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...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
-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
Image of page 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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

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)...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern