| View previous topic :: View next topic |
| Author |
Message |
Sarg338
Joined: 07 Feb 2008
Posts: 5143
|
 Posted: Thu Feb 24, 2011 8:00 pm Post subject: |
 |
|
| THABEAST721 wrote: | | The next section, related rates, will become your best friend muahahaha! |
If by best friend, you mean make me contemplate suicide when doing them, then yes, it will 
I already despise word problems like this. Doing them this was just makes me hate them more. |
|
| Back to top |
|
 |
ready2rock
Joined: 25 Aug 2007
Posts: 1738
Location: somewhere in this vast universe
|
| Posted: Thu Feb 24, 2011 9:54 pm Post subject: |
|
|
In the notation, it's saying that you take the derivative times the change in that variable, so the derivative of x is 1*dx. so if you have y=x, taking the derivative of both sides gives you dy=dx. If you divide both sides by dx, you get dy/dx=1, so the derivative of x is 1. This means that whenever you take the derivative with respect to y, you can write a dy/dx after it and you don't have to write anything after the x terms. This signifies writing a dy and dx after each term and then dividing the whole thing by dx.
Now, the reason that you never needed to do it this way before is because the equations have always been solved for y for you, but, especially in story problems, things aren't always that nice. So, it's sometimes easier to take the derivative then solve for dy/dx.
Hopefully that makes sense.
_________________
 
 
REREAD YOUR POSTS BEFORE YOU SUBMIT THEM!!!
Posts that win: this, this, this, this, this, this, this, this, this, this, this, this whole thread, this, these, this, this |
|
| Back to top |
|
|
bclare
Joined: 21 Jun 2008
Posts: 6048
Location: Boston
|
| Posted: Thu Feb 24, 2011 11:39 pm Post subject: |
|
|
Hey, this came up a few pages ago. Basically implicit differentiation is just the chain rule on y. Also, here's what I said already about the Implicit Function Theorem
| bclare wrote: | Shortcut for implicit differentiation (I'm not responsible if your teacher/professor doesn't accept this as work)
The Implicit Function Theorem
Set the equation as 0 = F(x,y)
then dy/dx = -(dF/dx) / (dF/dy)
ex: sin y = cos x
sin y - cos x = 0 = F(x,y)
dF/dx = 0 + sin x
dF/dy = cos y
so dy/dx = - sin x / cos y |
_________________
I'm back I suppose |
|
| Back to top |
|
|
Sarg338
Joined: 07 Feb 2008
Posts: 5143
|
| Posted: Fri Feb 25, 2011 4:49 pm Post subject: |
|
|
I don't even know how to start a problem about Rates of Change... seriously, I look at the back of the book and have no fucking clue how they even get close to that answer. Like for this problem:
| Code: | | Sand Pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height increases at a constant rate of 5 ft/min, at what rate is sand pouring from the chute when the pile is 10 ft high |
Seriously, How do i even begin a problem like that? Still haven't seen a single practical use I will need to use for any of this. |
|
| Back to top |
|
|
ThunderShade
Joined: 12 Jun 2008
Posts: 350
Location: Scotland
|
| Posted: Fri Feb 25, 2011 5:57 pm Post subject: |
|
|
| Sarg338 wrote: | I don't even know how to start a problem about Rates of Change... seriously, I look at the back of the book and have no fucking clue how they even get close to that answer. Like for this problem:
| Code: | | Sand Pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height increases at a constant rate of 5 ft/min, at what rate is sand pouring from the chute when the pile is 10 ft high |
Seriously, How do i even begin a problem like that? Still haven't seen a single practical use I will need to use for any of this. |
Look for as much information as possible. This is what I can find.
1. The shape of the sand will always be a cone.
2. The diameter is equal to the height.
3. The height increases at a constant rate of 5 ft/min.
First off, the shape of the sand is a cone (Point 1). Therefore the volume is given by V = pi*r^2*h/3. The diameter is equal to the height (Point 2) therefore d = h. d = 2r therefore 2r = h and thus r = h/2. Substituting this into the formula for volume gives V = pi*(h^2/4)*h/3 = pi*h^3/12. The question is asking for the rate at which the sand is pouring i.e. dV/dt. dV/dt = d/dt(pi*h^3/12) = (pi/12)*d/dt(h^3) = (pi/12)*3h^2*dh/dt (chain rule). We know that dh/dt = 5 (Point 3). Substituting this in gives dV/dt = (pi/12)*3h^2*5 = (5*pi/4)*h^2. The question wants dV/dt when h = 10. This gives dV/dt = (5*pi/4)*10^2 = 125*pi.
_________________
Last edited by ThunderShade on Fri Feb 25, 2011 6:52 pm; edited 1 time in total |
|
| Back to top |
|
|
Sarg338
Joined: 07 Feb 2008
Posts: 5143
|
| Posted: Fri Feb 25, 2011 6:29 pm Post subject: |
|
|
| ThunderShade wrote: | | The question is asking for the rate at which the sand is pouring i.e. dV/dt. dV/dt = d/dt(pi*h^3/12) = (pi/12)*d/dt(h^3) = (pi/12)*3h^2*dh/dt (chain rule). |
Lost me here. Mind explaining it or something? Not sure where you got pi/12 at.
So far I'm at V(t) = pi*(h^3 / 12) |
|
| Back to top |
|
|
ThunderShade
Joined: 12 Jun 2008
Posts: 350
Location: Scotland
|
| Posted: Fri Feb 25, 2011 6:51 pm Post subject: |
|
|
| Sarg338 wrote: | Lost me here. Mind explaining it or something? Not sure where you got pi/12 at.
So far I'm at V(t) = pi*(h^3 / 12) |
Just a small bit of re-organising.
V(t) = pi*(h^3)/12
= pi*(h^3)*(1/12)
= pi*(1/12)*(h^3)
= (pi/12)*(h^3)
_________________
|
|
| Back to top |
|
|
Sarg338
Joined: 07 Feb 2008
Posts: 5143
|
| Posted: Fri Feb 25, 2011 7:44 pm Post subject: |
|
|
| ThunderShade wrote: | | Sarg338 wrote: | Lost me here. Mind explaining it or something? Not sure where you got pi/12 at.
So far I'm at V(t) = pi*(h^3 / 12) |
Just a small bit of re-organising.
V(t) = pi*(h^3)/12
= pi*(h^3)*(1/12)
= pi*(1/12)*(h^3)
= (pi/12)*(h^3) |
ah, ok gotcha.
Well, after todays class, the teacher did about 6 problems or so, and it made more sense now. Hopefully I'll be able to do the homework better now. |
|
| Back to top |
|
|
GlassDragon
Joined: 24 May 2008
Posts: 1122
|
| Posted: Mon Feb 28, 2011 2:58 am Post subject: |
|
|
Hey anyone want to help me with some statistics?
So I get that the probability distribution function of a random variable X is:
f(x) = (3/16)x^2, for -c < x < c.
And I have to find c. So I integrate f(x) and get (1/16)x^3 taken from c to -c, which has to equal 1. So I get:
(1/16)c^3 - (1/16)(-c)^3 = (1/16)c^3 + (1/16)c^3 = (1/8)c^3
Set that equal to 1 and I get c = 2.
Am I correct in saying the distribution function is (1/8)x^3 for -2 < x < 2?
That would mean that the distribution function is negative for a negative x. Is that possible or did I do something wrong? I don't think probabilities are supposed to be negative.
_________________
|
|
| Back to top |
|
|
bclare
Joined: 21 Jun 2008
Posts: 6048
Location: Boston
|
| Posted: Mon Feb 28, 2011 7:53 am Post subject: |
|
|
Okay, let's take this one step at a time.
| GlassDragon wrote: | Hey anyone want to help me with some statistics?
So I get that the probability distribution function of a random variable X is:
f(x) = (3/16)x^2, for -c < x < c.
And I have to find c. So I integrate f(x) and get (1/16)x^3 taken from c to -c, which has to equal 1. So I get:
(1/16)c^3 - (1/16)(-c)^3 = (1/16)c^3 + (1/16)c^3 = (1/8)c^3
Set that equal to 1 and I get c = 2. |
Yep, sounds right.
| Quote: | | Am I correct in saying the distribution function is (1/8)x^3 for -2 < x < 2? |
Whoa whoa whoa. Let's just plug C = 2 into your original equation (don't want to sound patronizing, just emphasizing that)
| Quote: | the probability distribution function of a random variable X is:
f(x) = (3/16)x^2, for -c < x < c. |
and we get
f(x) = (3/16)x^2, for -2 < x < 2
Because of the x^2 term, f(x) will always be positive, you can check that f(x) < 1, and the integral showed that the sum of probabilities = 1, so we're all set here.
_________________
I'm back I suppose |
|
| Back to top |
|
|
Sarg338
Joined: 07 Feb 2008
Posts: 5143
|
| Posted: Mon Feb 28, 2011 7:11 pm Post subject: |
|
|
Well, I can safely say related rates are getting tons easier to do and understand right now. I was able to do 8/10 problems on the homework (the 2 I couldn't figure out were also the 2 most of the class asked about as well, so that makes me feel a bit better). So I'm definitely getting it down, so thanks to everyone for helping me out with this.
It seems like every new math section goes like that. Learn it in class, try to do homework, can't, rage, then learn it right when we start a new section.
I also had the brief thought to have a minor in mathematics, as I thought that would look really good next to my computer science major. Not sure if I fully want to make that jump yet though. |
|
| Back to top |
|
|
killzonedout
Joined: 12 Jul 2008
Posts: 3061
|
| Posted: Tue Mar 01, 2011 1:23 am Post subject: |
|
|
Kicking into infinite sequences and series, and damn... Calculus 1 and 2 made sense pretty much up until here. I don't even know where to start asking questions. :P
At least this is the last unit for the class, thank god.
_________________
twitter dot com |
|
| Back to top |
|
|
bclare
Joined: 21 Jun 2008
Posts: 6048
Location: Boston
|
| Posted: Tue Mar 01, 2011 1:27 am Post subject: |
|
|
| Sarg338 wrote: | | I also had the brief thought to have a minor in mathematics, as I thought that would look really good next to my computer science major. Not sure if I fully want to make that jump yet though. |
Depending on the school there's probably quite a bit of overlap between required classes, or at least electives, so that might not even be a whole lot more work to add that minor. I went to Boston University, and I know besides a lot of elective overlap they even have a Math/CS joint major (like a double major but it's only one major with courses in both subjects)
_________________
I'm back I suppose |
|
| Back to top |
|
|
Sarg338
Joined: 07 Feb 2008
Posts: 5143
|
| Posted: Tue Mar 01, 2011 1:57 am Post subject: |
|
|
| bclare wrote: | | Sarg338 wrote: | | I also had the brief thought to have a minor in mathematics, as I thought that would look really good next to my computer science major. Not sure if I fully want to make that jump yet though. |
Depending on the school there's probably quite a bit of overlap between required classes, or at least electives, so that might not even be a whole lot more work to add that minor. I went to Boston University, and I know besides a lot of elective overlap they even have a Math/CS joint major (like a double major but it's only one major with courses in both subjects) |
Well, these are the math courses I would need for Comp Sci.
| Code: | 201. Analytics and Calculus I (5)
251. Analytics and Calculus II (5)
260. Discrete Mathematical Structures (3)
313. Linear Algebra (3)
318. Probability (3) |
Can't seem to find any classes for Math Minors, only Majors. |
|
| Back to top |
|
|
THABEAST721
Joined: 26 Dec 2007
Posts: 2000
|
| Posted: Tue Mar 01, 2011 3:41 am Post subject: |
|
|
| Sarg338 wrote: | | bclare wrote: | | Sarg338 wrote: | | I also had the brief thought to have a minor in mathematics, as I thought that would look really good next to my computer science major. Not sure if I fully want to make that jump yet though. |
Depending on the school there's probably quite a bit of overlap between required classes, or at least electives, so that might not even be a whole lot more work to add that minor. I went to Boston University, and I know besides a lot of elective overlap they even have a Math/CS joint major (like a double major but it's only one major with courses in both subjects) |
Well, these are the math courses I would need for Comp Sci.
| Code: | 201. Analytics and Calculus I (5)
251. Analytics and Calculus II (5)
260. Discrete Mathematical Structures (3)
313. Linear Algebra (3)
318. Probability (3) |
Can't seem to find any classes for Math Minors, only Majors. |
Well chances are it won't be the same as WVU, but to get a minor in math up here, you need like through calculus 4, a class that like introduces you to proofs, and then 4 300 or higher level classes, and then either stats or discrete. That's not exactly what it is, but it's about what it is.
_________________
This does not leave my sig!!! |
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
|
Copyright © 2006-2024 ScoreHero, LLC
|
Powered by phpBB
|
![Inbox [ Login ]](templates/subSilver/images/icon_mini_message.gif){kind=icon}
Inbox [ Login ]
Search
Memberlist
Profile
Log in
