Query 17

#$&*

course Mth 174

017. `query 17Cal 2

*********************************************

Question:

Query problem 11.1.8 previously 11.1.10 Find a value of omega such that y '' + 9 y = 0 is satisfied by y = cos(omega * t).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y= cos(omega* t). Y’= -omega* sin(omega* t). Y’’’= -omega^2 * cos(omega* t). After substituting, we get -omega^2* cos(omega* t) + 9cos(omega* t) = 0. When we solve for omega, we get 3 and -3 as solutions.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

Substitute y = cos(omega * t) into the equation. The goal is to find the value of omega that satisfies the equation:

We first calculate y ‘’

y = cos(omega*t) so

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

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

Now substituting in y"""" + 9y = 0 we obtain

-omega^2*cos(omega*t) + 9cos(omega*t) = 0 , which can be easily solved for omega:

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

omega^2 = 9

omega = +3, -3

Both solutions check in the original equation.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating: OK

*********************************************

Question:

Query problem 11.1.18 (4th edition 11.1.14 3d edition 11.1.13) (formerly 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We first multiply by e^t to get P= (e^t) / (e^t + 1). We have {(e^t)’ (e^t + 1) - (e^t + 1)’ e^t} / (e^t + 1)^2 = {e^t (e^t + 1) - e^t * e^t} / (e^t + 1)^2 = e^t / (e^t + 1)^2. We substitute to get P(1-P)= (e^t) / (e^t +1) * { 1- e^t / (e^t +1)}. We keep simplifying to finally get (e^t) / (e^t +1)^2, which agrees with the dP/dt expression. Therefore, dP/dt = P(1-P).

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

RESPONSE -->

P= 1/(1+e^-t) ; multiplying numerator and denominator by e^t we have

P = e^t/(e^t+1) , which is a form that makes the algebra a little easier.

dP/dt = [ (e^t)’ ( e^t + 1) - (e^t + 1) ‘ e^t ] / (e^t + 1)^2, which simplifies to

dP/dt = [ e^t ( e^t + 1) - e^t * e^t ] / (e^t+1)^2 = e^t / (e^t + 1)^2.

Substituting:

P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)] . We simplify this in order to see if it is the same as our expression for dP/dt:

e^t/(e^t+1) * [1 - e^t/(e^t+1)] = e^t / (e^t + 1) - (e^t)^2 / (e^t + 1)^2. Putting the first term into a form with common denominator

e^t ( e^t + 1) / [ (e^t + 1) ( e^t + 1) ] - (e^t)^2 / (e^t + 1 ) ^ 2

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

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

= e^t / (e^t + 1)^2.

This agrees with our expression for dP/dt and we see that dP/dt = P ( 1 - P).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating: OK

*********************************************

Question:

Query problem (omitted from 5th edition but should be worked and self-critiqued) (4th edition 11.1.16 3d edition 11.1.15 formerly 10.1.1) match one of the equations y '' = y, y ' = -y, y ' = 1 / y, y '' = -y, x^2 y '' - 2 y = 0 with each of the solutions (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) )

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

B) y’ = -y can be matched with (II) y=cos(-x). C) y' = 1/y is matched with (V) y = sqrt(2x). D) y'' = -y can be paired with the solution (II) y = cos(-x). E) x^2y'' - 2y = 0 can be put with the solution (IV) y = e^x + e^-x.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

.

y '' = y would give us (x^2) '' = x^2. Since (x^2)'' = 2 we would have the equation 2 = x^2. This is not true, so y = x^2 is not a solution to this equation.

B) y' = -y and its solution can be (II) y = cos(-x)

C) y' = 1/y and its solution can be (V) y = sqrt(2x)

D) y'' = -y and its solution can be (II) y = cos(-x)

E) x^2y'' - 2y = 0 and its solution can be (IV) y = e^x + e^-x

If y = e^x + e^-x then y '' = (e^x) '' + (e^-x) '' = e^x + e^-x and this equation would give us

x^2 ( e^x + e^-x) - 2 ( e^x + e^-x) = 0.

This is not so; therefore y = e^x + e^-x is not a solution to this equation.

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.

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

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

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

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

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

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

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

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

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).

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

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

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

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

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

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating: OK

*********************************************

Question:

Query problem 11.2.8 (4th edition 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For the solution through (0,0), the graph is concave up from (0,0) to (3,4) and then goes to concave down. There is a horizontal asymptote when P=10 and an inflection point at (3,3). For the solution through (1,4), the graph is concave up from (-1,0) to (2,3), and then it is concave down when t is greater than 2. There is also an inflection point at (2,3) and a horizontal asymptote at P=10.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

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.

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

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

14:13:09

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating: OK

*********************************************

Question: Query problem 11.2.14 (was 11.2.10) (previously 10.2.6) slope field

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

There are both positive and negative slopes in Field I. The periodic odd function is sin(x). The field is for y’= sin(x). There is an even function on Field II, and the field is for y’=cos(x). Field III is negative for -x and positive for x. Therefore, y’= x e^(-x). For Field IV, the slope equation is: y’= x^2 e^(-x).The slopes are non-negative for Field V, and y’=e^(-x^2). For Field VI, all the slopes are positive and the field equation is y’= e^(-x).

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

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).

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

14:22:17

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

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

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.

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

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

14:25:41

describe the slope field corresponding to y' = sin x

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

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

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

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

14:28:17

describe the slope field corresponding to y' = cos x

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

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).

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

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

14:32:27

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

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

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.

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

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

14:36:20

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

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

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).

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

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

14:40:55

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

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

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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating: OK

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

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