Query_22

#$&*

course MTH-279

08/03 around 10:35 PM.

Query 22 Differential Equations*********************************************

Question: Find the values for which the matrix

[ t + 1, t; t, t + 1]

pictured as:

is invertible.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

note: 2x2 inverse is 1/det(A) [d, -b ; -c, a]

W(t) = ((t+1)*(t+1)) - ((t* t))

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

= 2t + 1

2t + 1 = 0

2t = -1

t = -1/2

---

all values but -1/2 because it makes W(t) / det(A) equal to zero

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Find the limit as t -> 0 of the matrix

[ sin(t) / t, t cos(t), 3 / (t + 1); e^(3 t), sec(t), 2 t / (t^2 - 1) ]

pictured as

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: Find A ' (t) and A ''(t), where the derivatives are with respect to t and the matrix is

A = [ sin(t), 3 t; t^2 + 2, 5 ]

pictured as

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A’(t) = [cost, 3 ; 2t, 0] & A’’(t) = [-sint, 0 ; 2, 0]

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Write the system of equations

y_1 ' = t^2 y_1 + 3 y_2 + sec(t)

y_2 ' = sin(t) y _1 + t y_2 - 5

in the form

y ' = P(t) y + g(t),

where P(t) is a 2 x 2 matrix and y and g(t) are 2 x 1 column vectors.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

P(t) = [t^2, 3 ; sint, t]

g(t) = [sect; -5]

y(t) = [y1 ; y2]

[y1’ ; y2’] = [t^2, 3; sint, t] * [y1 ; y2] + [sect; -5]

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: If

A '' = [1, t; 0, 0]

with

A(0) = [ 1, 1; -2, 1]

A(1) = [-1, 2; -2, 3 ]

then what is the matrix A(t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A’’ = [1, t ; 0, 0] -> A’ = [1t+A, t^2/2 + B ; C, D] -> A = {t^2 + At + E, t^3 + Bt + F ; Ct + G, Dt + H]

A(0) = [0^2/2 + A(0) + E, 0^3 + B(0) + F ; C(0) + G, D(0) + H]

A(0) = [E, F ; G, H] = [ 1, 1 ; -2, 1]

E = 1, F = 1, G = -2, H = 1

-----

A(1) = [1^2/2 + A(1) + 1, 1^3/6 + B(1) + 1 ; C(1) - 2, D(1) + 1]

A(1) = [3/2 + A, 7/6 + B; C - 2, D + 1] = [-1, 2; -2, 3]

3/2 + A = 1

A = -1 - 3/2 -> -2/2 - 3/2, A = -5/2

7/6 + B = 2

B = 2 - 7/6 -> 12/6 - 7/6 = 5/6, B = 5/6

C-2 = -2

C = 0

D + 1 = 3

D = 2

----

So,

A = [t^2/2 + 5/2t + 1, t^3/6 + 5/6t + 1; -2, 2t + 1]

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Find the matrix A(t), defined by

A(t) = integral( B(s) ds, s from 0 to t),

where

B = [ e^s, 6s; cos(2 pi s), sin(2 pi s) ].

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

B = [ int(0 to t) e^s ds, int(0 to t) 6s ds; int(0 to t) cos(2pi(s)) ds, int (0 to t) sin (2pi(s)) ds]

A(t) = [e^t-1, 3t^2 - 0 ; sin(2pi(t)) - 0/2pi , -cos(2pi(t))/2pi - (-1)/2pi

A(t) = [e^t - 1, 3t^2; sin(2pi(t))/2pi, -cos(2pi(t))/2pi + 1/2pi]

Confidence rating: 2,3

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

Given Solution:

Self-critique (if necessary): OK

Self-critique rating: OK

"

Self-critique (if necessary):

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

Self-critique rating:

critiqued_deferred student work 140804d

08-04-2014

Query_22

#$&*

course MTH-279

08/03 around 10:35 PM.

Query 22 Differential Equations*********************************************

Question: Find the values for which the matrix

[ t + 1, t; t, t + 1]

pictured as:

is invertible.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

note: 2x2 inverse is 1/det(A) [d, -b ; -c, a]

W(t) = ((t+1)*(t+1)) - ((t* t))

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

= 2t + 1

2t + 1 = 0

2t = -1

t = -1/2

---

all values but -1/2 because it makes W(t) / det(A) equal to zero

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Find the limit as t -> 0 of the matrix

[ sin(t) / t, t cos(t), 3 / (t + 1); e^(3 t), sec(t), 2 t / (t^2 - 1) ]

pictured as

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: Find A ' (t) and A ''(t), where the derivatives are with respect to t and the matrix is

A = [ sin(t), 3 t; t^2 + 2, 5 ]

pictured as

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A’(t) = [cost, 3 ; 2t, 0] & A’’(t) = [-sint, 0 ; 2, 0]

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Write the system of equations

y_1 ' = t^2 y_1 + 3 y_2 + sec(t)

y_2 ' = sin(t) y _1 + t y_2 - 5

in the form

y ' = P(t) y + g(t),

where P(t) is a 2 x 2 matrix and y and g(t) are 2 x 1 column vectors.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

P(t) = [t^2, 3 ; sint, t]

g(t) = [sect; -5]

y(t) = [y1 ; y2]

[y1’ ; y2’] = [t^2, 3; sint, t] * [y1 ; y2] + [sect; -5]

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: If

A '' = [1, t; 0, 0]

with

A(0) = [ 1, 1; -2, 1]

A(1) = [-1, 2; -2, 3 ]

then what is the matrix A(t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A’’ = [1, t ; 0, 0] -> A’ = [1t+A, t^2/2 + B ; C, D] -> A = {t^2 + At + E, t^3 + Bt + F ; Ct + G, Dt + H]

A(0) = [0^2/2 + A(0) + E, 0^3 + B(0) + F ; C(0) + G, D(0) + H]

A(0) = [E, F ; G, H] = [ 1, 1 ; -2, 1]

E = 1, F = 1, G = -2, H = 1

-----

A(1) = [1^2/2 + A(1) + 1, 1^3/6 + B(1) + 1 ; C(1) - 2, D(1) + 1]

A(1) = [3/2 + A, 7/6 + B; C - 2, D + 1] = [-1, 2; -2, 3]

3/2 + A = 1

A = -1 - 3/2 -> -2/2 - 3/2, A = -5/2

7/6 + B = 2

B = 2 - 7/6 -> 12/6 - 7/6 = 5/6, B = 5/6

C-2 = -2

C = 0

D + 1 = 3

D = 2

----

So,

A = [t^2/2 + 5/2t + 1, t^3/6 + 5/6t + 1; -2, 2t + 1]

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Find the matrix A(t), defined by

A(t) = integral( B(s) ds, s from 0 to t),

where

B = [ e^s, 6s; cos(2 pi s), sin(2 pi s) ].

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

B = [ int(0 to t) e^s ds, int(0 to t) 6s ds; int(0 to t) cos(2pi(s)) ds, int (0 to t) sin (2pi(s)) ds]

A(t) = [e^t-1, 3t^2 - 0 ; sin(2pi(t)) - 0/2pi , -cos(2pi(t))/2pi - (-1)/2pi

A(t) = [e^t - 1, 3t^2; sin(2pi(t))/2pi, -cos(2pi(t))/2pi + 1/2pi]

Confidence rating: 2,3

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

Given Solution:

Self-critique (if necessary): OK

Self-critique rating: OK

&#Good responses. Let me know if you have questions. &#