1b-2009-exam_3_practice_solutions

1b-2009-exam_3_practice_solutions - Math 1B Exam#3 Practice...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 1B — Exam #3 Practice Solutions Fall Program for Freshmen 2009 Fred Bourgoin 1. Use Euler’s method with step size 0.1 to estimate y (0 . 3), where y is the solution to the initial-value problem y 0 = x + xy , y (0) = 1. Solution. It is easier to organize things in a table. x y ( x ) y 0 ( x ) y 0 (0) = 0 y (0) = 1 0 0.1 y (0 . 1) 1 + (0 . 1)(0) = 1 y 0 (0 . 1) 0 . 2 0.2 y (0 . 2) 1 + (0 . 1)(0 . 2) = 1 . 02 y 0 (0 . 2) 0 . 404 0.3 y (0 . 3) 1 . 02 + (0 . 1)(0 . 404) = 1 . 0604 2. A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour? Solution. If we let y ( t ) be the amount of salt in the tank at time t (in minutes), the problem can be translated as the differential equation dy dt = [rate in] - [rate out] = (0 . 03)(25) - y 5000 25 = 0 . 75 - 0 . 005 y with initial value y (0) = 20. The equation is separable. dy dt = 0 . 75 - 0 . 005 y = ⇒ - 200 dy dt = y - 150 = dy y - 150 = - dt 200 = Z dy y - 150 = Z - dt 200 = ln | y - 150 | = - t 200 + C = ⇒ | y - 150 | = Ce - t 200 = y - 150 = Ce - t 200 = y = 150 + Ce - t 200 Now plug in the initial condition: y (0) = 150 + C = 20 = C = - 130 . Hence, the solution is y ( t ) = 150 - 130 e - t 200 . 3. The initial size of a population is 100. After 2 days, the population size has increased to 300.
Image of page 1

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

View Full Document Right Arrow Icon
1. Give the differential equation for a natural growth model, then solve it. Show your steps.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the 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