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 -- 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...
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 teh 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)
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.
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?
