This is lab 2 in CS 252. You can look what we are doing here PDF

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This is for the class lecture note in CS 252.
This is very useful for the class preparation and exams and etc....

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Lab 2 Problem 1) It only shows upper half or bottom half.

Problem 2)

It shows whole direction field where dplane does not.

Problem 3)

4000y = x. At b= 0.5, maximum value of y is 0.14 Therefore value of x = 4000 * 0.14 =560. 560 miles At b = 1.0, maximum value of y is 1 X = 4000 * 1.0 = 4000. 4000 miles. At b = 1.5, as s approaches to infinity, v approaches to 0.55

X’ = 4000* ds/dt * v 0.0061

S = √ 4000 𝑡 0.0061

ds/dt = √ 4000

0.0061

X’ = 4000* √ 4000 * 0.55 = 2.72 Therefore, the terminal velocity of when b = 1.5 is 2.72 mi/sec. When b = 2.0, it approaches to 1.46. 0.0061

X’ = 4000* √ 4000 * 1.46 = 7.21. Thus terminal velocity is 7.21 mi/sec.

Problem 4)

At x = 0, then y = 0, and V(0) = 1.35 4000*ds/dt * 1.35 = 6.7 miles/sec

Problem 5) a)

Therefore, Equation (D) can be written as equation above.

It happens when v = 0. Therefore, y is maximum at y = 1.

b)

Thus, y approaches to infinity as v approaches to zero. c)

Problem 6) System

According to the graph, escaping velocity exist between v = 1 and v = 1.5

Next Solve equation

Therefore, equation can be written as above. If y = 0, then V(0) = V0. If we plug 0 to y and V0 to V, then we get 100 ∗ v0 + 104 ln|1 + 0.01𝑣0| − 1 = 𝐶 Now v = 1.355. It is almost not changed, but slightly goes up. if 100 ∗ v0 + 104 ln|1 + 0.01𝑣0| > 1, then C > 0. If C < 0, then escape velocity cannot exist.

According to the graph, y = 1.45 when x = 0. Thus if we convert y and x to v and y. v= 1.45 if y = 0. 0.0061

Release velocity = X’ = 4000* √ 4000 * 1.45 = 7.16 mi/sec

Problem 7) K = 5 with new seed. If we calculate with new seed, we get: System

Solving equation:

Draw graph

V0 = 5.69 mi/sec (My matlab does not work. I do not know why.)...