She said, "An oval has four vertices". That is the definition she gave us.
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She said, "An oval has four vertices". That is the definition she gave us.
Partial Gamification
Banned
I'm thinking circles can be inscribed and circumscribed, respectively, inside and outside any oval. These circles have four points of intersection, along the minor and major axes of the oval. It will have horozontal and vertical bulges because it is closed and convex.
These points have the same tangent, and a circle has a parallel second derivative of the tangent; thus, there are at least four points that the oval will have parallel: T // T''
Instead of the ellipse, just show all points on a circle have the parallels in question, and there has to be a property in your material that is what you are being asked to demonstrate. Something about curvature of the oval; showing it has; a base, an apex, and sides less than that vertical length (because it will be mathemagically oriented in that fashion, by design).
cpp_is_king
Member
Ok I think I've got it. Been years since differential geometry, took me a long time messing around on Mathworld and Wikipedia to figure this out, but here we go.
r(t) is your curve
r'(t) = T [T = tangent]
r''(t) = k N [ k = curvature, N = normal]
r'''(t) = k N' + k' N (chain rule)
Now, at each of the 4 vertices of the curve, the derivative of curvature is 0. That's just the definition of a vertex. So, for each of the 4 verticies, we have
r'''(t) = k N'
However, N' = t B - k T [B = binormal vector, t= torsion]
The torsion is identically 0 at all points on the curve if and only if the curve is contained in a single plane. Since we're talking about a 2 dimensional oval, this is obviously the case. So for this oval, at each of the 4 vertices, we have:
N' = -k T
r'(t) = T
r'''(t) = k N' = -k^2 T
Thus, r'''(t) = -k^2 r'(t)
Since r'''(t) is a scalar multiple of r'(t), r'''(t) is parallel to r'(t) at each vertex.
QED
r(t) is your curve
r'(t) = T [T = tangent]
r''(t) = k N [ k = curvature, N = normal]
r'''(t) = k N' + k' N (chain rule)
Now, at each of the 4 vertices of the curve, the derivative of curvature is 0. That's just the definition of a vertex. So, for each of the 4 verticies, we have
r'''(t) = k N'
However, N' = t B - k T [B = binormal vector, t= torsion]
The torsion is identically 0 at all points on the curve if and only if the curve is contained in a single plane. Since we're talking about a 2 dimensional oval, this is obviously the case. So for this oval, at each of the 4 vertices, we have:
N' = -k T
r'(t) = T
r'''(t) = k N' = -k^2 T
Thus, r'''(t) = -k^2 r'(t)
Since r'''(t) is a scalar multiple of r'(t), r'''(t) is parallel to r'(t) at each vertex.
QED
schroddycat
Member
This is what I saw, a so called riddle, on FB and I'm interested to hear from GAF on the answers:
There was a man spent $8 on a chicken, sold it for $9. Then, he bought back the chicken for $10 and subsequently sold it to another person for $11. How much did he earn in total?
Intriguing to see how the answer can vary from 0 to 3 and even losses. The interpretation I saw involved algorithm to cost accounting. What say you math-GAF?
There was a man spent $8 on a chicken, sold it for $9. Then, he bought back the chicken for $10 and subsequently sold it to another person for $11. How much did he earn in total?
Intriguing to see how the answer can vary from 0 to 3 and even losses. The interpretation I saw involved algorithm to cost accounting. What say you math-GAF?
systemfehler
Member
Partial Gamification
Banned
I like it.
How much he earned cannot be quantified. The lessons of "chicken-flipping" are priceless. On top of that, he earned a little self respect for being able to pluck a profit from the poultry.
youngwerther
Banned
I feel the need to skip trig today for personal reasons.
Here is what we are scheduled to cover
Polar Coordinates
Rose curves
Limacons
Lemniscates
Cardioids
Tangents to polar curves
complex numbers in polar form
DeMoivre's Theorem
Am I OK to learn this on my own later on or should I go to class?
Here is what we are scheduled to cover
Polar Coordinates
Rose curves
Limacons
Lemniscates
Cardioids
Tangents to polar curves
complex numbers in polar form
DeMoivre's Theorem
Am I OK to learn this on my own later on or should I go to class?
Partial Gamification
Banned
Personal Reasons are a fair reason enough. These topics lay the foundation for many "higher" applications. There are many resources out there but you will need to do the legwork.
From http://wheresthemath.com/parents/math-help/ : http://patrickjmt.com/
Khan Acadamy, Wolfram, etc.
To anyone: is there something like MathJax to format mathematics in the forum, or a way to do it with the Bulletin Board Code?
nicoga3000
Saint Nic
OK Math GAF - time to put your geometry skills to use!
I have been trying to come up with a general form for the volume of the shape shown below for a while. I can't...do it. I can get something drawn up in AutoCAD, but that's not helping my current dilemma. I'm going to try to walk you through what you see...
Top
Pretty easy - this is the top view of the shape in question. Although it's not noted, the bottom is the same. The eye is for the front/back views on the left.
Side
Yeah, pretty simple. This is the side view.
Back
The back is easier to explain. The back side has width b and height h2. The bottom sloped section tapers down to the front side with width a and height h1.
Front
After understanding the back view, the front view makes more sense. It looks busy because it is. The sloped surfaces aren't hidden so there are a lot of lines. It should be relatively clear from the rest of the drawings what's going on though.
I've also tried to do a rendition of the shape in 3D below it. It's a bit wonky because it's hard as shit to draw, but hopefully it makes the shape a bit clearer as well.
Also, the only stipulations are that b > a and h2 > h1.
For what it's worth, this is the soil profile under transient loadings of a ringwall spread foundation. I design these all the time, but we conservatively assume that h2 is distributed across the entire area. The problem is that using that for a loading gives insanely high structural requirements. I really think this can cut down on that, but the calcs are just dumb and I can't figure out the geometry.
Good luck?
E: Sorry for the "meh" image. The scanner did OK, but I had to convert it to a clipped file via cutting tool since I can't save the PDF as an image.
I have been trying to come up with a general form for the volume of the shape shown below for a while. I can't...do it. I can get something drawn up in AutoCAD, but that's not helping my current dilemma. I'm going to try to walk you through what you see...
Top
Pretty easy - this is the top view of the shape in question. Although it's not noted, the bottom is the same. The eye is for the front/back views on the left.
Side
Yeah, pretty simple. This is the side view.
Back
The back is easier to explain. The back side has width b and height h2. The bottom sloped section tapers down to the front side with width a and height h1.
Front
After understanding the back view, the front view makes more sense. It looks busy because it is. The sloped surfaces aren't hidden so there are a lot of lines. It should be relatively clear from the rest of the drawings what's going on though.
I've also tried to do a rendition of the shape in 3D below it. It's a bit wonky because it's hard as shit to draw, but hopefully it makes the shape a bit clearer as well.
Also, the only stipulations are that b > a and h2 > h1.
For what it's worth, this is the soil profile under transient loadings of a ringwall spread foundation. I design these all the time, but we conservatively assume that h2 is distributed across the entire area. The problem is that using that for a loading gives insanely high structural requirements. I really think this can cut down on that, but the calcs are just dumb and I can't figure out the geometry.
Good luck?
E: Sorry for the "meh" image. The scanner did OK, but I had to convert it to a clipped file via cutting tool since I can't save the PDF as an image.
cpp_is_king
Member
i dont agree with your image of side. That edge is ony L for a 2d trapezoidal cross section taken within the boundaries of the line a.
Also, on the image of top, is this a cutout of an isosceles triangle? So angles on left are same, and angles on right are same?
if not, then i believe we have insufficient information. Any of the following conditions (in isolation) should be enough:
a) Let's say that L intersects a at P, and L intersects b at Q, in the top view. The length of the segment P to either of the endpoints of the line a, and the length of the segment Q to either of the endpoints of the line b.
b) P and Q are BOTH the midpoints of lines a and b, respectively.
c) Let's say that in the top view, the upper unlabelled edge is called E and the lower unlabelled edge is called F. The lengths of E and F.
d) Projecting the lines E and F further to the point where they intersect results in an isosceles triangle.
Treefingers
Member
I think it's the side view, not a diagram of the side face. So if you looked at it from the side you would see exactly that.
nicoga3000
Saint Nic
The top is an isosceles trapezoid.
And for the side image, I'm pretty sure it's correct.
Yes, sorry. These are all views (hence the eyeball!).
cpp_is_king
Member
You're right, that clarifies. OP also clarified that top is isosceles trapezoid.. I wasnt sure because it looks close but not quite in the diagram. But that helps, i'll have a look when I get out of this meeting.
nicoga3000
Saint Nic
Yeah, given that I know what the design assumptions are that we make, it's super clear to me. But it's miserably hard to portray, haha. Glad you "get it" now!
Also, if there's any idea on how to determine the centroid along L, that would be awesome, too.
My problem is that I can't figure out the equations for the planes that cut that shape. I have a feeling for how to solve this problem, but like any multiple integral problem, the issue comes from determining the planes in question. D:
Treefingers
Member
Spent 15 mins on this so there could be something wrong with it and it probably could be done more easily but here's a huge picture of it (sorry it's so big lol): https://dl-web.dropbox.com/get/Public/2012-10-25%2016.39.04.jpg?w=c7a8fd74. Ignore the stuff about x and theta.
So I get V = L [ (b-a)(h2/3 + h1/6) + (2a/3)(h2 - h1) + a(h1) ]
Basically I divided the volume into three wedges and a rectangular prism, then used the formulas for their volumes.
So I get V = L [ (b-a)(h2/3 + h1/6) + (2a/3)(h2 - h1) + a(h1) ]
Basically I divided the volume into three wedges and a rectangular prism, then used the formulas for their volumes.
If everything is right angles and symmetrically sloped, just extend it to a pyramid using the distance between the faces H2B and H1A and their relative dimensions. (Draw some triangles.)
Find the volume of the full pyramid. A pyramid's volume is 1/3 (area of base) (height).
Find the volume of the top part of the pyramid (from H1A to the tip) using the same formula.
Subtract the volume of the smaller pyramid (from H1A to the tip) from the entire pyramid.
If it's not symmetrically sloped, define the area of the cross section at any point along H and then integrate over H. H would be a line from the center of one face to the center of the other face. L is basically useless.
Edit: Is the top guaranteed to be level? Treefingers has this covered with 3 wedges of cheese subtracted from the block.
Find the volume of the full pyramid. A pyramid's volume is 1/3 (area of base) (height).
Find the volume of the top part of the pyramid (from H1A to the tip) using the same formula.
Subtract the volume of the smaller pyramid (from H1A to the tip) from the entire pyramid.
If it's not symmetrically sloped, define the area of the cross section at any point along H and then integrate over H. H would be a line from the center of one face to the center of the other face. L is basically useless.
Edit: Is the top guaranteed to be level? Treefingers has this covered with 3 wedges of cheese subtracted from the block.
nicoga3000
Saint Nic
Yes - the top is always level, the bottom is always linear. I'm not at the office at the moment, so I don't have my working files with me. I should be able to check any results first thing in the morning though!
Thank you!
cpp_is_king
Member
Your solution is different than mine. FWIW, I simplified your answer and it simplifies down to:
L(a+b)(2h_2 + h_1)/6
I get
L(a+b)(Lh_2 - h_2 - h_1) / 2
Our answers are fairly similar though, only the third factor is incorrect. I used a completely different method though, basically integration of the area of cross-sectional rectangles along the L-axis. I will review my answer to see if there are any mistakes.
The bolded should not stand as is, right? You get a dimension error since you have a [length]^2 - [length]. Additionally, the final answer will have a term with [length]^4
Treefingers
Member
@cpp_is_king Going over my calculations again I made an error where I calculated the volume of the third wedge. So the volume I now get is
V = L [b(2h_2 + h_1) + a(2h_1 + h_2)]/6
It seems right now but maybe not? I wanna try integrating but I don't quite remember how to so I'll have to refresh my memory about that.
V = L [b(2h_2 + h_1) + a(2h_1 + h_2)]/6
It seems right now but maybe not? I wanna try integrating but I don't quite remember how to so I'll have to refresh my memory about that.
cpp_is_king
Member
Hah, I went over my calculations again as well and I got the same thing you just got, using my original method. So it appears that is the correct answer.
I'll show the work later, but it's fairly difficult as far as integration goes. Not the actual integration itself, but just figuring out how to create the appropriate integral. I think your method is probably easier, but at least we confirmed each others' work.
I find the strange symmetry of the answer interesting. For example, you can re-write the answer as:
(L/6) * [(a+b)(h1+h2) + (a,b).(h1,h2)]
Where the second term is the vector dot product. Probably means nothing, just found it interesting.
cpp_is_king
Member
Here's how I did the volume problem:
I felt like 4 images was enough, so I didn't actually show how to solve the integral. But it's pretty easy, just multiply out the terms, and you get a y^2 term, a y term, and a constant term, all of which are trivial to integrate. After some simplification, you get the answer above.
I felt like 4 images was enough, so I didn't actually show how to solve the integral. But it's pretty easy, just multiply out the terms, and you get a y^2 term, a y term, and a constant term, all of which are trivial to integrate. After some simplification, you get the answer above.
Having never had formal education in vector algebra, I am getting crushed by this problem:
"Find the minimum distance between two lines in a 3D space (ie. x,y,z), as well as the midpoint between the two closest points."
I've managed to figure out that I need to make parametric equations for the lines, and must somehow compute the least squares (a new concept for me).
I'm not sure why it's giving me so much trouble, but after 10 hours on this, I am hoping somebody here might be able to explain at least a bit about how to do this. I found a Matlab solution that works for any number of lines at this link, but had difficulty understanding it after it normalized the vectors.
"Find the minimum distance between two lines in a 3D space (ie. x,y,z), as well as the midpoint between the two closest points."
I've managed to figure out that I need to make parametric equations for the lines, and must somehow compute the least squares (a new concept for me).
I'm not sure why it's giving me so much trouble, but after 10 hours on this, I am hoping somebody here might be able to explain at least a bit about how to do this. I found a Matlab solution that works for any number of lines at this link, but had difficulty understanding it after it normalized the vectors.
Are you familiar with the Dot product?
With it you find all the equation of all the vectors that are perpendicular to one of the lines, with a simple substitution you find all the vectors the are perpendicular and, when centered at the line, reach (intersects) the second line. Derive the vector length equation to find the minimum.
With it you find all the equation of all the vectors that are perpendicular to one of the lines, with a simple substitution you find all the vectors the are perpendicular and, when centered at the line, reach (intersects) the second line. Derive the vector length equation to find the minimum.
CoffeeJanitor
Member
Hey man if you have any problems don't hesitate to ask...
thanks. didnt know this thread existed until now, i'll certainly be back!
Don't worry, not even Mathematicians like such equations. Simple is beautiful.
nicoga3000
Saint Nic
@cpp_is_king - I'm very excited to check through your work tomorrow and compare it to my model. If it checks out, you and Treefingers will have contributed to some HUGE improvements in design at my office!
Speaking of which, how do I access your file Treefingers? It says it's unavailable...
Speaking of which, how do I access your file Treefingers? It says it's unavailable...
cpp_is_king
Member
Don't worry, not even Mathematicians like such equations. Simple is beautiful.
Pshh, I love long equations. Although to your credit, I guess it's more accurate to say that I I think it's awesome going through pages and pages of equations with complicated expresisons and simplifications and then ending up with something simple and beautiful. Like the whole e^(i*pi) = -1 thing. If you write it as:
e^(i*pi) + 1 = 0
then you've got a single equation relating the 5 most important constants in all of mathematics. That's pretty amazing!
GremlinFool
Member
I'm a logician myself, so I don't think going through plumes of equations is particularly rewarding. Computability axioms, on the other hand...
nicoga3000
Saint Nic
cpp_is_king and treefingers - your equations check out. A+
Now for the centroid, haha.
E: Duh, integrate h
wydy and divide it by the original result. :|
You guys rock.
Now for the centroid, haha.
E: Duh, integrate h
wydy and divide it by the original result. :| You guys rock.
Greetings, Gaf.
I'm currently enrolled in Introduction to Logic and I was performing well on each assignment and exam.
We recently started to delve into Natural Deduction, and the Rules of Implication and Rules of Replacement. I'm struggling to efficiently practice this methodology – which is infuriating – as I have had no pitfalls up to this point.
Below follows an example with its answer already filled in.
Are there any websites or tools out there to assist one with learning and practicing the Natural Deduction method? I've attempted to search through Google several times and didn't find any satisfying results.
I'm currently enrolled in Introduction to Logic and I was performing well on each assignment and exam.
We recently started to delve into Natural Deduction, and the Rules of Implication and Rules of Replacement. I'm struggling to efficiently practice this methodology – which is infuriating – as I have had no pitfalls up to this point.
Below follows an example with its answer already filled in.
Are there any websites or tools out there to assist one with learning and practicing the Natural Deduction method? I've attempted to search through Google several times and didn't find any satisfying results.
Partial Gamification
Banned
MIT Open Courseware (from page 110) It is a powerpoint presentation but this last section walks through an example.
cpp_is_king
Member
Edit: i fail at the internet

(following from e.) ...assigning to each (u,v) the point (0, 0, 1) where the line through (0,0,1) and (u,v,0) intersects the unit sphere, as pictured above.
Exercise 1. Derive the formula given in Example 1(e) for the parametrization of the unit sphere.
The formula to be derived is:
http://www.math.poly.edu/courses/projective_geometry/chapter_three/img31.gif
(only the z-coordinate in the derived formula should be u^2+v^2-1 in the numerator.
My attempt at a solution uses the parametrization of a unit circle, as seen here:
http://mathnow.wordpress.com/2009/11/06/a-rational-parameterization-of-the-unit-circle/
Although I can see the relation between the two, I cannot figure out how to arrive at the solution.
(following from e.) ...assigning to each (u,v) the point (0, 0, 1) where the line through (0,0,1) and (u,v,0) intersects the unit sphere, as pictured above.
Exercise 1. Derive the formula given in Example 1(e) for the parametrization of the unit sphere.
The formula to be derived is:
http://www.math.poly.edu/courses/projective_geometry/chapter_three/img31.gif
(only the z-coordinate in the derived formula should be u^2+v^2-1 in the numerator.
My attempt at a solution uses the parametrization of a unit circle, as seen here:
http://mathnow.wordpress.com/2009/11/06/a-rational-parameterization-of-the-unit-circle/
Although I can see the relation between the two, I cannot figure out how to arrive at the solution.
cpp_is_king
Member
not sure i understand what part d is asking. a is a line, how can it enclose an area? or does it mean the area of the entire ellipse?
edit: nvm i get it, a *is* the ellipse
edit: nvm i get it, a *is* the ellipse
Greetings, Gaf.
I'm currently enrolled in Introduction to Logic and I was performing well on each assignment and exam.
We recently started to delve into Natural Deduction, and the Rules of Implication and Rules of Replacement. I'm struggling to efficiently practice this methodology which is infuriating as I have had no pitfalls up to this point.
Below follows an example with its answer already filled in.
Are there any websites or tools out there to assist one with learning and practicing the Natural Deduction method? I've attempted to search through Google several times and didn't find any satisfying results.
I'm not exactly sure what you are asking, but there doesn't exist any computer tools/websites that will really help with doing proofs. It's something you just kind of have to practice and develop an intuition about.
cpp_is_king
Member
I just need help on parts a and d.
Not sure if I did this right, because the notation is still confusing me a little bit, but for part a, I simply calculated p'(theta) as dp / d theta and then you get:
p(theta) = cos(theta) x + sin(theta) y
p'(theta) = -sin(theta) x + cos(theta) y
Then, I plugged these two equations into the expression given for a(theta) and arrived at:
a(theta) = (x, y)
The problem already states that p(theta) is tangent to the curve at points (x, y), so... done?
Maybe I'm still confused about what the problem is asking though, because it seems like there should be something else.
Not really sure about part d
My professor told me to think of the angles as being different and the lines to the new angles form a triangle, where r is the hypotenuse, and an equation for the triangle is: cos(\phi - \theta) = p(\phi)/r
and then we somehow figure it out from there...but yeah the whole thing on these two parts just confuses me, and apparently my professor as well.
cpp_is_king
Member
Actually maybe i got it. The area should just be the integral of the radius from 0 to 2pi. So if you have some expression for the length of the radius at a given theya, you'd be done.
But you do have such an expression, because you have expressions for the x and y coordinates of an arbitrary point on the curve with the answe to part a. So take Sqrt(x^2 + y^2) and see what you get
Partial Gamification
Banned
I think she is referring to:
... from the right triangle: O P_0 P ("oh" "pee-naught" "pea")
Apologies for poor scan and if I am just stating the obvious.
r_0 is the radius you need (and yet it does seem like this would be the root of the difference between the squares of x and y -according to the diagram). Yet, the angle measure needed, in this "triangle breakdown," is theta-naught and not theta. All I can really say is that I feel this is what your teacher was speaking of but I haven't worked it to see how this would (or if it would at all) change what cpp said.
I am inquiring about any tools to practice proofs or websites that offer lessons that break down the principles and rules further, as my book is fairly vague.
I am not looking for any sort of "calculators" or "shortcuts," as that would defeat the purpose of enrolling in the course.
Partial Gamification
Banned
https://oli.cmu.edu/jcourse/webui/guest/join.do?section=logic
Its a whole course online, free. You don't have to register, but your work will not be saved.
The site takes a little digging to find what you need, Modus Tollens is on Page 92, and Natrual Deduction is not a topic in its own right. There are examples at the bottom of most of the course pages (that I looked at).
Your best bet might be heading to the librabry and checking out a book full of problems to work, ideally one with solutions for reference.
As a tip to you or anyone else that doesn't know (maybe you do):
google searches can be limited by adding site:.edu (or whatever .de // .org // neogaf.com)
and this will limit searches to the given restriction (file: works too).
I looked briefly and my search terms brought up a lot of syllabi for
"Natural Deduction"+"Practice Exercises" site:.edu
Okay I did that, but I still did not arrive at the correct answer...I don't know if I'm doing it wrong, but I get \sqrt(p(\theta)^2 + p'(\theta)^2) when I do \sqrt(x^2+y^2), also I'm not sure where to take it from there.
Partial Gamification
Banned
That is the area differential and does produce the given integral. Good insight.
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