Assignment 17

course Mth 174

ᦊѷϢassignment #017

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017. `query 17

Cal 2

04-09-2007

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13:05:41

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 -->

y = cos(omega*t)

y' = -omega*sin(omega*t)

y"" = -omega^2*cos(omega*t)

Substituting in y"" + 9y = 0:

-omega^2*cos(omega*t) + 9cos(omega*t) = 0

-omega^2*cos(omega*t) = -9cos(omega*t)

omega^2 = 9

omega = +3, -3

Both solutions check in the original equation.

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13:17:17

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 -->

p= 1/(1+e^-t) = e^t/(e^t+1)

dP/dt = e^t/(e^t+1)^2 (using quotient rule)

Substituting:

P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)]

P(1-P) = e^t/(e^t+1) * [(e^t+1)-e^t)/(e^t+1)]

P(1-P) = e^t/(e^t+1) * [1/(e^t+1)]

P(1-P) = e^t/(e^t+1)^2

Therefore, P(1-P) = dP/dt

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13:17:27

how did you show that the given function satisfies the given equation?

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RESPONSE -->

See previous response.

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13:17:40

What is the derivative dP/dt?

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RESPONSE -->

See previous response.

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13:18:00

Does P(1-P) simplify to the same expression? If you have already shown the details, show them now.

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RESPONSE -->

See previous response.

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13:18:17

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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13:32:48

which solution(s) correspond to the equation y'' = y and how can you tell?

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RESPONSE -->

Only solution IV corresponds to y"" = y

y = e^x + e^(-x)

y' = e^x - e^(-x)

y"" = e^x + e^(-x)

Thus, y = y""

Solution I, y = cos x, y' = -sin x, y""= -cos x, not = y

Solution II, y = cos(-x), y' = sin(-x), y"" = - cos(-x), not = y

Solution III, y = x^2, y' = 2x, y"" = 2, not = y

Solution V, y = sqrt (2x), y' = 1/sqrt(2x), y"" = -1/sqrt[(2x)^3], not = y

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13:38:53

which solution(s) correspond to the equation y' = -y and how can you tell

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RESPONSE -->

None of the solutions correspond to y' = -y

Solution I: y = cos x, y' = -sin x, not = -y

Solution II: y = cos(-x), y' = sin x, not = -y

Solution III: y = x^2, y' = 2x, not = -y

Solution IV: y = e^x + e^(-x), y' = e^x - e^(-x), not = -y

Solution V: y = sqrt(2x), y' = 1/sqrt(2x), not = -y

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13:45:52

which solution(s) correspond to the equation y' = 1/y and how can you tell

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RESPONSE -->

Solution V corresponds to y' = 1/y

y = sqrt(2x), y' = 1/sqrt(2x), y' = 1/y

None of the other solutions correspond (see previous responses).

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13:49:50

which solution(s) correspond to the equation y''=-y and how can you tell

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RESPONSE -->

Solutions I and II correspond to y"" = -y

Solution I: y = cos x, y' = -sin x, y"" = -cos x, y"" = -y

Solution II: y = cos(-x), y' = sin(-x), y"" = -cos(-x), y"" = -y

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13:54:43

which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell

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RESPONSE -->

Solution III corresponds to x^2*y"" - 2y = 0

Solution III: y = x^2, y' = 2x, y"" = 2

Substituting: x^2*(2) - 2(x^2) = 2x^2 - 2x^2 = 0

None of the other solutions correspond.

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14:12:55

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 -->

Solution through (0,0): concave up from (0,0) to about (3,3) then concave down for t>3, point of inflection at (3,3), horizontal asymptote at P=10. Graph appears to be P=0 for all t<0. Since givens include P>=0, no graph for negative P.

Solution through (1,4): concave up from (-1,0) to about (2,3) , then concave down for t>2, point of inflection at (2,3), horizontal asymptote at P=10. Graph appears to be P=0 for at t<-1. Since givens include P>=0, no graph for negative P.

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14:13:09

Query problem 11.2.10 (was 10.2.6) slope field

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14:22:17

describe the slope field corresponding to y' = x e^-x

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RESPONSE -->

Slope field corresponding to y' = x e^-x: Slope field III: slope increasingly negative for x<0, slope increasing positive (at decreasing rate) to about x=1, then decreasing positive, slope appears to have a limit of 0 for large x.

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14:25:41

describe the slope field corresponding to y' = sin x

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RESPONSE -->

Slope field for y' = sin x: Slope field I: Slope ranges from 0 to 1 to 0 to -1 to 0 repeatedly, with slope = 0 at (0,0), period = 2pi

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14:28:17

describe the slope field corresponding to y' = cos x

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RESPONSE -->

Slope field corresponding to y' = cos x: slope field II. Same as slope field for y' = sin x, except slope = 1 at x = 0 (cyled shifted pi/2 to left).

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14:32:27

describe the slope field corresponding to y' = x^2 e^-x

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RESPONSE -->

Slope field corresponding to y' = x^2 e^-x: Slope field IV: Slope increasingly positive as x decreases from 0, slope 0 at x=0, slope increasingly positive (at decreasing rate) from x=0 to x=2, then decreasingly positive, appears to approach 0 for large x.

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14:36:20

describe the slope field corresponding to y' = e^-(x^2)

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RESPONSE -->

Slope field corresponding to y' = e^-(x^2): Slope field V: slope maximum at x= 0, decreasing positive as |x| icreases, slope appears to approach 0 as |x| gets large (slope is very small with |x|>2).

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14:40:55

describe the slope field corresponding to y' = e^-x

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RESPONSE -->

Slope field corresponding to y' = e^-x: Slope field VI: Slope increasing positive as x decreases, slope 1 at x=0, slope approaches 0 for large positive x (slope very small for x>about 2).

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14:46:22

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Good work throughout. Let me know if you have questions.