course Mth 174 A couple of my answer were not inserted correctly. I am not sure why. I may have hit the spacebar or something without meaning to. But I do feel like I have grasped section 11.1 really well. And as usual, I need to practice reading and interpreting graphs, but 11.2 was not very difficult to grasp. xΐΛτΆΏ»ιvκΖ{½]ω}assignment #017
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18:40:07 Query problem 11.1.10 Find a value of omega such that y '' + 9 y = 0 is satisfied by y = cos(omega * t).{}{}Give your solution.
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RESPONSE --> My first step in this was to find y''. y'' = omega*cos(omega*t) therefore omega * cos(omega * t) + 9 * cos(omega * t) = 0 My next step was to factor out cos(omega * t) to get cos(omega * t) * (omega + 9) = 0 I set each term equal to zero: cos(omega * t) = 0 and omega + 9 = 0 This means that when omega = -9 the function = 0. I also concluded that when omega * t = pi/2 it equals zero also, but only because cos(pi/2) is equal to zero.
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19:34:48 Query problem 11.1.14 (3d edition 11.1.13) (was 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P)
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RESPONSE --> I worked on this problem for longer than usual. My first mistake was that I was unsure about the derivative of 1 / (1+e^-t). This is the same as (1 + e^-t) ^ -1. So I found the derivative to be -1 * (1 + e^-t) ^ -2. Also, by the quotient rule, the derivative is e^-t / (1 + e^-t) ^ 2.
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19:34:51 how did you show that the given function satisfies the given equation?
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RESPONSE -->
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19:35:16 What is the derivative dP/dt?
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RESPONSE --> (e ^ -t) / (1 + e ^ -t) ^2
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19:43:20 Query problem 11.1.16 (3d edition 11.1.15) (was 10.1.1) match equation with solution (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) )
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RESPONSE --> .
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20:36:42 which solution(s) correspond to the equation y'' = y and how can you tell?
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RESPONSE --> y = x^2 corresponds when sqrt(2) = x. y'' of x^2 = 2. So for 2 = x^2, x must equal sqrt(2). Also, e^x + e^-x corresponds. The second derivative of this is itself, so y'' = y.
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20:39:38 which solution(s) correspond to the equation y' = -y and how can you tell
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RESPONSE --> The only one I found that corresponds was y = x^2. y' = 2x. 2x = -(x^2) Therfore when x = -2, this works. 2(-2) = -4 and -(-2*-2) = -4.
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20:41:44 which solution(s) correspond to the equation y' = 1/y and how can you tell
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RESPONSE --> I only found one that corresponds. It is y = x^2 when x = (1/2) ^ (1/3) 2(1/2) ^ (1/3) = 1 / (1/2) ^ (1/3) ^ 2
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20:47:01 which solution(s) correspond to the equation y''=-y and how can you tell
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RESPONSE --> Both y = cos(x) and y = cos(-x) correspond with this equation. y'' of the first is -cos(x) and y'' of the second is -cos(-x). For the first: -cos(x) = cos(x), it is this simple. For the second: -cos(-x) = -cos(-x)
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20:48:03 which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell
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RESPONSE --> I found y = x^2 to correspond with this for all x. The sceond derivative is 2. Therefore x^2(y'') - 2y = 0 is equal to 2x^2 - 2x^2 = 0
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20:59:11 Query problem 11.2.5 (3d edition 11.2.4) The slope field is shown. Plot solutions through (0, 0) and (1, 4).{}{}Describe your graphs. Include in your description any asymptotes or inflection points, and also intervals where the graph is concave up or concave down.
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RESPONSE --> Obviously I have two points: (0, 0) and (1, 4) The solution at (0, 0) is a horizontal line. the solution at (1, 4) is a line that starts at y = 0 and is concave up to (1, 4) then changes to concave down when y = 10
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21:07:00 Query problem 11.2.10 (was 10.2.6) slope field
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RESPONSE -->
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21:14:27 describe the slope field corresponding to y' = x e^-x
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RESPONSE --> Graph III. It's slopes to the left of the y-axis start out as zero and then slowly change into undefined as it gets closer to the y-axis.
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21:15:24 describe the slope field corresponding to y' = sin x
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RESPONSE --> This graph actually looks like the sin function itself.
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21:15:52 describe the slope field corresponding to y' = cos x
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RESPONSE --> Similar to the behavior of the sin graph, this slope field acts simlarily to its actual graph.
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21:17:42 describe the slope field corresponding to y' = x^2 e^-x
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RESPONSE --> Graph V. From left to right it has slopes of zero, then as it gets closer to the y-axis it rises, then levels off backto zero on the opposite side as it began to rise.
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