course Mth 174 ????O-?????n??assignment #017
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12:13:11 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 --> I first found the second derivative of cos(omega*t): ( cos(omega*t) )'' = -2 omega cos(omega*t) I then substituted this into the equation along with y: -2 omega cos(omega*t) + 9(cos(omega*t)) = 0 I then solved for omega: 9 cos(omega*t) = 2 omega cos(omega*t) 9 = 2 omega omega = 9/2
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12:22:30 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 --> First I found the derivative of P: P' = e^(-t) / ( 1 + e^(-t) )^2 I then substituted this into the equation given: e^(-t) / ( 1 + e^(-t) )^2 = 1/( 1 + e^(-t) ) * ( 1 - 1/( 1 + e^(-t) ) ) I noted that 1 = ( 1 + e^(-t) ) / ( 1 + e^(-t) ) and substituted: e^(-t) / ( 1 + e^(-t) )^2 = 1/( 1 + e^(-t) ) * ( ( 1 + e^(-t) ) - 1 ) / ( 1 + e^(-t) ) e^(-t) / ( 1 + e^(-t) )^2 = 1/( 1 + e^(-t) ) * ( e^(-t)/( 1 + e^(-t) ) e^(-t) / ( 1 + e^(-t) )^2 = e^(-t) / ( 1 + e^(-t) )^2
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12:22:35 how did you show that the given function satisfies the given equation?
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RESPONSE --> See above.
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12:23:09 What is the derivative dP/dt?
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RESPONSE --> dP/dt = e^(-t) / (1 + e^(-t))^2
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12:24:16 Does P(1-P) simplify to the same expression? If you have already shown the details, show them now.
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RESPONSE --> P(1-P) simplifies like this: 1/( 1 + e^(-t) ) * ( 1 - 1/( 1 + e^(-t) ) ) = 1/( 1 + e^(-t) ) * ( ( 1 + e^(-t) ) - 1 ) / ( 1 + e^(-t) ) = 1/( 1 + e^(-t) ) * ( e^(-t)/( 1 + e^(-t) ) = e^(-t) / ( 1 + e^(-t) )^2
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12:31:28 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 --> a) y'' = y. Equation: y = e^x + e^(-x) b) y' = -y. Equation: none. c) y' = 1/y. Equation: y = 'sqrt(2x) d) y'' = -y. Equation: y= cos(-x) and y = cos(x) e) x^2 y'' - 2y = 0. Equation: y = x^2
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12:31:45 which solution(s) correspond to the equation y'' = y and how can you tell?
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RESPONSE --> I checked by taking derivatives and substituting.
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13:23:35 which solution(s) correspond to the equation y' = -y and how can you tell
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RESPONSE --> I couldn't find a solution that solved this equation.
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13:27:06 which solution(s) correspond to the equation y' = 1/y and how can you tell
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RESPONSE --> I found that y = 'sqrt(2x) corresponded to this equation. I checked by taking derivatives: y' = (2x)^(-1/2) or 1 / 'sqrt(2x)
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13:28:15 which solution(s) correspond to the equation y''=-y and how can you tell
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RESPONSE --> I found two solutions to this equation, y= cos(x) and y = cos(-x). I checked by taking derivatives: y = cos(x) y' = -sin(x) y'' = -cos(x) y = cos(-x) y' = sin(-x) y'' = -cos(-x)
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13:29:46 which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell
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RESPONSE --> I found that y = x^2 solved this equation. I checked by taking derivatives and substituting. y' = 2x y'' = 2 x^2(2) - 2(x^2) = 0
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13:41:30 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 --> For the solution through (0,0), the line is on the x axis, no changes. For the solution through (1,4), the line starts on the x-axis and curves upward towards 10. There seems to be an inflection point about y=5 where the slope changes from positive to negative. It looks like there are asymptotes at the x-axis and at y=10. The graph seems to be concave up from x= -2 to x= 1, and concave down from x=1 to x=2.
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13:42:37 Query problem 11.2.10 (was 10.2.6) slope field
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RESPONSE --> a) Graph III b) Graph I c) Graph II d) Graph IV e) Graph V f) Graph VI
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13:44:57 describe the slope field corresponding to y' = x e^-x
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RESPONSE --> For negative numbers, the slope goes from very negative to 0. When x is positive, the slope becomes slightly positive for small values of x, moving towards 0 for large values of x.
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13:47:48 describe the slope field corresponding to y' = sin x
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RESPONSE --> Quite simply the slope corresponds to a recurring curve going from 0 to 1 to 0 to -1 to 0. The antiderivative of this is -cos(x), so it matches graph I.
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13:57:58 describe the slope field corresponding to y' = cos x
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RESPONSE --> Basically the same curve as the sin(x) curve, but shifted to the left. The antiderivative is sin(x), so it matches graph II.
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14:02:46 describe the slope field corresponding to y' = x^2 e^-x
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RESPONSE --> For negative values of x, the slope is large and becomes smaller as it moves towards 0 where it becomes 0. For positive values of x, it becomes somewhat positive then shrinks towards 0. I believe it matches graph IV.
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14:04:04 describe the slope field corresponding to y' = e^-(x^2)
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RESPONSE --> For large negative or positive values of x, the value of the slope is basically 0. Near 0 on either side, the value of the slope is slightly positive, so it matches graph V.
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14:08:14 describe the slope field corresponding to y' = e^-x
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RESPONSE --> For large negative values of x, the slope is positive and large, moving towards 0 as x moved toward 0. For positive values of x, the slope shrinks toward 0 very quickly. This matches graph VI.
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14:08:36 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Interesting, especially the slope fields.
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