Assignment 17 Query

course Mth 174

????O-?????n??assignment #017

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017. `query 17

Cal 2

12-01-2007

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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

(cos(omega t) ' = -omega sin ( omega t), the derivative of which is -omega ( omega cos(omega t) ) = -omega^2 cos(omega t), not -2 omega cos(omega t).

So omega ^2 = 9 and omega = 3.

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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.

Good. Check against the following:

Field I alternates positive and negative slopes periodically and is symmetric with respect to the y axis, so slope at –x is negative of slope at x, implying an odd function. The periodic odd function is sin(x). So the field is for y ‘ = sin(x).

Field II similar but slope at –x is equal to slope at x, implying an even function. The periodic even function is cos(x). So the field is for y ‘ = cos(x).

Field III is negative for negative x, approaching vertical as we move to the left, then is positive for positive x, slope increasing as x becomes positive then decreasing for larger positive x. The equation is therefore y ‘ = x e^(-x).

Field IV has all non-negative slopes, with near-vertical slopes for large negative x, slope 0 on the y axis, slope increasing as x becomes positive then decreasing for larger positive x. These slopes represent the equation y ‘ = x^2 e^(-x).

Field V has all non-negative slopes, about slope 1 on the x axis, slopes approaching 0 as we move away from the x axis. This is consistent with the equation y ‘ = e^(-x^2).

Field VI has all positive slopes, with slope approaching zero as x approaches infinity, slope approaching infinity (vertical segments) for large negative x. The field therefore represents y ‘ = e^(-x).

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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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This looks good. See my notes. Let me know if you have any questions. &#