course Mth 174 æ ôëžþU„ØÖŠ¦ºæŒ‰J£¡×‰ë°D¸Óûþï¾assignment #018
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21:54:09 Query problem 11.3.4 (was 10.3.6) Euler y' = x^3-y^3, (0,0), `dx = .2, 5 steps
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22:15:40 what is your estimate of y(1)?
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RESPONSE --> To find this I went from P(0) to P(5). P(0): x = 0 and y = 0. We already know that the next x will be x(0) + deltax = 0 + 1/5 = 1/5. To find the next y value I needed to find delta y. deltay = slope * delta x. The slope is dy/dx = x^3 - y^3, which at this point is (0) ^ 3 - 0 = 0. So at P(1), x = 1/5 and y = 0. To find P(2) I went through the same process. I will do this without all of the commentary. The next x value is (1/5) + (1/5) = 2/5. The slope is (1/5)^3 - (0)^3 = (1/125). delta y = (1/125) * (1/5) = 1/625. So at P(2), x = 2/5 and y = 1/625. The next x value is 2/5 + 1/5 = 3/5. The slope is (2/5)^3 - (1/625)^3 = .06399 (approx). delta y = [(2/5)^3 - (1/625)^3] * (1/5) = .012799. The new y value is y + delta y = .014399 (approx.) So at P(3), x = 3/5 and y = .014399 (approx) The next x value is 3/5 + 1/5 = 4/5. The slope is (3/5)^3 - .014399^3 = .21599. delta y = .21599 * 1/5 = .043199. The next y value is y + delta y = .014399 + .043199 = .057598 (approx) So at P(4), x = 4/5 and y = .057598 (approx). The next x value is 1. The slope is (4/5)^3 - (.057598)^3 = .511808. delta y = .511808 * 1/5 = .102361 (approx). The final y value is .057598 + .102361 = .1599597 (approx)
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22:17:30 Describe how the given slope field is consistent with your step-by-step results.
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RESPONSE --> I am not real sure other than at the points of x and y's that I came up with do seem to match the slope field graph, but I do not know of a way to concludethis exactly.
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22:18:36 Is your approximation an overestimate or an underestimate, and what property of the slope field allows you to answer this question?
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RESPONSE --> This part of the slope field graph is concave up, so I would say that my answer is an underestimate.
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22:35:41 Query problem 11.3.10 (was 10.3.10) Euler and left Riemann sums, y' = f(x), y(0) = 0
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23:06:19 explain why Euler's Method gives the same result as the left Riemann sum for the integral
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RESPONSE --> I do not really know why exactly. I thought I had an answer but when I started to type it into my work, it sounded rediculous and I could not prove anything.
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23:20:26 Query problem 11.4.19 (3d edition 11.4.16) (was 10.4.10) dB/dt + 2B = 50, B(1) = 100
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23:31:33 what is your solution to the problem?
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RESPONSE --> My solution: I first set it up to where each side of the equation had one variable. dB/dt + 2B = 50. I set dB/dt = 50 - 2B. Then I divided by 2 to get (1/2)( dB / dt) = 25 - B. I rearranged it to (1 / (25 - B)) dB = 2 dt. I then took the integral of both to be ln abs(25 - B) = 2t + C. Which is abs(25 - B) = e^2t+C = e^2t * e^C. Which equals abs(25 - B) = A * e^2t, where A = e^C. -> 25 - B = +-A * e^2t = -B = D * e^2t - 25, where D = +-A. So, B = -D * e^2t + 25. If B(1) = 100, then 100 = -D * e^2 + 25 -> -75 = D*e^2 -> D = -75 / e^2. I used D as a substitute for +- A so I wouldn't get confused. But in the end it didn't help. This is as far as I got and couldn't come up with an exact answer.
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23:31:36 What is the general solution to the differential equation?
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23:31:38 Explain how you separated the variables for the problem.
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23:31:40 What did you get when you integrated the separated equation?
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23:42:23 Query problem 11.4.40 (3d edition 11.4.39) (was 10.4.30) t dx.dt = (1 + 2 ln t ) tan x, 1st quadrant
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RESPONSE --> I did well at setting this up, but I didn't turn in an answer becuase I couldn't get past finding the integrals for each side. I had it set up (1 / tan(x)) dx = (1 + (2ln(t)/t)) dt. I spent a good while trying to find the integrals, but I guess I'll have to go back and practice up.
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23:42:25 what is your solution to the problem?
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23:42:27 What is the general solution to the differential equation?
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23:42:29 Explain how you separated the variables for the problem.
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23:42:31 What did you get when you integrated the separated equation?
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23:42:32 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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