Horatorobor
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What are some good programs to experiment with parametric curves and surfaces?
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The Math Help Thread
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Stumpokapow
listen to the mad man
Finding Square Area from Perimeter
The square has perimeter 2pi, thus length/width pi/2, thus area = (pi^2)/4 or (pi)(pi)/4. Remember when you square a fraction you square both the numerator and denominator.
Finding Circle Radius from Circle Area
Area of a circle = pi r^2
pi r^2 = (pi)(pi)/4 <-- this is the equivalence of the two areas.
r^2 = pi/4
r = sqrt(pi)/2
Finding Circle Circumference from Circle Radius
Circumference = 2 pi r. We have r, which is sqrt(pi)/2
2 * pi * sqrt(pi) / 2 = pi * sqrt(pi)
Answer
The answer is B.
You can certainly find the radius that way, and will get the same value.
Consider equation (1),
(1) pi * r^2 = s^2,
which says that the areas of the circle and the square are the same.
You can do what you said: square s = (pi / 2) first to get the area of the square (which is pi^2 / 4, be careful), set the equation pi * r^2 = pi^2 / 4, then solve for the radius r.
----
What I did was the following:
Go ahead and divide both sides of equation (1) by pi, and then take the square root, so that we can write the radius r in terms of the side length s. I had called this equation (3):
(3) r = sqrt(s^2 / pi) = s / sqrt(pi).
When you take the square root of both sides of an equation, r = -sqrt(s^2 / pi) is a possible solution, but because of the physical meaning that r is supposed to have (it's a radius--a length), we discard this solution.
Cheers,
spelltropy
Member
this is gonna be a really dumb question having completed my linear algebra module, but i've started studying for the exam. is one of the requirements of a field the fact that the additive identity doesn't equal the multiplicative identity (1=/=0)?
The identities usually need to be distinct, but I have seen some books that drop that requirement in the definition. Note that if they are the same then the identity is the only element of the field.
spelltropy
Member
that's what stumped me, but I guess my lecturer decided to include it in the definition. Cheers!
Splash Wave
Member
I sincerely appreciate the help, y'all. As you can probably tell, I'm more than a little rusty.
Stumpokapow
listen to the mad man
I'm having a brain fart and I just want someone to tell me "Yes, you are right"
I have a series of the form:
5r^0 + 100r + 5r^2 + 100r^3 ...
I am looking for the sum of this series (r<1, this is convergent)
Obviously for the form
5r^0 + 5r + 5r^2 + 5r^3 ..., the sum is 5/(1-r)
My thought is that this is correct:
Let u = r^2
My series is the sum of two series:
sigma(i=0->inf) 5u^i = 5 sigma u^i = 5*(1/(1-u)) = 5/(1-r^2)
and
sigma(i=0->inf) (100r)u^i = 100r * sigma u^i = 100r*(1/(1-u)) = 100r/(1-r^2)
So the whole sum is
(5 + 100r)/(1-r^2)
Am I missing anything? Thanks.
I have a series of the form:
5r^0 + 100r + 5r^2 + 100r^3 ...
I am looking for the sum of this series (r<1, this is convergent)
Obviously for the form
5r^0 + 5r + 5r^2 + 5r^3 ..., the sum is 5/(1-r)
My thought is that this is correct:
Let u = r^2
My series is the sum of two series:
sigma(i=0->inf) 5u^i = 5 sigma u^i = 5*(1/(1-u)) = 5/(1-r^2)
and
sigma(i=0->inf) (100r)u^i = 100r * sigma u^i = 100r*(1/(1-u)) = 100r/(1-r^2)
So the whole sum is
(5 + 100r)/(1-r^2)
Am I missing anything? Thanks.
FortuneFaded
Member
A way to check is to sum both sides by (1-r^2) and look for cancellations. Looks right to me though.
Stumpokapow
listen to the mad man
Thanks. I haven't done any series stuff in a long time, this is just an ancillary sanity check in a model specification in a seriously unrelated topic and I don't want to look like an idiot when someone reads it and tells me I made an elementary math error.
MoG EclipsE
Member
Stats question:
The amount of cereal packed into a cereal box is supposed to be 500g. Either too high
or too low a population mean is considered bad (either the company is wasting money
by giving away free product, or customers will complain about being shortchanged),
so the t-value is required to be in the middle 95%. A sample of 20 boxes is taken, and
the sample mean is 480 and sample standard deviation 2.23.
(a) What is the t-value for our sample?
(b) What is the range of t-values that the company considers acceptable?
(c) Will the company find this t-value acceptable?
No idea how to go about this. Do I use the formula for t-distribution? I know there is a table, but I am not sure how to use it and what degrees of freedom this has. Thanks for any guidance!!
The amount of cereal packed into a cereal box is supposed to be 500g. Either too high
or too low a population mean is considered bad (either the company is wasting money
by giving away free product, or customers will complain about being shortchanged),
so the t-value is required to be in the middle 95%. A sample of 20 boxes is taken, and
the sample mean is 480 and sample standard deviation 2.23.
(a) What is the t-value for our sample?
(b) What is the range of t-values that the company considers acceptable?
(c) Will the company find this t-value acceptable?
No idea how to go about this. Do I use the formula for t-distribution? I know there is a table, but I am not sure how to use it and what degrees of freedom this has. Thanks for any guidance!!
It has been a while since I studied stats, so I don't remember the theory behind t test (other than we don't use z test cause we often don't know the population standard deviation). Maybe another GAFer can explain why it works to you.
(a) We recall that the t value is defined to be,
t = (sample mean - supposed population mean) / (sample standard deviation / sqrt(sample size)).
Hence, here,
t = (480 - 500) / (2.23 / sqrt(20)) = -40.1088.
If |t| > t* (t* is a critical value, which we will find next), then we can reject the null hypothesis that the population mean is 500. In other words, we can conclude that the population mean is actually something other than 500.
On the other hand, if |t| <= t*, then we accept the null hypothesis, i.e. the population mean is indeed 500.
(b) Recall that the t distribution curve is bell-shaped, and symmetric about t = 0. So the middle 95% refers to the area under the curve from t = -t* to t = t* being equal to 0.95. The question is, for which value of t* (this is the critical value) will this occur?
This is where a table like the one below comes in handy:
t distribution table
The alpha value (two-tailed) refers to the area outside the interval [-t*, t*]. For one-tailed, the alpha value is the area outside the interval [-infinity, t*]. Here, we would want alpha to equal to 1 - 0.95 = 0.05.
Since we have 20 samples, the number of degrees of freedom is 20 - 1 = 19. So we look at df = 19 and alpha = 0.05 (two-tailed), and see that t* = 2.0930.
In other words, the area under the t distribution curve from t = -2.0930 to t = 2.0930 is 0.95 (when df = 19). The interval [-2.0930, 2.0930] gives us the values of t that the company deems ok.
(c) Clearly, our sampling (for which t = -40.1088, way outside the interval) says no, the population mean weight is not 500 g. Angry customers will run amok.
Stumpokapow
listen to the mad man
The t-distribution is very similar to the z-distribution but has a longer tail; using the t-distribution when you only know the sample standard deviation (rather than the population standard deviation) basically gives you a "penalty" in your confidence interval to reflect the greater uncertainty.
I mean, basically the z-distribution is normal while the t-distribution is normal-ish. Worth noting that if you didn't know the population standard deviation but you were certain that the assumption of normality should be maintained, you'd be able to use the z-distribution. But the assumption of normality is extremely strong and likely not true for most real-world datasets.
(Historical trivia: The t-distribution is typically called the "Student's T-Distribution"--Student was working for Guinness when he developed it and was not allowed to publish under his actual name!)
MoG EclipsE
Member
Thank you very much for the help! I was really having trouble understanding how to use the table, but this makes sense! Thanks again!
Multiply the first term by 1 = sin(x)/sin(x) and the second term by 1 = cos(x)/cos(x). (We assume that x is a value such that neither sin(x) nor cos(x) equals to 0.)
We see that,
sin(x)^2 / (1 + cot(x)) + cos(x)^2 / (1 + tan(x))
= sin(x)^3 / (sin(x) + cos(x)) + cos(x)^3 / (cos(x) + sin(x))
= (sin(x)^3 + cos(x)^3) / (sin(x) + cos(x)).
Recall that a^3 + b^3 = (a + b) * (a^2 - ab + b^2). Hence,
(sin(x)^3 + cos(x)^3) / (sin(x) + cos(x))
= sin(x)^2 - sin(x)cos(x) + cos(x)^2
= 1 - 1/2*sin(2x).
Now differentiate with respect to x.
Thank you
Could someone help with a directional derivative problem?
It seems I can't use the Duf = |grad f||u|cos θ equation, I believe it is because θ must be between 0 and π from what I've seen of the dot product, but I'm not sure.
I need to explain why I can't use θ = -π/6 and what value of θ should I be using?
If you look at the image you can see the questions I need to answer written in pen. Thanks.
It seems I can't use the Duf = |grad f||u|cos θ equation, I believe it is because θ must be between 0 and π from what I've seen of the dot product, but I'm not sure.
I need to explain why I can't use θ = -π/6 and what value of θ should I be using?
If you look at the image you can see the questions I need to answer written in pen. Thanks.
Biggest-Geek-Ever
Member
In Duf = |∇f||u|cos θ the angle refers to the angle between ∇f and u. The direction θ given in the problem is referring to the angle u makes with the positive x axis.
When given an angle direction like in this case, your unit vector u = <cos θ, sin θ> = <cos -π/6, sin -π/6> = < √3/2, -1/2 >. Taking the dot product of this and your gradient vector at the given point will give you the directional derivative.
Ahh thanks that makes a lot more sense, any idea on the second question in the pic, what alteration makes it true? The professor wants a number.
Biggest-Geek-Ever
Member
Since to use Duf = |∇f||u|cos θ we need to know the angle between the two vectors, we have to go about determining that angle. We know the angle u makes with the x-axis, so if we can find the angle ∇f makes with the x-axis, we can find the interior angle we desire.
You found the gradient vector at the given point is <5/8, -1/2>, so using the trigonometric property tan θ = y/x, we can solve for the angle.
tan θ = -1/2 / 5/8
θ = arctan ( -1/2 / 5/8 ) = -arctan (4/5)
Basic geometry tells us this angle is bigger than -π/6, so the angle between ∇f and u is -arctan(4/5) - (-π/6). Thus, the equation for our directional derivative becomes:
Duf = √41/6 * cos (-arctan(4/5) - (-π/6))
That looks kind of messy, but you can numerically check to see it's indeed the same answer as finding the gradient using u = <cos θ, sin θ>.
Edit: if you'd like to see a picture of the geometry involved:
Orbis Tabula
Member
Okay, so who here has any experience with integer programming? Because I've got a problem I just can't solve and I'm beating myself up because it seems like it should be simple. But I just can't get it right.
I'm trying to write a constraint for a IP that works like this: If variable a is bigger than variable b by a certain amount, we'll call it x, then you should set a fourth variable, y, to 1. If this is not true, then set y to 0. Written out better, it's like this.
IF b+x<=a THEN y=1
IF b+x>a THEN y=0
Now how can I write that as one or more linear constraints? I need to get rid of the IF/THENs and have one or two linear inequalities such that they follow the above rule. And I just can't crack it.
I'm trying to write a constraint for a IP that works like this: If variable a is bigger than variable b by a certain amount, we'll call it x, then you should set a fourth variable, y, to 1. If this is not true, then set y to 0. Written out better, it's like this.
IF b+x<=a THEN y=1
IF b+x>a THEN y=0
Now how can I write that as one or more linear constraints? I need to get rid of the IF/THENs and have one or two linear inequalities such that they follow the above rule. And I just can't crack it.
What's wrong with if/then statements?
y = (b+x <= a) ? 1 : 0
Orbis Tabula
Member
Well, the point is that it's all supposed to be solvable with elementary row operations and stuff like that. It's all linear Algebra. It has to be in the form of "some linear sum of variables <=, >=, or = some other linear sum of variables."
shouldn't the inequality comparison return 1 or 0 based on the result? can't you just leverage that
y = (b+x <= a)
you don't even need to use the ternary at that point, or am I still missing the problem?
Orbis Tabula
Member
I think the word "programming" is a bit of a misnomer here. It has nothing to do with computers. I'm trying to make a series of linear equations to model something. But it's all paper stuff (I mean, a computer could so it, it's just math. But this has nothing to do with "programming" as its usually understood).
For example, if I were to come up with a rule "at most one of four binary variables may be equal to 1", to put that in a linear program, you would use a linear equation like:
a+b+c+d<=1.
So you can express logical statements with simple linear inequalities. That's what I'm trying to do here. The only operators I can really use are +, -, =, <=, or >=.
Edit: You can multiply, but you can't multiply variables together or you lose linearity. In my first example, a, b, and y are variables, and x was not. So x*a is allowed, but not a*b or a/b.
Splash Wave
Member
EskimoJoe, you will want to read on logical constraints. Some notes from MIT go over the same type of problem as yours; see p. 278 in the link.
If I understand the notes correctly, you need to know a priori how large (b + x) - a can get, so that the multiplicative constant B that penalizes the binary variable (y in the notes, but it's 1 - y for your problem) is "large enough." Hopefully you have an upper bound for b and a lower bound for a in your problem.
triplestation
Member
Gonna start calculus II soon
Any words of advice?
Any words of advice?
Stumpokapow
listen to the mad man
Assuming Calc II is integral calculus from Riemann Sums through Integration by Parts (which is a pretty typical setup):
- Going in, learn all the trigonometric identities inside and out
- Going in, be extremely comfortable with taking derivatives
In terms of useful course content:
- FTOC FTOC FTOC FTOC FTOC FTOC Fundamental Theorem of Calculus.
- Remember that indefinite integrals need +C added at the end. If your prof doesn't drill this in enough or you don't understand conceptually why, drill it in.
- U-substitution is the single most important anything you'll ever learn ever
- Get lots of practice doing "rotation about the axis" / "solid revolution" problems; profs love them on exams.
Calc II is generally considered a "weed-out" course, as it's the last math class needed for a variety of BSc pathways (typically not the last for Physics or CompSci, but the last for most other BSc programs). Lots of people tank. Attend the lab/discussion section/whatever, do the practice problems, take the homework seriously.
For what it's worth, I consider myself pretty good at math since I was young, and obviously I've gone on to do a lot more math since, but Calc II is the single hardest I ever worked at a math course. Take it seriously. I got an A in the end
the second time I took the course *priceisrightfail.mp3*
triplestation
Member
I can't thank you enough for this great advice
I hear differential equations and linear algebra can be pretty tough too but thats worry for a later time ._.
spelltropy
Member
Analysis is the absolute worst, diff eqns and linear algebra are a walk in the park in comparison...
Orbis Tabula
Member
Thank you so much! This will be a big help. I have an easy upper bound in this problem, so it should work.
I agree. I was always told to fear diff eq. but in the end, it turned out to be pretty easy for me. Same with algebra. But analysis sucks a lot. But anyway, Stump's advice for Calc II is spot on. I can't think of anything to add.
Draw a picture to visualize what's happening in a problem. Calculus is very much a visual subject. With practice, you will be able to do this in your head.
Other than that, I'd say get excited for the subject? Integration, along with differentiation, forms the ground for other topics; the idea behind Riemann integration gives rise to numerical integration techniques, and integration by parts is the key step for finite element methods.
You can also prove the formulas for the area of a circle and the volume of a sphere, maybe even the circumference of the circle and the surface area of the sphere (ironically, these are harder). You know, things that we took as facts in high school.
Topology is the worst for me; I give it another try every year, but still don't understand a thing or know why it's useful, and am just happy the Euclidean space is enough for most things. Functional analysis comes next.
Glad to hear that, no problem.
Orbis Tabula
Member
See, for me topology was really hard, but I still liked it. It was enjoyable and fun, even when I struggled through it. Analysis wasn't. It was hard and annoying. When I struggled through topology and solved something, I felt accomplished. Analysis never really did that for me.
themagicalkitsune
Member
Topology can be beautiful. Though when we got to the algebraic side of it, shit got real. I was like this most of the time.
Right now I'm reading up on Laurent / Taylor series.
Say I have a function 1/(z-1)^2 * 1/(z + 2) and I want to write it in terms of powers of z. Noting the singularities at z = 1 and z = - 2, can I just take the maclaurin expansions of each fraction within the domains |z| < 1, 1 < |z| < 2, 2 < |z| and multiply them together to get the series I want? (depending on the domain, each series may change)
The defined coefficients in the Laurent expansion of a function are kind of cumbersome to work with.
Right now I'm reading up on Laurent / Taylor series.
Say I have a function 1/(z-1)^2 * 1/(z + 2) and I want to write it in terms of powers of z. Noting the singularities at z = 1 and z = - 2, can I just take the maclaurin expansions of each fraction within the domains |z| < 1, 1 < |z| < 2, 2 < |z| and multiply them together to get the series I want? (depending on the domain, each series may change)
The defined coefficients in the Laurent expansion of a function are kind of cumbersome to work with.
Yeah, that should be just what you want to do if I remember it correctly.
Downfall Hitler rant parody on topology ^_^: https://www.youtube.com/watch?v=SyD4p8_y8Kw
Haha, yeah, I've seen that before. It's funny cause it's so true; bad terminologies make the subject harder than it should be.
chancellor
Member
Was signposted here from a thread that I made
Otakumegane
Member
Hopefully there isn't more math to learn post differential calculus for my Electrical engineering major.
Some Vector Calculus application problems came up today and I had to write down all the equations and get familiarized with the concept all over again.
Some Vector Calculus application problems came up today and I had to write down all the equations and get familiarized with the concept all over again.
RadioHammer500
Banned
im about to start statistics in my math class, is this generally easier or harder then probability?
chancellor
Member
Thanks!
Hexa
Member
Assuming mph is for meters per hour.
So there is Car a going forward and and car b going back. Lets say that car b starts 100 ahead of car a just for the scenario, though this number won't matter in the end.
Their positions can then be modeled by:
a=30t
b=100-30t
And the distance between them can be modeled by the difference:
a-b=60t-100
Set a-b=d
t=(d+100)/60
So t tells you the points that that they are a-b meters apart. Set a-b equal to the distance you want, for both positive and negative to get when they begin being within that range and and when they leave being in that range, and then subtract.
So t1=(d+100)/60 and t2=(-d+100)/60
I'll use T for total time.
T= t1-t2=2d/60=d/30
And then you can just plus whatever time you want in to get the time they're within that range.
The math requirements for an EE major tend to look something like:
-Differential calculus
-Integral calculus
-Calc III stuff (sequences & series, more involved vector calc, random shit)
-Linear algebra (fun with matrices!)
-Differential equations (where you learn methods for finding solutions to equations that have a function's derivatives in them, basically) (and by "methods" I mean "silly hacks that work in specific cases")
And they might require some other random thing or another, like statistics,
EE degrees usually require basically getting a math minor.
The relative speed is 60mph, which is 26.8224 meters per second.
Which means that you get .03728 seconds per meter. Multiply that by the distance in meters and you get your seconds.
50m*.03728s/m means that after coming within 50m of each other, they'll spend about 1.86 seconds with the other car ahead before they pass.
If by "within 50 meters" you're including the time after they pass by each other, you'd double that figure (since "within 50 meters" then deals with a distance of 100m).
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