Homework 4 work area
Collaborate on HWK 4 here.
Section 8.4 #29 Does anybody have any thoughts about the solution to the second part of number 29 and the proof for #30? For the positive definite case and negative definite case, finding the determinate seems sufficient. I'm not sure how to show the indefinite case.
(rekblad 9/18) for #29 part 2 showing Q is indefinite, isn't it enough to just find two vectors that show Q > 0 and Q < 0 and also show that Q!=0 for x!=0 ? (Actually, on second thought, I think Q indefinite => Q may = 0 for some x!=0)
(Response to rekblad and top poster) Since prob. 29 only asks to show that it is indefinite, I believe using two vectors showing Q>0 and Q<0 is sufficient. --Rayala 20:40, 19 September 2010 (UTC)
Problem 18 on page 146 Do I have the time rate of change equations correct:
Y1' = 48/100Y1 + 16/400Y2 - 64/100Y1 Y2' = 64/100Y1 - 64/100Y2
I am not sure on the 48/100Y1 portion of equation 1.
Reply --Maloner 21:12, 18 September 2010 (UTC) I dont think you are correct. I think everything should have 400 in the denominator, except for maybe the 48gpm pure water inlet. I am not sure what to do with that inlet stream?
Reply-- --Rickey I think the only thing that the 48 gpm of pure water going into tank 1 is good for is helping you know that at all times, there are 400 gallons of fluid in tank 1. Also I think, like Maloner said, you want 400 in the denominator for everything (you should see that there are always 400 gallons in each tank at all times). Also, since Y1'= the rate of change of fertilizer in the tank, you don't need the 48/100Y1 term in there at all, because that stream is not delivering any fertilizer. Rate of fertilizer coming in from a given stream = (rate of gallons of fluid coming in) * (# lbs of fertilizer/gallons of fluid) = (rate of gallons of fluid coming in) * ( concentration of fertilizer in that stream). Please let me know if I'm completely wrong here, but this is how I've started the problem.
Reply -- I think you are absolutely right.
Reply -- (jmarin) So what you are saying is that the 48 gallon/min inlet at T1 and the outlet at T2 should not be included in the equations? I think they should be included since the rate of change of the tanks is the inflow - outflow and in T1 you have an inflow of 48 gpm and in T2 you have an outflow of 48 gpm. So I included them in the equations as follows:
y'1 = -64/400y1 + 16/400y2 + 48/400 y'2 = 64/400y1 - 16/400y2 - 48/400
I'm not completely sure about that either but its how I worked mine.
Reply-- --Rickey You DO need to care about the pure water coming in, but only for the denominators' sake in your Y1' equation (to know that there are 400 gal in each tank all of the time). I second what Rayala says below. So just get rid of that 48/400 term in your Y1' equation.
Reply -- based on the example problems I don't think you need to address the initial fertilizer concentration. I'm not positive about this, but none of the examples did.
Reply --cslabau I believe the initial fertilizer mass is only needed in the end to evaluate the constants c1 and c2 - they are your initial conditions. Also, I agree with Rickey that there is no term for the pure water inlet in the fertilizer mass balance of T1. The fertilizer 'density' would equal zero and the term will disappear...
Reply ---Rayala 23:11, 20 September 2010 (UTC) Jmarin, as cslabau and Rickey point out, y'1 is the rate of change of fertilizer in T1. Since the 48gpm is pure water, there is no fertilizer being added and thus not part of y'1. However, 48gpm in the outlet of T2 does include fertilizer so it is part of y'2. Using your equations and this information, your equations should be:
y'1 = -64/400y1 + 16/400y2 y'2 = 64/400y1 - 16/400y2 - 48/400y2
Need help with P356 #29 and 30. This is my understanding of #29-> Positive definiteness of Prob23: [x1 x2]^T * [4 Sqrt(3), Sqrt(3) 2] * [x1 x2] >0 for all X(vector) not equal to 0(vector). So, 4 > 0, and the det([4 Sqrt(3), Sqrt(3) 2]) >0. Therefore, it is positive definite. And for Prob19: [x1 x2]^T * [1 12, 12 -6] * [x1 x2], 1 > 0 and det ([1 12, 12 -6]) < 0. Therefore, it it indefinite.
Response: I might be wrong and just being nitpicky but prob. 29 says "....A necessary and sufficient condition for positive definiteness is that all the "principal minors" are positive". It doesn't explicitly say that if det < 0 it is indefinite or negative. I think the way to approach the 2nd part of Prob. 29 is to use examples as rekblad did above. --Rayala 20:40, 19 September 2010 (UTC)
I am also having difficulty with P356 #29 & 30. I have found the Eigen values and Eigen vectors, placed the Eigenvectors into matrix, and solved the diagonal matrix. I don't know where to go from here??? Any help??
Response: For 29, rekblad has a good suggestion for the 2nd part and other posters have used the hint from the problem regarding determinants. For 30, using Theorem 5 and equation (10) [ page 353] from the book provide us good direction on this problem. Also, you could use your answer to 30 to solve 29 in case anyone skipped 29. --Rayala 20:40, 19 September 2010 (UTC)
Words from Bell: I like what I'm seeing here! This is what the rhea is for.
Here are a couple hints. Problem 29 only asks you to show two things. First, you need to show that the matrix from problem 23 is positive definite using the determinant conditions that are given. Second, you need to show that the matrix for problem 19 is indefinite. For this, you'll need to find values to plug in that make it positive, and other values that make it negative. (That's not hard because you can see terms with positive coefficients and terms with negative coefficients in the sum for 19.)
Problem 30, however, asks you to think a little deeper. The key is to use the change of variables to express the quadratic form Q as
$ Q=\sum_{j=1}^n \lambda_j y_j^2. $
The signs of those eigenvalues λj say it all.
Question: For p.355 #13: Is it just me, or are x = Py not eigenvectors of A? The back of the book says that x = [-2; 0], [0; -1] but I get the eigenvectors of A to be [1; 0], [0; 1]. Any thoughts? Thanks.
Note: If a is an eigenvector for an eigenvalue r, so is any non-zero constant times a. Your eigenvectors are just constants times the book's eigenvectors, and that's ok. --Steve Bell 19:53, 20 September 2010 (UTC)
Reply: You can verify that [-2, 0] and [0, -1] are eigenvectors of A using the equation Ax = lambda*x...which also works for [1, 0] and [0, 1].--Wmitche 17:39, 20 September 2010 (UTC)
Reply: I am also getting [1, 0], [0, 1]. I think it might not matter b/c x1 is a free variable for one eigenvalue, and x2 a free variable for the other...? And then maybe they picked [-2, 0] and [0, -1] b/c of the second part of the problem?
