query 22

#$&*

course Mth 279

4:44 pm 8/6

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

Question:  Find the values for which the matrix

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

pictured as:

is invertible.

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Your solution: 

 A matrix is invertible if W is not equal to 0

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

2t + 1 = 0

t = -1/2

matrix is invertible on the interval (- infinity, -1/2)u(-1/2, infinity)

 

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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

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Your solution: 

 

 To find the limit of the matrix, take the limit of each element

[1, 0, 3;

1, 1, infinity]

 

 

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2t / (t^2 - 1) approaches zero as t approaches zero. The numerator approaches zero, the denominator approaches -1.

*@

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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

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Your solution: 

 Taking derivative of each element

 A' = [cos(t), 3;

2t, 0]

A'' = [-sin(t), 0;

2, 0]

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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.

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Your solution: 

 P(t) = [t^2, 3;

sin(t), t]

g(t) = [sec(t); -5]

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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)?

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Your solution: 

 

 Not sure if this was the way I was supposed to do it

A' = [t, 1/2 t^2; 0, 0] + [C1, C2; C3, C4]

A = [1/2 t^2, 1/6 t^3; 0, 0] + [C1*t, C2*t; C3*t, C4*t] + [D1, D2; D3, D4]

A(0) = [ 1, 1; -2, 1] = [ 0, 0; 0, 0] +[ 0, 0; 0, 0] + [D1, D2; D3, D4]

D1 = 1

D2 = 1

D3 = -2

D4 = 1

A(1) = [-1, 2; -2, 3 ] = [1/2, 1/6; 0, 0] + [C1, C2; C3, C4] + [1, 1; -2, 1]

-1 = 1/2 + C1 + 1

C1 = -5/2

2 = 1/6 + C2 + 1

C2 = 5/6

-2 = 0 + C3 - 2

C3 = 0

3 = 0 + C4 + 1

C4 = 2

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

-2, 2t + 1]

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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) ].

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Your solution: 

 integral [e^s ds] from 0 to t

e^t- e^0 = e^t - 1

integral[ 6s ds] from 0 to t

3t^2 - 0 = 3t^2

integral[ cos(2pi*s) ds] from 0 to t

1/(2pi) sin(2pi*t) - 0 = 1/(2pi) sin(2pi*t)

integral[sin(2pi*s) ds] from 0 to t

-1/(2pi) cos(2pi*t) + 1/(2pi) cos(0) = -1/(2pi) cos(2pi*t) + 1/(2pi)

A = [e^(t) - 1, 3t^2;

1/(2pi) sin(2pi*t), -1/(2pi) c os(2pi*t) + 1/(2pi)]

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#