Assignment 17

course MTH 174

òµÑ÷²Ê®®DܶH|yF±´óéØ­€assignment #017

017. `query 17

Cal 2

11-09-2008

......!!!!!!!!...................................

16:09:45

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.

......!!!!!!!!...................................

RESPONSE -->

y = cos (`omega * t)

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

y'' = -`omega^2 * cos (`omega t)

y'' = -9y so

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

- `omega^2 = - 9

`omega = +/- 3

.................................................

......!!!!!!!!...................................

16:22:42

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)

......!!!!!!!!...................................

RESPONSE -->

a)

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

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

So P = (1 + e^-t)^-1

P' = -1 * (1 + e^-t)^-2 * (-e^-t)

P' = e^-t *(1 + e^-t)^-2

P' = e^-t *(1 + e^-t)^-2

P' = (1 + e^-t - 1) / (1 + e^-t)^2

P' = [(1 + e^-t) - 1] / (1 + e^-t)^2

P' = (1 + e^-t) / (1 + e^-t)^2 - 1 / (1 + e^-t)^2

P' = 1 / (1 + e^-t) - 1 / (1 + e^-t)^2

b) lim t-> infinity 1 / (1 + e^-t) = 1

.................................................

......!!!!!!!!...................................

16:25:12

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

......!!!!!!!!...................................

RESPONSE -->

I took the derivative of P substituting it for dP/dt in the equation dP/dt = P(1 - P). I also substituted P into the equation and simplified both sides to ensure they were the same.

.................................................

......!!!!!!!!...................................

16:31:11

What is the derivative dP/dt?

......!!!!!!!!...................................

RESPONSE -->

P' = e^-t * (1 + e^-t)^-2

P'' = e^-t * -2* (1 + e^-t)^-3 * -e^-t + (1 + e^-t)^-2 * -e^-t

P'' = -2e^-2t (1 + e^-t)^-3 + -e^-t * (1 + e^-t)^-2

.................................................

......!!!!!!!!...................................

16:33:12

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

......!!!!!!!!...................................

RESPONSE -->

P(1-P) simplifies to dP/dt as found by taking the derivative of 1/(1 + e^-t). I am not sure how this relates to the second derivative, the derivative of dP/dt that was asked in the last question.

.................................................

......!!!!!!!!...................................

16:53:06

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) )

......!!!!!!!!...................................

RESPONSE -->

Well, I started off by taking he first and second derivatives of each equation I - IV

y = cos (x)

y' = -sin (x)

y'' = -cos (x)

y = cos (-x)

y' = sin (-x)

y'' = -cos(-x)

y = x^2

y' = 2x

y'' = 2

y = e^x + e^-x

y' = e^x + - e^-x

y'' = e^x + e^-x

y = `sqrt(2x) = (2x)^(1/2)

y' = 1/2 * (2x)^(-1/2) = 1 / ( 2* `sqrt(2x) )

y' = -1/4 * (2x)^(-3/2) = -1 / ( 4* `sqrt(2x)^3 )

Next, I compared each differential equation to the values of y, y', and y'' to see if the solution fit.

a) y'' = y, it seems that y = e^x + e^-x fits.

b) y' = -y, no solution seems to fit.

c) y' = 1/y, no solution seems to fit.

d) y'' = -y, y = cos (x) and y = cos (-x) fit.

e) x^2 * y'' - 2y = 0, y = x^2 fits

The function y = cos(x) has derivative y ‘ = -sin(x), which in turn has derivative y ‘’ = - cos(x). If we plug these functions into the equation y ‘’ = -y we get –cos(x) = - ( cos(x)), which is true, so this function satisfies the differential equation y ‘’ = - y.

The function y = cos(-x) has derivative y ‘ = sin(-x), which in turn has derivative y ‘’ = - cos(-x). If we plug these functions into the equation y ‘’ = -y we get –cos(-x) = - (cos(x)), which is true, so this function satisfies the differential equation y ‘’ = - y.

The function y = x^2 has derivative y ‘ = 2 x , which in turn has derivative y ‘’ = 2. Plugging into the equation x^2 y ‘’ – 2 y = 0 we get x^2 * 2 – 2 ( x^2) = 0, or 2 x^2 – 2 x^2 = 0, which is true.

The function y = e^x + e^(-x) has derivative y ‘ = e^x – e^(-x), which in turn has derivative y ‘’ = e^x + e^(-x). It is clear that y ‘’ and y are the same, so this function satisfies the differential equation y ‘’ = y.

The function y = sqrt(2x) has derivative y ‘ = 2 * 1 / (2 sqrt(2x)) = 1 / sqrt(2x) and second derivative y ‘’ = -1 / (2x)^(3/2). It should be clear that y ‘ = 1 / sqrt(2x) is the reciprocal of y = sqrt(2x), so this function satisfies the differential equation y ‘ = 1 / y.

.................................................

......!!!!!!!!...................................

16:53:22

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

16:53:25

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

16:53:26

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

16:53:28

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

16:53:34

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:27:14

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.

......!!!!!!!!...................................

RESPONSE -->

The slope is 0 at (0,0) and is 2.4 at (1,4)

There is an asymptote at x = 0

For the graph throuh (1,4)

Using P' = 0.1P(10 - P):

P' = P - .1P^2

P'' = 1 - .2P

Points of inflection:

0 = 1 - .2P

-1 = -.2P

1 = .2P

P = 5, so there is a point of inflection at P = 5

From P = 0 to P= 5, the graph is concave up, from P=5 to P = 10, the grapih is concave down.

There is another asymptote at P = 10.

.................................................

......!!!!!!!!...................................

17:44:46

Query problem 11.2.10 (was 10.2.6) slope field

......!!!!!!!!...................................

RESPONSE -->

y' = xe^-x corresponds to the slope field that looks like it could be a graph of -xe^-x - e^-x + C.

y' = sin (x) has a slope field that looks like it could be a graph of -cos (x) + C

y' = cos (x) has a field that looks like it could be a graph of sin (x) + C

y' = x^2 e^-x has a field that looks like it could be a graph of -x^2 * e^-x - x * e^-x + e^-x + C

y' = e^-x^2 has a field that looks like it could be a graph of that has slope 0 and then increases (being concave up), passes a point of inflection becoming concave down, and then approaches a slope of zero again.

y = e^-x has a field that looks like it could be a graph of -e^-x + C

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).

.................................................

......!!!!!!!!...................................

17:44:53

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:44:55

describe the slope field corresponding to y' = sin x

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:44:58

describe the slope field corresponding to y' = cos x

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:45:00

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:45:03

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:45:05

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

17:45:21

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

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