This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 8 9 104321 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 40F c) at about 15 hours 2 4 6 8 10 12 14 16 18 2010 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
 Fall '06
 Mohanty
 matlab

Click to edit the document details