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The Math Help Thread
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The exact value is -12, so your answer seems all right. In order for us to check your answer, you have to tell us how you approximated the volume under the surface f. (Are you using one of the four corners or the midpoint of each subdomain? Are you using the volume of a rectangular box or some other shape?) Like in 1D calculus, there are many ways to do the approximation.
Oh right, I forgot about that part. I was told to use the lower right corners.
Basically what I did was made a graph, with the intervals given by R for x and y, which makes a rectangle. Then I divided it into 2 along the x-axis, into 3 along the y-axis, then plotted each of the lower right corners. Then I found delta A by measuring the area of the subrectangles.
Then I plugged each of the coordinates given by the lower right corners into the function, added them all up, and multiplied by delta A, which gave -20.
It doensn't seem difficult, it's just that the negative answer was throwing me off because I thought I was calculating a volume.
Basically what I did was made a graph, with the intervals given by R for x and y, which makes a rectangle. Then I divided it into 2 along the x-axis, into 3 along the y-axis, then plotted each of the lower right corners. Then I found delta A by measuring the area of the subrectangles.
Then I plugged each of the coordinates given by the lower right corners into the function, added them all up, and multiplied by delta A, which gave -20.
It doensn't seem difficult, it's just that the negative answer was throwing me off because I thought I was calculating a volume.
The negative area/volume isn't a problem; we can think of them as positive area/volume under the x-axis/xy-plane.
I think the approximate answer is -12 (matches the exact answer by coincidence). If we choose the lower right corners, then we'd be evaluating f at these six points:
f(2, -1) = -1
f(4, -1) = -3
f(2, 0) = 1
f(4, 0) = 1
f(2, 1) = -1
f(4, 1) = -3.
Since each subdomain has an area of (2)(1) = 2, the approximate volume is given by,
2 * (-1 - 3 + 1 + 1 - 1 - 3) = 2 * -6 = -12.
uncleniccius
Member
"A small box is pushed along a floor. The floor is modelled as a rough horizontal plane and the box is modelled as a particle. The coefficient of friction between the box and the floor is . The box is pushed by a force of magnitude 100 N which acts at an angle of 30º with the floor, as shown in Figure 1.
Given that the box moves with constant speed, find the mass of the box."
I have never had to find the mass before, and I'm confused because it doesn't say it starts moving at this exact mass.. any help would be appreciated.
Given that the box moves with constant speed, find the mass of the box."
I have never had to find the mass before, and I'm confused because it doesn't say it starts moving at this exact mass.. any help would be appreciated.
Duebrithil
Member
I recommend writing Newton's second law for both X and Y axis. It should be a pretty straightforward answer once you've done that.
uncleniccius
Member
Will that apply when it isn't accelerating?
I've resolved vertically and horizontally and am now stuck.
Duebrithil
Member
Yup, if it isn't accelerating it just means a = 0. You should have something similar to this, and then it's just a matter of solving for M.
uncleniccius
Member
I'm not completely following the notation on yours, should I use the value I resolved for it horizontally as the f value?
Duebrithil
Member
From the Y axis you obtain N, which you'll introduce into the X axis through the force of friction (N·Mu, where mu is the friction coefficient). Then you just have to solve for M, with g = 9.8Mm/s^2 and F = 100 N
I'm probably not explaining myself very well, sorry about that
uncleniccius
Member
Thank you for your help!
I think I get it.. I'll have a go! I really don't like Mechanics much though
spelltropy
Member
Two quick linear algebra qs here, I understand the first one, but not sure how to write it out, while I'm not sure about the second one:
Trying to show that linear equations having no solutions implies that the first n columns of the augmented m x (n+1) matrix are not linearly combined in the (n+1)th column.
It feels obvious but I'm not sure how to express it rigorously..
The second q goes on from this: trying to show that the Rank of the augmented matrix above being greater than the Rank of the matrix created by removing the last column from the augmented matrix implies the lowest non-zero row of the row reduced form of the augmented matrix is (0,0,0,...,1).
Have no idea what to do for this one, would love any help
Trying to show that linear equations having no solutions implies that the first n columns of the augmented m x (n+1) matrix are not linearly combined in the (n+1)th column.
It feels obvious but I'm not sure how to express it rigorously..
The second q goes on from this: trying to show that the Rank of the augmented matrix above being greater than the Rank of the matrix created by removing the last column from the augmented matrix implies the lowest non-zero row of the row reduced form of the augmented matrix is (0,0,0,...,1).
Have no idea what to do for this one, would love any help
Omega Ultimus
Member
Two quick linear algebra qs here, I understand the first one, but not sure how to write it out, while I'm not sure about the second one:
Trying to show that linear equations having no solutions implies that the first n columns of the augmented m x (n+1) matrix are not linearly combined in the (n+1)th column.
It feels obvious but I'm not sure how to express it rigorously..
The second q goes on from this: trying to show that the Rank of the augmented matrix above being greater than the Rank of the matrix created by removing the last column from the augmented matrix implies the lowest non-zero row of the row reduced form of the augmented matrix is (0,0,0,...,1).
Have no idea what to do for this one, would love any help
First Question:
It's implied that the last row (in order for the matrix to not yield a solution) has zeros for every column except for the last (which equals some #). If this occurs, then the matrix is inconsistent. You cannot write it as a set of solutions.
Second Question:
Building off the First Question, if the augmented matrix has a higher rank (more non-zero rows), this implies an inconsistent matrix. Why? Because, the last row has zeros in each column of the (un-augmented) matrix (corresponding to a variable) equaling a # in (the b column vector).
Did this make sense?
EDIT: Here is an example:
[1 2 3 | 3]
|0 1 2 | 3|
[0 0 0 | 1]
spelltropy
Member
trying to wrap my head around the first one: so basically, if there are no solutions to the linear equations, then the final row of the augmented matrix must be (0,0,..,1)? And then 1 can't be expressed as a linear combination of {0,...,0}, so the final column as a whole can't be expressed as a linear combination of the first n columns? Am i on the right track?
Omega Ultimus
Member
Yup, that's it. Your final variable in that (0,0..., 1) column has no value that allows 0 to equal 1, so there is no set of solutions (because there is no value for that variable that can be used to solve the other equations).
The other answer to your first question seems to assume that the augmented matrix has been row-reduced, which is not how I understood the problem. If you would like to do this directly from the unreduced augmented matrix, remember that the augmented matrix [ A | b ] represents the linear system Ax=b, where x=(x1, x2, ..., xn)^T. If we index the columns of A by A=[ A1 | A2 | ... | An ], then matrix multiplication givesTwo quick linear algebra qs here, I understand the first one, but not sure how to write it out, while I'm not sure about the second one:
Trying to show that linear equations having no solutions implies that the first n columns of the augmented m x (n+1) matrix are not linearly combined in the (n+1)th column.
It feels obvious but I'm not sure how to express it rigorously..
The second q goes on from this: trying to show that the Rank of the augmented matrix above being greater than the Rank of the matrix created by removing the last column from the augmented matrix implies the lowest non-zero row of the row reduced form of the augmented matrix is (0,0,0,...,1).
Have no idea what to do for this one, would love any help
Ax = A1x1 + A2x2 + ... + Anxn = b.
In other words, if the system has a solution x, then we have explicitly written the (n+1)th column, b, as a linear combination of the first n columns, A1, ..., An.
Hypertrooper
Member
I just want to let you guys I know that I passed the exam. I feel sad about it because they cut two of the 11 tasks, so the points you could get were cut down from 68 to 44... So to pass the exam, you needed 34 first, but after the cutting down, 20 were needed to pass. I got 26.5 points. As you can see now I am sad about it. But what should I say, I was lazy and therefore I have to repeat some of the content during the last two weeks of my break.
But I want to thank all of you guys for your help with some exercises! Without you, I would have not even been closed to pass it!
But I want to thank all of you guys for your help with some exercises! Without you, I would have not even been closed to pass it!
I've heard horror stories about European exams and the high fail ratio from my best mates (although these were when they were in undergrad). You should be happy, congratulations!
Charmicarmicat
Banned
I have a question about plotting log-linear graphs in Excel, for a paper I'm currently working on.
So I have two values, N and t. I have to find the natural logarithm for N, and plot this against the value for t (not the log of t).
The problem I'm encountering is Excel wont let me set the natural logarithm value of e, only proper numbers. Do I use the value of e for the logarithmic scale in Excel, or will that not work?
Alternatively, is there a better, fairly user-friendly graphing program out there that will allow me to do what I want?
So I have two values, N and t. I have to find the natural logarithm for N, and plot this against the value for t (not the log of t).
The problem I'm encountering is Excel wont let me set the natural logarithm value of e, only proper numbers. Do I use the value of e for the logarithmic scale in Excel, or will that not work?
Alternatively, is there a better, fairly user-friendly graphing program out there that will allow me to do what I want?
I'm a bit confused. Isn't the number e expressed as exp(1) in Excel?
Stumpokapow
listen to the mad man
LightOfTruth
Member
Hey guys, I could really use help with this problem:
Here I need to solve for R in terms of L and c. Also, at the bottom of the picture I wrote something wrong. I said c, which equals 0.5, is the arc-length of each semi-circle, but I really meant to say each quarter circle. My bad. I'm not given a number for L so that can just remain a variable.
Can anyone see the solution? I wasn't able to figure it out without introducing an angle. I'd appreciate any help.
Here I need to solve for R in terms of L and c. Also, at the bottom of the picture I wrote something wrong. I said c, which equals 0.5, is the arc-length of each semi-circle, but I really meant to say each quarter circle. My bad. I'm not given a number for L so that can just remain a variable.
Can anyone see the solution? I wasn't able to figure it out without introducing an angle. I'd appreciate any help.
Helpless Hometown Motor
Member
I think I'm going crazy trying to figure out what's going wrong here, but I have no clue what's going on.
So I had a question that said to evaluate this integral using Green's theorem:
I did that and I got -16π.
Now, to check my work, I did the integral using the other line integral method, where I parameterized the circle as:
x=4cos(t)
y=4sin(t)
I took, the partial derivatives, substituted everything, and crunched the integral on my calculator to get:
+16π.
I have no idea why the two are off by a negative sign. I've checked my work for both problems and can't find any mistake.
Is there a difference between doing line integrals with Green's theorem vs doing line integrals using the other method, or should they get the same answer (or maybe I'm just stupid)? (And thanks in advance!)
So I had a question that said to evaluate this integral using Green's theorem:
I did that and I got -16π.
Now, to check my work, I did the integral using the other line integral method, where I parameterized the circle as:
x=4cos(t)
y=4sin(t)
I took, the partial derivatives, substituted everything, and crunched the integral on my calculator to get:
+16π.
I have no idea why the two are off by a negative sign. I've checked my work for both problems and can't find any mistake.
Is there a difference between doing line integrals with Green's theorem vs doing line integrals using the other method, or should they get the same answer (or maybe I'm just stupid)? (And thanks in advance!)
boviscopophobic
Member
They should be the same, so I'd check for a sign error in the line integral computation -- maybe in taking the derivative of the parameterization?
Helpless Hometown Motor
Member
Ha you're right! Looks like I left off some parenthesis when doing it the second way.
Thanks for the help.
Helpless Hometown Motor
Member
Stumpokapow
listen to the mad man
I'm not extremely confident in my answer here given the specification of the question but:
Z intersects the circle such that the E1-center-Z angle is a 90 degree right triangle--you should be able to see why.
The circumference of a circle = 2 pi r--little r being radius, not big R in your problem. So if the circumference of half the circle (2c) = 1, then the so the circumference of the circle is 4c = 2.
Thus 2 = 2 pi r. So r = 2 / (2 pi) = 1/pi. We also know that a = r in this case, since any line from the center of a circle to the circle is a radius of the circle. You now have a.
a, L, and R form a right triangle, so a^2 + L^2 = R^2, thus R = sqrt(a^2 + L^2) = sqrt((1/(pi))^2 + L^2) = sqrt(1/((pi^2)) + L^2)
If you want to solve in terms of c without using the value "0.5" at all, you would do:
4c = 2pi r
r = 4c/2pi = 2c/pi
R^2 = (2c/pi)^2 + L^2
R = sqrt((2c/pi)^2 + L^2)
AKingNamedPaul
I am Homie
I stopped understanding math after third grade. I just stopped in here to feel stupid.
spelltropy
Member
Cheers for the help earlier Therion & Omega - I ended up using both of your suggestions, just Omega's for a different part of the proof.
I've got another linear algebra question I wouldn't mind some help with:
I can't really figure this one out. For n=2, the determinant is a2-a1. This is equivalent to the rhs where the only i,j values are 1,2, so the product just ends up being a2-a1?
For n=3, writing out the determinant using cofactors, I got a3^2a2-a2^2a3-a1a3^2+a1^2a3+a1a2^2-a1^2a2. What would the product from the rhs be in this case? [e/ never mind, i think this was straightforward and the product is just (a3-a1)(a2-a1)(a3-a2). Wouldn't mind help with the induction part though!]
Wouldn't mind some help with this and the whole induction part afterwards and sorry for the shitty writing out above, i'll be happy to rewrite it in latex or something and screenshot if that would help.
I've got another linear algebra question I wouldn't mind some help with:
I can't really figure this one out. For n=2, the determinant is a2-a1. This is equivalent to the rhs where the only i,j values are 1,2, so the product just ends up being a2-a1?
For n=3, writing out the determinant using cofactors, I got a3^2a2-a2^2a3-a1a3^2+a1^2a3+a1a2^2-a1^2a2. What would the product from the rhs be in this case? [e/ never mind, i think this was straightforward and the product is just (a3-a1)(a2-a1)(a3-a2). Wouldn't mind help with the induction part though!]
Wouldn't mind some help with this and the whole induction part afterwards
and sorry for the shitty writing out above, i'll be happy to rewrite it in latex or something and screenshot if that would help.Cheers for the help earlier Therion & Omega - I ended up using both of your suggestions, just Omega's for a different part of the proof.
I've got another linear algebra question I wouldn't mind some help with:
Here is my solution:
I didn't write down the general case, but you should be able to see what to do from the 3 x 3 case above. Just let me know if you need clarification.
Edit: Sorry, I don't think my proof works for the general n x n matrix. I falsely assumed that the (n - 1) x (n - 1) cofactor is a Vandermonde matrix, but it's not except when n = 3. Back to the drawing board...
spelltropy, I found a proof online, hope that's ok by you.
It starts out similarly to mine (add a scalar multiple of row to other rows), but shows that we also need to add a scalar multiple of column to another (the next one) in order to get a Vandermonde submatrix. I personally don't like hard-to-see tricks like this; I wonder if there was an easier, more natural way to solve the problem.
https://proofwiki.org/wiki/Vandermonde_Determinant#Proof_1
It starts out similarly to mine (add a scalar multiple of row to other rows), but shows that we also need to add a scalar multiple of column to another (the next one) in order to get a Vandermonde submatrix. I personally don't like hard-to-see tricks like this; I wonder if there was an easier, more natural way to solve the problem.
https://proofwiki.org/wiki/Vandermonde_Determinant#Proof_1
dr3upmushroom
Banned
Find and sketch each function according to the conditions given on the same set of axes below.
Label all intercepts as points, indicate the end behavior and check with a graphing device.
1. The function f(x) is a degree 4 polynomial with zeros at x = −1, 2, 4 and x = 6 that satisfies
f(1) = −6
So here's my problem?
The way I understand it is, that it translates to f(x)=(x+1)(x-2)(x-4)(x-6) right?
But I get confused with the f(1)=-6 part... do I plug in the 1 into the formula to get the -6... Which doesn't work out that way.
Label all intercepts as points, indicate the end behavior and check with a graphing device.
1. The function f(x) is a degree 4 polynomial with zeros at x = −1, 2, 4 and x = 6 that satisfies
f(1) = −6
So here's my problem?
The way I understand it is, that it translates to f(x)=(x+1)(x-2)(x-4)(x-6) right?
But I get confused with the f(1)=-6 part... do I plug in the 1 into the formula to get the -6... Which doesn't work out that way.
Multiply the whole thing by a constant that will adjust it to being equal to -6 when f(1). As you can see, it won't change the fact that the function is a zero at x=-1, 2, 4, 6.
The_Inquisitor
Member
f(1) = -6 is a boundary condition I believe. You need to shift the equation s.t f(1) = 6. I don't know if this may help but recall for a line y = mx +b, where m is some constant. So in your case, @ 1 ->
-6 = 2*-1*-3*-5*C
C = -6/30 = -1/5
So your final equation would be f(x) = -1/5(x+1)(x-2)(x-4)(x-6).
Disclaimer: This may be wrong.
dr3upmushroom
Banned
The_Inquisitor
Member
Glad we could help. Verify on Wolfram alpha/ your calculator to verify for yourself.
dr3upmushroom
Banned
actually I'm still a bit confused, doing the steps does give me -1/5 but the point doesn't come up on Desmos... It's only when I make the 1/5 a positive, that it does show the point (1,-6).
Omega Ultimus
Member
It's just a sign error. 1/5=C is the value you want.
The_Inquisitor
Member
Because I am very tired after doing circuit analysis all evening.
It's actually -6/-30 = 1/5 There are 3 negative signs on the right side, so it cancels with the left side negative.dr3upmushroom
Banned
Because I am very tired after doing circuit analysis all evening.
yea, just saw that too. I reworked the problem.
Thanks Math Gaf!
It's asking you to find the transpose of the matrix A. The more common/correct notation is A^T (put the letter T as a superscript). The prime/apostrophe was probably there to follow Matlab's notation.
Certinty
Member
Ah that makes sense then, thanks a lot. Much appreciated.
nicoga3000
Saint Nic
This should be a lot easier for me than it is...Anyone care to help?
I need the zeros of this equation:
I need the zeros of this equation:
This should be a lot easier for me than it is...Anyone care to help?
I need the zeros of this equation:
I'm not sure if we can; you will want to use Mathematica or other software. In general, it's difficult to write down the closed form of the solution for quartic equations, more so with variables like in your equation.
I'm a little puzzled, though. I see that you're working with beams, so R_a and w_a likely denote the reaction force and the deflection at node a. If so, x has the unit of length, and the dimensions in your terms end up being inconsistent: the slope theta_a is dimensionless, x^2/(EI) * R_a/2 is dimensionless, but the last two terms have the dimension of m^4/(Pa * m^4) = 1/(Pa).
Partial Gamification
Banned
This should be a lot easier for me than it is...Anyone care to help?
I need the zeros of this equation:
That is no equation, that is an expression.
Set it equal to zero.
(-)theta_a = [-12*L*(R_a)*(x^2) + 4*L*(w_a)*(x^3) - (w_a -w_b)*(x^4)] / (24*L*E*I) {I think the signage got mixed in my head but it gives a polynomial to solve}
theta_a = [12*L*(R_a)*(x^2) - 4*L*(w_a)*(x^3) + (w_a -w_b)*(x^4)] / (24*L*E*I)
or better yet, a nice polynomial to solve:
0= 12*L*(R_a)*(x^2) - 4*L*(w_a)*(x^3) + (w_a -w_b)*(x^4) - theta_a* (24*L*E*I)
R_a, w_a, w_b, theta_a, L, E, and I are assumed to be constants.
nicoga3000
Saint Nic
I an indeed working with beams.
The equation is actually a "simplified" slope equation from the manual of stress and strain (Roark 7th ed). Theta is in radians, x^2/EI comes out to 1/#, the stuff in the parenthesis comes out to #; it all simplifies out to radians. I'm actually equating to zero to find the max/min for the maximum deflection of the beam.
Ok, I see why I was confused. The w_a and w_b aren't nodal deflections, but parameters for a distributed load over the beam, with dimensions of force/length. Then, all four terms are have the same dimension (dimensionless/radians in small deformation theory).
nicoga3000
Saint Nic
You are correct! I should have specified, but I wasn't sure it mattered.
As for the final answer, I'm going to tinker around with it in Mathcad. I'll try out what Partial Gamification posted, too. Getting a general solution for this may prove to be more challenging than I initially thought.
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