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Reply --[[User:Maloner|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 --[[User:Maloner|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--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-- --[[User:Rickey|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. |
Revision as of 10:45, 19 September 2010
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)
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.
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.
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??
