first came upon it was just a looping pattern (like "a b c d e a b c d e a "). So is the solution to just create a circle out of the pattern, so I may compare points? I guess so. Maybe, I don't know what I'm asking really. Blechchhch

Yes, use modulo.

Using your 10 wheel example, say x is 2 and you want to find the colours that are 3 spots away. So you'll calculate (x+3)mod10 and (x-3)mod10. With the first you'll get:
5mod10 = 5
With the second you'll get:
-1mod10=9

So your mod term will be the size of the system (10 colours, mod 10, 6 colours, mod 6). You'll want to start your numbering system at 0 instead of 1.

 

[Kieli]

I have two more questions to ask.

1) Suppose your given a position function r(t) for a spaceship. The loading station for the spaceship is located at point <x, y, z>. The question is asking at what time the captain should shut off the engine to float into the space station.

I don't need a solution to this problem (hence why I did not provide any values or functions). To do this question, I essentially have to find a time such that the velocity function is parallel to the vector from the origin to the loading station AND the position of the spaceship lies along the vector from the origin to the loading station?

2) For the second question, there is a river that runs straight from West to East. It is "y" meters wide. The speed of the water along the river can be described by a function f(x) where "x" represents the distance in meters you are from the north bank (your starting point). This means the speed of the water is variable. You are a swimmer wanting to swim straight downwards. How many points east of your starting point will you end up once you reach the south bank of the river.

To do this question, could I find the average value of the function f(x) and then multiply it by the time that it takes for the person to travel across the river? Even though the river has variable speeds depending on how far you are from the north bank, I could just treat the river as having one constant speed (which simplifies the problem down to high-school level physics).

Both these questions are copyright of the Steward textbook (so please don't sue me).

 

[luoapp]

I have two more questions to ask.

1) ...To do this question, I essentially have to find a time such that the velocity function is parallel to the vector from the origin to the loading station AND the position of the spaceship lies along the vector from the origin to the loading station?

Not necessarily the origin, but some point on the trajectory.

2) ...To do this question, could I find the average value of the function f(x) and then multiply it by the time that it takes for the person to travel across the river? Even though the river has variable speeds depending on how far you are from the north bank, I could just treat the river as having one constant speed (which simplifies the problem down to high-school level physics).

Yes, but how can you find the average?

 

[Kimosabae]

Hi, trying to understand some things posited in a book I'm currently re-reading called "Love and Math" by Edward Frenkel.

I'm having trouble completely understanding his articulation of Modulo Arithemetic and subsequently, Modulo Primes.


I understand the notion, fundamentally, but after using the simple example of reading clocks (Modulo 12) I get confused when he suggests that:

"Likewise, we can do addition modulo any natural number N. Consider the set of all consecutive whole numbers between 0 and N-1,

{0,1,2,....., N - 2, N - 1}

If N=12, this is the set of all possible hours. In general, the role of 12 is played by Number N, so that it's not 12 that takes us back to 0, but N"



Where does "N -2" come in in regards to the set of integers that are infinite? When do the whole numbers stop and then pick up with "N - 2"? I don't get it.

 

[Partial Gamification]

Hi, trying to understand some things posited in a book I'm currently re-reading called "Love and Math" by Edward Frenkel.

I'm having trouble completely understanding his articulation of Modulo Arithemetic and subsequently, Modulo Primes.


I understand the notion, fundamentally, but after using the simple example of reading clocks (Modulo 12) I get confused when he suggests that:

"Likewise, we can do addition modulo any natural number N. Consider the set of all consecutive whole numbers between 0 and N-1,

{0,1,2,....., N - 2, N - 1}

If N=12, this is the set of all possible hours. In general, the role of 12 is played by Number N, so that it's not 12 that takes us back to 0, but N"



Where does "N -2" come in in regards to the set of integers that are infinite? When do the whole numbers stop and then pick up with "N - 2"? I don't get it.

N is the modulo, it would be finite. Any multiple of 12 will be equivalent to 0 mod(12).
The whole numbers don't stop, N is one of those positive integers.

"Likewise, we can do addition modulo any natural number N. Consider the set of all consecutive whole numbers between 0 and N-1,

{0,1,2,....., N - 2, N - 1}

He's just saying pick an N bigger than five (or five) that is a natural number.
Say N is six, then the set is { 0 ,1 ,2 ,3, 4, 5 }

Think of the division having a remainder (or not); with fractions:

7/6 = 1 + 1/6; 7 is equivalent to 1 mod(6)
10/6 = 1 + 4/6; 10 is equivalent to 4 mod(6)

61/6 = 10 + 1/6; 61 is equivalent to 1 mod(6)
70/6 = 11 + 4/6; 70 is equivalent to 4 mod(6)

6/6 = 1 + 0/6; 6 is equivalent to 0 mod(6)
600/6 = 100 + 0/6; 600 is equivalent to 0 mod(6)


Also, { 1, 2, 3, 4, ... } is the infinte set of Natural (Whole) numbers;
but {1, 2, 3, ..., N - 1, N } is the set from 1 to N, a finite set, N bigger than 5 (or five) b/c N-1 is listed too.
edit: its really a notation thing but if N= 4, then N-1= 3 and the set would look like {1,2,3,3,4}, just from how its written in the "N-notation."

 

[Kimosabae]

I'm sorry, Gam, I still don't get it. Your examples I understand just fine (or maybe I don't). I still don't get what Frenkel is presenting.

"He's just saying pick an N bigger than five (or five) that is a natural number.
Say N is six, then the set is { 0 ,1 ,2 ,3, 4, 5 }"

But he states "if N=12"

12 - 1 = 11. 12 - 2 = 10.

{0,1,2,....., N - 2, N - 1}

Where are you getting 5 from? What am I missing?

I really appreciate the help btw

 

[Partial Gamification]

I'm sorry, Gam, I still don't get it. Your examples I understand just fine (or maybe I don't). I still don't get what Frenkel is presenting.



But he states "if N=12"

12 - 1 = 11. 12 - 2 = 10.



Where are you getting 5 from? What am I missing?

I really appreciate the help btw

N can be any Natural (Whole) Number. Twelve is an example, I chose six.

If N = 12
then { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 } is the set of modulo 12.
The clock just references the circular nature of the modulo arithmetic.

1 is equivalent to 1 mod(12)
2 " " 2 mod(12)
3 " " 3 mod(12)
...
10 is equivalent to 10 mod(12)
11 is equvalent to 11 mod(12)
12 is equivalent to 0 mod(12)
...
24 is equivalent to 0 mod(12) [2*12=24]
...
36 is equivalent to 0 mod(12) [3*12=36]
37 is equivalent to 1 mod(12)
38 " " 2 mod(12)
39 " " 3 mod(12)

The clock just serves a visual of the cyclical nature of the way this works, except twelve is zero. You could have any number of divisions of a circle, 360 degrees for example but it can be arbitrary.


If I write {0, 1,2,3, ..., M-2, M-1}, that's the set from zero to M-1. M can really be anything but here it should be greater than six as to not have repeat elements in this set. The first few elements are listed to show the pattern of the set, some sets are too large and too random to list like that but this is easy enough to recognize the next element is one lager than its previous.

edit: five just came as to not have a repeated element, so M-2 is "4" or more because "3" is listed in the set already. Eh, that seems confusing...

He's just saying pick an N bigger than five (or five) that is a natural number.
Say N is six, then the set is { 0 ,1 ,2 ,3, 4, 5 }

Just from the way it was written, {1,2,3, ..., N-2, N-1}
You want N-2 > 3
so N-2+2 > 3+2
N > 5

 

[Treefingers]

I'm sorry, Gam, I still don't get it. Your examples I understand just fine (or maybe I don't). I still don't get what Frenkel is presenting.



But he states "if N=12"

12 - 1 = 11. 12 - 2 = 10.



Where are you getting 5 from? What am I missing?

I really appreciate the help btw

{0, 1, 2, ..., N-2, N-1} is just the set of integers from 0 to N-1. Is the .... throwing you off? it's just notation that's used so as to not have to write out all of the numbers in the set. The ... doesn't mean include all the integers in there, just the ones in between 2 and N-2.

But regardless of notation it all means the same, the set of integers from 0 to N-1.

So N-1, N-2, N-3, etc are included in the set until you get to 0.

If N = 12, then the set consists of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. It would usually be written as {1, 2, ..., 10, 11} or something similar just so you don't have to write {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.

 

[Kimosabae]

Treefingers:

I get set notation.

{0, 1, 2, ..., N-2, N-1}

So if the N = 6 here, this would look like:

{0,1,2,3, N - 2 (4), N - 1(5)}

Correct?

 

[Partial Gamification]

Treefingers:

I get set notation.

{0, 1, 2, ..., N-2, N-1}

So if the N = 6 here, this would look like:

{0,1,2,3, N - 2 (4), N - 1(5)}

Correct?

Yes, but technically its written wrong; ( N-2(4) = N-8 ) and ( N-1(5) = N-5 )
I see what you mean; (N=6) => (N-2 = 4) and (N-1=5).

Where does "N -2" come in in regards to the set of integers that are infinite? When do the whole numbers stop and then pick up with "N - 2"? I don't get it.

I understand the notion, fundamentally, but after using the simple example of reading clocks (Modulo 12) I get confused when he suggests that:

"Likewise, we can do addition modulo any natural number N. Consider the set of all consecutive whole numbers between 0 and N-1,

You just pick an N, N-1 is one less than N.
If N = 10503, then the set {0,1,2, ..., N-2, N-1} is {0,1,2, ..., 10501, 10502}

N can be any number; well almost, you don't want to see repeated elements or a funky ordering. Like if N =1 then the set {0,1,2,..., N-2, N-1} is {0,1,2, -1, 0}. N has to be 'so big' in order for the set to be ascending and without repeated elements.

The original set was {0,1,2, ..., N-2, N-1}
N-2 has to be greater than or equal to three.
N-2 >= 3
add two to both sides:
N >= 5
That's where the five comes from, its just a result form how the set was written.

Apologies for making this more confusing than it needs to be. If you still have a question, restate it and I'm sure a more concise answer will arise.

 

[Kimosabae]

You just pick an N, N-1 is one less than N.
If N = 10503, then the set {0,1,2, ..., N-2, N-1} is {0,1,2, ..., 10501, 10502}


Apologies for making this more confusing than it needs to be. If you still have a question, restate it and I'm sure a more concise answer will arise.

I'm pretty sure I get it now. Seeing it demonstrated this way makes it clearer to me that it is a finite set of numbers in respect to the modulo (as you mentioned before) and not some arbitrary list of numbers that magically convert to some specific modulo reference at some arbitrary point on an infinite number line (or something).

And I appreciate the patience and thoroughness being displayed. I'm working diligently to improve an innumeracy that stems from my childhood, so I'm a bit slow to pick up on fundamental mathematical concepts. No need to apologize as I'm easily confused.

 

[Buggy Loop]

nvm

 

[Buggy Loop]

Wow, i think i need an update in geometry. I have the solution but i dont understand it.

In a physics problem i have this kind of triangle at one point. I have this info.

AE = 1.414m
angle B in BAC = 25 degrees
angle B in the BEC triangle is 45 degrees.

They found the BE dimension (altitude) by doing this.

AE cot ( 45 + 25 ) = AE tan 20 = 0.515m

But how i really dont understand the bolded part.

Can someone explain to me please?

 

[battousai11]

so you'll have a 90-70-20 triangle if you draw the altitude (EBA) so to find BE while knowing AE, the quickest way would be tangent of 70 (opposite/adjacent)

so you have tan(70) = AE/BE;
BE = AE*cot(70) = 0.515

Alternatively, you can do the tangent of 20 (opposite/adjacent)
tan(20) = BE/AE;
BE = AE*tan(20) = 0.515

Hope that helps!

 

[Buggy Loop]

so you'll have a 90-70-20 triangle if you draw the altitude (EBA) so to find BE while knowing AE, the quickest way would be tangent of 70 (opposite/adjacent)

so you have tan(70) = AE/BE;
BE = AE*cot(70) = 0.515

Alternatively, you can do the tangent of 20 (opposite/adjacent)
tan(20) = BE/AE;
BE = AE*tan(20) = 0.515

Hope that helps!

Oh yes that helps. Damn i feel dumb. My mind is so rusted with geometry lol. Thanks a lot!

 

[Tater Tot]

Calc 2 question. Consider the region bounded by the curve e^(-x), y = 0, x = 0, and x = -1. Calculate the volume of the solid obtained by rotating the bounded region about the x-axis.


So I know V=(pi)(r)^2
I let r = e^(-x) it becomes e^(-2x)
pi&#8747; e^(-2x) u=-2x (-1/2)du=-2
(-1/2)pi&#8747; e^(-2x) from x = 0, and x = -1

(-1/2)pi[1-e^(3)] = -pi/2+((pie^(3))/(2))

Does this seem right?

 

[DEATH™]

Calc 2 question. Consider the region bounded by the curve e^(-x), y = 0, x = 0, and x = -1. Calculate the volume of the solid obtained by rotating the bounded region about the x-axis.


So I know V=(pi)(r)^2
I let r = e^(-x) it becomes e^(-2x)
pi&#8747; e^(-2x) u=-2x (-1/2)du=-2
(-1/2)pi&#8747; e^(-2x) from x = 0, and x = -1

(-1/2)pi[1-e^(3)] = -pi/2+((pie^(3))/(2))

Does this seem right?

I'm a bit rusty at my calculus, but there's something off with that... you did a u substitution and you didn't change your bounds from x to u... Let me review real quick about that but I gotta review about other things too...


-----------

Help guys... I'm kinda lost about node voltage analysis, particularly supernodes... I am confused on what values should I put in my matrix to solve since I seemed getting dependent equations on that...

 

[CO_Andy]

The answer is 4, but i'd like to know how this is the answer. Wolfram Alpha isn't helping here

 

[YianGaruga]

The answer is 4, but i'd like to know how this is the answer. Wolfram Alpha isn't helping here

If you insert the function f(x) =x^4 into the definition of m_tan you get

m_tan = ((x+h)^4 - x^4)/h. Using the binomial theorem on (x+h)^4 gives you x^4 + 4hx^3 + 6 h^2x^2 +4h^3x +h^4.
The rest is in spoilers if you want just a hint and not the full solution

Spoiler

Putting that back into the definition gives you 4x^3 + 6hx^2 +4h^2x. Letting h->0 gives you just 4x^3. Evaluating it at P(1,1) gives you 4. I love explaining derivatives with examples like this because it's relatively easy to see what happens.

 

[About to go HAM]

Hey Gaf,

I'm a senior in chemistry. With research and a full credit load, I'm struggling this quarter to get stuff done.

Linear algebra is getting pushed to the back and I'm suffering because of it.

Can you help me understand what I need to do for this linear algebra assignment? I'm kind of in a panic over this. Its due next week, but this reads like gibberish to me...

http://faculty.wwu.edu/benyia/math204/assignment1math204winter2014.pdf

 

[harriet the spy]

Hey Gaf,

I'm a senior in chemistry. With research and a full credit load, I'm struggling this quarter to get stuff done.

Linear algebra is getting pushed to the back and I'm suffering because of it.

Can you help me understand what I need to do for this linear algebra assignment? I'm kind of in a panic over this. Its due next week, but this reads like gibberish to me...

http://faculty.wwu.edu/benyia/math204/assignment1math204winter2014.pdf

I'd like to help you, but barring giving you all the answers, I am not sure where to start. The good thing about linear algebra is that most of it derives from very few principles. If you can learn those principles and at least get some intuition behind them, things should become easy.

I'll say that much: unless my eyes are broken, something is either wrong or missing in problem 4.