Matlab - ass#2

# Matlab - ass#2 - 8 9 10-4-3-2-1 1 2 3 4 t y y = k(A y A = 1...

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Hoon Shin Math 20D Section: B52 Assignment #2 Exercise 2.1 a) b) c) Exercise 2.2 A) -2 0 2 4 6 8 10 -4 -3 -2 -1 0 1 2 3 4 x y y ' = (exp( - x) - y) (exp( - x) + 1 - 2 y) B) Plotting the field, allows us to understand easier and faster instead of solving the equation by hand. Also, by clicking on to the point that we are looking for, it allows us to see the solution near that point. Exercise 2.3

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-2 0 2 4 6 8 10 -4 -3 -2 -1 0 1 2 3 4 x y y ' = x - y If the initial value was no exactly (1,1), it would give slightly off answer but it would not affect that much as long as it is near point (1,1). Exercise 2.4 -2 0 2 4 6 8 10 -4 -3 -2 -1 0 1 2 3 4 tx y y ' = y - 2
Different from exercise 2.3, the results will be far different if one did not click on to the point (0,2) because the change (rate) is rapid. Exercise 2.5 0 1 2 3 4 5

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Unformatted text preview: 8 9 10-4-3-2-1 1 2 3 4 t y y ' = k (A - y) A = 1 k = 3 “A” changes the height of the graph where “k” changes the rate. Comparing to real life, A is representing equilibrium solution where temperature has no change in a time. On the other hand, “k” is representing the rate of change which determines how quick the object hits the equilibrium solution. Exercise 2.6 a) dy/dt = k ( A - y ) , y(0) = -7 b) “A” should be 40°F c) at about 15 hours 2 4 6 8 10 12 14 16 18 20-10 10 20 30 40 50 t y y ' = 0.3 (36 - y) d) about an 2.8 hours Thus 15 – 2.8 = 12.2 hours will be saved 2 4 6 8 10 12 14 16 18 20 5 10 15 20 25 30 35 t y y ' = 0.3 (68 - y)...
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