Applications of ode and matrices soluti0n
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Transcript of Applications of ode and matrices soluti0n
As we know
Y (0) = y°
Put t = 0 in
Y (0) = Aekt
Y (0) = A
So A= y°
Exponential Growth Example Solution:
we have y0 = 500
t= 3.5
The number of cells doubles every 3.5 hours, so after 3.5 hours there should be 1000 cells.
Therefore after 12 hours there will be cells in the dish.
HALF LIFE:
Example 1 solution
Example 2 solution
Falling object example
Hence v(0)=0
So
C=0.
Put again in v
So
V= -9.81t
Encryption example
18 5
A = 4 0
18 21
13 0
B = 4 -2
-1 3
67 -21
X = A B = 16 -8
51 27
52 -26
The message that you would pass on to the other person is the stream of numbers 67, -21, 16, -8, 51, 27, 52, -26.
DECRYPTION
67 -21
X = 16 -8
51 27
52 -26
B-1 = 0.3 0.2
0.1 0.4
18 5
A = X B-1 = 4 0
18 21
13 0
The receiver then takes the matrix and breaks it apart into values 18, 5, 4, 0, 18, 21, 13, 0 and converts
each of those into a character according to the numbering scheme. 18=R, 5=E, 4=D, 0=space, 18=R,
21=U, 13=M, 0=space.
Triangle area
Evaluate that determinant. I'll expand on column 1.
-2 2 1
1 5 1 = + (-2) 5 1 - 1 2 1 + 6 2 1
6 -1 1 -1 1 -1 1 5 1
= -2 ( 5 + 1 ) - 1 ( 2 + 1 ) + 6 ( 2 - 5 ) = -2 ( 6 ) - 1 ( 3 ) + 6 ( -3 ) = -12 - 3 - 18 = -33.
It is possible that you will get a negative determinant, like we did here. Don't worry about that. The sign is determined by the order you put the points in and can be easily changed just by switching two rows of the determinant. Area, on the other hand, can't be negative, so if you get a negative, just drop the sign and make it positive. Finally, divide it by 2 to find the area.
| -33 | = 33 33 / 2 = 16.5, which was the area.