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The Math Help Thread
- Thread starter -COOLIO-
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Just finished my calculus 2: trig substitution test. Never felt so useless in my life, I know there is something to learn from this though. To all the members who got good grades in Calc 2 could you please post some tips on what you did to be successful, I would greatly appreciate it.
I'm still confused but I don't think you can square the entire quantity together like that.
because 1 and y are the two radii. in your work you've said that there is one radius that is 1-y.
I think that is a different function if you use that as the radius. It has a negative slope.
Actually, I'm hopelessly confused, because I rotated the same region about the x-axis and got pi/3.
I really don't know what's wrong with this problem.
You set it up wrong. Washer method is pi(R^2-r^2)dy. In this instance R=1 and r=y.
y=1 is already in terms of x, it means that for any x value y will be 1. So R=1 and r=4sqrtx. The volume would be the integral of pi(1-16x)dx. Lower limit of integration is 0, while the upper is when 1 equals 4sqrtx, so 1/16
Rotating an area around the y-axis isn't necessarily going to give you the same volume as when you rotate it around the x-axis. The volume when rotating around the y-axis is 2pi/3 and the method you used is the correct one.
Partial Gamification
Banned
No it doesn't represent the graph you are right. Its the difference of each radius squared individually, I had it wrong. I was essentially squaring one value and woudl have ended up with a cone. It makes sense to have to get each circle's area and then subtract them in order to get thee area of the washer. Ring is bad to use too.
Sorry to keep blowing up the thread, but there was a notation problem. I actually meant that the function is y = x^(1/4)
The answer in the back of the book is pi/3.
So pi*int(1)^2 - pi*int(x^1/4)^2
pi*int(1) - pi*int(x^(1/16))
pi*(x - 16x^(17/16)/17) from 0 to 1
pi* (1 - 16/17)
pi * ( 17/17 - 16/17)
pi/17
???
fireside said:
Now that I hear someone else say it, it seems obvious that you won't necessarily get the same area. Thanks.
(x^1/4)^2 = x^(1/2).
A graph allways helps me.
So each disc's[ reallly, each ring, its like a stack of records with a cone-shape missing from the middle] area is pi*radius^2 and that's what you are integrating.
Now I am confused. Isn't the volume just cylinder's minus the cone's, which is pi-1/3 pi = 2/3 pi?
Memento_Mori15
Member
Having a huge brain fart here. I'm trying to see how they derived this in a bucket sort proof:
Expectation of[(n_i)^2]
=summation of j=1 to n (1/n) + summation of 1<j<n summation of 1<k<n where k doesn't equal k (1/n^2)
=n * 1/n + n(n-1) 1/n^2
How did they get the n and the n(n-1)?
Expectation of[(n_i)^2]
=summation of j=1 to n (1/n) + summation of 1<j<n summation of 1<k<n where k doesn't equal k (1/n^2)
=n * 1/n + n(n-1) 1/n^2
How did they get the n and the n(n-1)?
In the first summation you have n '1/n's. Ie, you're summing '1/n' n times. For the second term you are likewise getting n times the sum I=1...n (bar k) of '1/n^2'. When i=k we skip that one term, so you end up with (n-1) '1/n^2'. So in total you'll have n(n-1) '1/n^2'
Make sense?
Memento_Mori15
Member
Yes, thank you. It's pretty obvious, but I was thinking it some other way and was confusing myself. Thanks again.
smokeandmirrors
Banned
With a regular language L, is it safe to assume the prefixes and suffixes of that language are regular as well? Or am I making too many assumptions on what the prefixes or suffixes might be?
I've got a really rookie F=ma sign thing that I can't seem to get my head around. Defining positive as the direction of movement, a force of 200N is acting on a particle in the direction of movement and a force 'B' is introduced acting against movement. The acceleration of the particle is -2.4ms^-2 and the resultant force is -1920N. Now, my intuitive thought here would be to make 200-B = -1920, since those are the positive/negative directions I established. Am I wrong in doing this, and if so why should it be the opposite?
I'm not really able to understand the question that you are trying to ask here.
Particle Kinetics -
Since the question states that initially a force of 200N was applied on a particle, thus particle initially had a positive acceleration, and velocity indicating the direction of motion, but then a Force B was applied, resulting in a net resultant force of 1920N in opposite direction of the motion. That means force B was definitely applied in opposite direction ( must be negative considering your sign convention). Now velocity is positive, but acceleration is negative so particle will come to rest after some time, and then start moving in opposite direction with increasing velocity and constant acceleration.
If you are still not able to understand why force B is negative, considering your sign convention.
Just assume that B was applied in the same direction as 200N force. That means,
F1 + F2 = Fn (resultant force)
200 +B = -1920
Therefore, B = -2120 ( negative sign clearly indicated that particle is under retardation motion until it comes to rest).
I took Calc 2 last semester. Easy A for me.
My suggestion to do great in this course would be that you should be able to "play with graphs". You should be able to imagine the imagine the graph of a function just when someone gives you a function. For this practice the graphs of all basic poly, trig, log, exp functions. A rough sketch of a function just after looking at the function (not actually reading the question) will help you a lot.
Plus you should be good in remembering formulas/ techniques of differentiation and integration. Do this and you should be fine.
Yeah, this was pretty much how I understood it, it seems like a wrong answer was marked as correct and not changed later. Thanks!
campfireweekend
Member
So I know the process involved in most of this problem, I just don't know how to find a shared normal for both equations. Help?
Edit: Wait, all I have to do is find the cross product of the two direction vectors given by the lines, right?
Take the cross product to find the orthogonal vector of the required plane that contains these lines.
To find the shortest distance between two skew lines - Find a vector PQ using the symmetric equation given of two lines (Just find the direction vector from the points P and Q that these given lines pass through). Now, find the unit vector along the normal vector that we calculated earlier in the first part, and take the scalar projection(dot product) of vector PQ along the direction of unit vector.
Khan academy. Thorough understanding of the fundamental theorem of calculus.
Memento_Mori15
Member
Patrickjmt with a combination of a bit of khan. It's important to understand rather than to memorize.
Has anyone here dealt with the Discrete Math textbook by Kenneth Rosen?
My class is using it currently and I have been having a hell of a time understanding how to solve recurrence relations. The textbook doesn't make them any easier to understand, and it seems to be lacking clear examples.
My class is using it currently and I have been having a hell of a time understanding how to solve recurrence relations. The textbook doesn't make them any easier to understand, and it seems to be lacking clear examples.
cpp_is_king
Member
That's the book I used many years ago in college, although I'm sure it's a newer edition now. I remember it being a fine book, nothing obviously wrong with it, but it's probably better if you just post some specific examples that you're having trouble with.
The book we are using is the 7th edition, but I assume it hasn't changed much.
Here's two sample problems that I had difficulty with.
The solving through iteration doesn't really click with me.Find the solution to recurrence relation using iteration.
a= -a(n-1) + n - 1, a(0) = 7
My attempt at this was the formula: B(k) = B(k-1) + (B(k-1) * (0.07/12)) - 100k
I tried to make these as clear as I could, but if they don't make sense the way I typed them let me know. I can upload some pictures of the exact problems from the book.
cpp_is_king
Member
I'm not sure what "solving through iteration" means, but I'm going to assume that it means solving by plugging the formula into itself a few times and trying to notice a pattern.
Assuming that's what it means, let's see if we can get "formulas" for a(1), a(2), a(3), a(4), a(5), a(6) in terms of a(0). That's the key: Trying a bunch of values for n, and getting everything in terms of ONLY the initial condition, a(0).
a(1) = -a(0) + 1 - 1 = -a(0)
a(2) = -a(1) + 2 - 1 = 1 - a(1) = 1 + a(0)
a(3) = -a(2) + 3 - 1 = 2 - a(2) = 2 - (1 + a(0)) = 1 - a(0)
a(4) = -a(3) + 4 - 1 = 3 - a(3) = 3 - (1 - a(0)) = 2 + a(0)
a(5) = -a(4) + 5 - 1 = 4 - a(4) = 4 - (2 + a(0)) = 2 - a(0)
a(6) = -a(5) + 6 - 1 = 5 - a(5) = 5 - (2 - a(0)) = 3 + a(0)
Cutting out all the crap and writing this more cleanly, the pattern becomes more clear:
a(1) = 0 - a(0)
a(2) = 1 + a(0)
a(3) = 1 - a(0)
a(4) = 2 + a(0)
a(5) = 2 - a(0)
a(6) = 3 + a(0)
So, if n is even, the second term is positive, otherwise negative. That's easy to achieve by using (-1)^n
The first term is simply Floor[n/2]
So, our hypothesis is that a
= Floor[n/2] + (-1)^n*a(0)You can then prove this by induction
Laurentius
Member
Can anybody nudge me in the right direction for proving that that the Petersen graph is hypohamiltonian? I did okay in proving that the Petersen graph is NOT Hamiltonian, but struggling to prove that it's hypohamiltonian. Didn't have much luck Googling it either. (Found info about how it's the smallest hypohamiltonian graph, but not much in the way of proofs.)
Partial Gamification
Banned
Peterson graph is the smallest snark, new word to me. You know the graph (G) is nonHamiltonian, is G-v Hamiltonian for all v in V?
from that second link, (Herz et al. 1967; Bondy and Murty 1976, p. 61) is the reference of the proof in question.
from goooglebooks
Its in French but old enough and widely cited, I bet you could find a translation. cite your sources in your work.
Laurentius
Member
Peterson graph is the smallest snark, new word to me. You know the graph (G) is nonHamiltonian, is G-v Hamiltonian for all v in V?
from that second link, (Herz et al. 1967; Bondy and Murty 1976, p. 61) is the reference of the proof in question.
from goooglebooks
Its in French but old enough and widely cited, I bet you could find a translation. cite your sources in your work.
Thank you for the reference, I will look into that.
how would I write "B is not a necessary condition for A" in symbols?
I tried "~( B ⊃ A )" but that doesn't give me a truth table result for the argument that makes sense.
The entire argument that I'm supposed to symbolize and create a truth table for is this:
This seems obvious to me like an argument where the premises are not enough to prove the conclusion, yet when I symbolize it and make a truth table, I find no instance where both premises are true and the conclusion is false.
I symbolized it as:
Hopefully this post makes any sense I'm drunk
edit: I think some books have "->" where mine has "⊃"
I tried "~( B ⊃ A )" but that doesn't give me a truth table result for the argument that makes sense.
The entire argument that I'm supposed to symbolize and create a truth table for is this:
This seems obvious to me like an argument where the premises are not enough to prove the conclusion, yet when I symbolize it and make a truth table, I find no instance where both premises are true and the conclusion is false.
I symbolized it as:
Hopefully this post makes any sense I'm drunk
edit: I think some books have "->" where mine has "⊃"
Partial Gamification
Banned
I'm a little confused with the notation, are you describing a superset [ ⊃ ]? Also, is the arrow supposed to be a right-double-arrow [ => / ⇒ ]? Some modern logic books are using the single-arrow [ -> / →] and I am unfamiliar with this practice.
As a side, anyone know why one is preferred to the other or what motivated the relatively recent changes?
The if-then statement is the obvious response but I'm not sure if it meets the criteria. These types of questions always have me second-guessing and over/under-thinking the problem.
"If monsters talk, then they are intelligent beings. But talking is not a necessary condition for monsters to be intelligent beings. Thus, monsters are intelligent beings."
The bold seems like a starting point.
A: monsters can talk
M: monsters are intelligent beings
A | M | A ⇒ M
T | T | T
T | F | F
F | T | T
T | F | T
The table reflects not being able to talk and still being an intelligent monster but I start to wonder if there is also the need to represent the whole of the given text.
Let M* be the necessary condition for monster intelligence, so:
A ∧ M* ⇒ M** [with M** as monsters that can talk and are intelligent, M** subset of M]
It makes me think there might be some sort of identity relation at play where you end up with, essentially, "monsters that talk, and are, are intelligent" is equivalent to "monsters are intelligent."
I don't think its right but
( ( A ⇒ M ) ∧ ( ¬A ∧ M* ⇒ M ) )
∴M
edit: I think the bold might work better as " A or M* " I'm in an internal debate feedback loop on this whole problem.
edit2: yeah, I don't know if that works.... A or M* is equivalent to ¬(¬A⇒M*)
( A ⇒ M ) ∧ ¬(¬A⇒M*) is equivalent to ¬( ( A ⇒ M ) ⇒ (¬A⇒M*) )
Here, there might just be an issue with introducing [M*] but it all could just be wrong.
are there more natural numbers than even natural numbers?
are there more natural numbers than there are fractions?
are there more real numbers than there are fractions?
are there more points inside a square than there are points on a line segment?
are there more natural numbers than prime numbers?
does two dimensional space have the same cardinality as 3 dimensional space?
1. no
2. no
3. uncertain
4. no
5. no
6. uncertain
are there more natural numbers than there are fractions?
are there more real numbers than there are fractions?
are there more points inside a square than there are points on a line segment?
are there more natural numbers than prime numbers?
does two dimensional space have the same cardinality as 3 dimensional space?
1. no
2. no
3. uncertain
4. no
5. no
6. uncertain
cpp_is_king
Member
What do you think?
FortuneFaded
Member
I've got a counting question. I have the answer but I'm wondering why it isn't this other answer my friend came up with.
A lot of 140 semiconductor chips is inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements.
c) how many samples of five contain at least one nonconforming chip?
So my friend said it was
(10 choose 1) * (139 choose 4)
since every combination would include at least 1 bad chip, but the answer is
(10 choose 1) * (130 choose 4) + (10 choose 2) * (130 choose 3) ect to (10 choose 5)(130 choose 0)
Any explanation on why my friend is wrong would be great as I'm not quite getting it.
A lot of 140 semiconductor chips is inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements.
c) how many samples of five contain at least one nonconforming chip?
So my friend said it was
(10 choose 1) * (139 choose 4)
since every combination would include at least 1 bad chip, but the answer is
(10 choose 1) * (130 choose 4) + (10 choose 2) * (130 choose 3) ect to (10 choose 5)(130 choose 0)
Any explanation on why my friend is wrong would be great as I'm not quite getting it.
If you need to do this with prop. logic:
Let "p" stand for "Monsters talk" and "q" for " Monsters are intelligent beings".
Then the first premise can be written as "p->q".
Now let's reword the second premise:
But talking is not a necessary condition for monsters to be intelligent beings
But Monsters talking is not a necessary condition for monsters to be intelligent beings.
So it is false to say that, if monsters do not talk, then they are not intelligent beings.
So it is false to say that, if monsters are intelligent beings, then they talk.
So ~ (q->p).
Necessary conditions are the consequents of true conditionals, not the antecedent. In the first premise, "p" is sufficient, rather than necessary, for "q".
So we have
p->q
~(q->p)
Therefore:q
This is valid, because the second premise is equivalent to " q ^ ~ p". So it asserts that "q" is true, as weird as it sounds. And therefore the conclusion indeed follows. However, it also really depends on how you want to interpret the second premise. In predicate calculus we can bring out some nuances of the second premise and better support the intuitive interpretation that this is not a valid argument.
Exuro: Your friend's solution is wrong because it allows for some combinations to occur twice or more.
Your friend's solution states that, among the bad chips, you can choose any one and group them with any other of the remaining chips, even bad ones.
So let's say you take bad chip A. Among the remaining 139 you can pick bad chip B and three good ones.
But you can also say you pick bad chip B from the ten and then from the remaining 139 you pick chip A and the same three good ones. Now you got this combination twice already. The actual solution avoids this by only picking bad chips from the pool of 10.
Hey guys, need a little help. Tell me if I'm right or wrong.
I have a parabola and a line equation
y=-x^2+6x
y=x+6
I need to find their intersection. I found it to be (2,8) (3,9)
I need the max value of the parabola which I found to be (3,9)
Also, I need to find the intercepts on the parabola which I found to be (6,0) (0,0)
Could someone tell me if I'm right? Also, does every parabola must have a Y intercept? because if I put 0 in the equation, I get Y=0
Which would mean that 2 intercepts meet at the same spot?
I have a parabola and a line equation
y=-x^2+6x
y=x+6
I need to find their intersection. I found it to be (2,8) (3,9)
I need the max value of the parabola which I found to be (3,9)
Also, I need to find the intercepts on the parabola which I found to be (6,0) (0,0)
Could someone tell me if I'm right? Also, does every parabola must have a Y intercept? because if I put 0 in the equation, I get Y=0
Which would mean that 2 intercepts meet at the same spot?
YianGaruga
Banned
I'm not very good with combinatorics so the explanation might be lacking, but I thought about it and I think yours must be wrong because it doesn't make sense when you turn it around.
From left to right I'd say it makes sense at first, but if you write it like this
(139 choose 4) *(10 choose 1)
it becomes impossible to interpret. 4 from which 139? 1 out of which 10? Do we even have those 10 left?
The problem is a standard example of the hypergeometric distribution and that's why the result has to look like the answer.
For example, drawing EXACTLY X=1 nonconforming chip in a sample of 5 has (10 choose 1) * (130 choose 4) possibilities. This would probably be your answer as well.
Now you just have to calculate all the possibilities for X=1, X=2, X=3 independently and sum them up to get to the right answer.
Summing up the possibilities in that combinatorics problem as explained by YianGaruga would be my approach as well. Your friend's theory that every combination of five chips would have at least one bad chip can clearly be proven to be false with an example, so his approach to the problem is wrong from the get-go.
That's all correct. As for your first question, what is the general from of a parabolic function? What happens if you let x=0?
Jayhawk: What example would that be? His friend's solution counts some combinations more than once, but provided the first group of then is thee group of bad chips, then any of his combinations do indeed include at least one bad chip. Or am I off?
Yeah, I misunderstood what his friend meant since I just woke up. The combinations you're interested in would have at least one bad chip. In terms of proving his friend to be incorrect, YianGaruga was alluding to it with the reversal of the equation but (139 choose 4) * (10 choose 1) has the flaw where a specific bad chip can be picked in both groups and that is not supposed to be happening. That is why separating it into a summation of (10 choose X) * (130 choose (5-X) ) where X is 1..5 makes more sense to me.
Well, no disagreement really, but I think he accounts for no single chip picked twice by writing 139, rather than 140. The flaw is that he doesn't account for multiple countings of combinations with at least two bad chips.
A specific chip can be picked twice in his scenario because let's look at the example where in the (139 choose 4) part, three bad chips are chosen. How are those chosen bad chips not still in the (10 choose 1) pool? The flaw you pointed out and the flaw I point out are just different perspectives at the same issues behind the math I think. lol
Don't Laugh
Banned
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4. You would have to look at the actual data on the chart they're referring to and get the power index for the 4th place band and then multiply that by 26902.
5. This is a subjective question, and should match your opinion. Just back it up with an argument pointing to the numbers.
Don't Laugh
Banned
.
I couldn't find power index either, so that stats problem seems to be dated. That is something you bring up to the instructor/professor.
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