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Unformatted text preview: MAT 127: Calculus C, Fall 2009 Solutions to Midterm I Problem 1 (20pts) (a; 8pts ) Show that the function y ( x ) = 2e x is a solution to the initialvalue problem y  6 y + 5 y = 0 , y = y ( x ) , y (0) = 2 , y (0) = 2 . Compute y ( x ) and y ( x ) to check that the differential equation is satisfied: y ( x ) = 2e x = y ( x ) = 2e x = y ( x ) = 2e x = y  6 y + 5 y = 2e x 6 2e x + 2 5e x = 0 e x = 0 . check It remains to check that the initial condition are satisfied: y (0) = 2e = 2 check , y (0) = 2e = 2 check . (b; 10pts ) Find the general solution y = y ( x ) to the differential equation y  6 y + 5 y = 0 , y = y ( x ) . The associated polynomial equation is r 2 6 r + 5 = 0 ( r 1)( r 5) = 0 . So the two roots are r 1 , r 2 = 1 , 5 and the general solution is y ( x ) = C 1 e x + C 2 e 5 x (c; 2pts ) What is the relation between the solution given in (a) and the general solution in (b)? The solution in (a) is the C 1 = 2, C 2 = 0 case of the general solution. Problem 2 (20pts) A bacteria culture grows at a rate proportional to its size. It contained 125 cells at 8pm and 250 cells at 8:20pm. (a; 12pts ) Find an expression for the number of cells in the culture t minutes after 8pm. Let y ( t ) be the number of cells in the culture t minutes after 8pm; so y (0) = 125. Since the growth rate is proportional to y , y ( t ) satisfies the exponential growth equation: y ( t ) = y (0)e rt = 125e rt , where r is the relative growth rate. By the last assumption, y (20) = 125e r 20 = 250 e r 20 = 2 20 r = ln 2 r = (ln 2) / 20 . Thus, y ( t ) = 125e (ln 2) t/ 20 = 125 2 t/ 20 Note: In this case, it is not necessary to specify the units for either y ( t ) or t , but specifying them is fine also. Either of the two expressions for y ( t ) in the box is fine. (b; 8pts ) When will the bacteria culture reach 2,000 cells? We need to find t so that y ( t ) = 125e (ln 2) t/ 20 = 2000 e (ln 2) t/ 20 = 16 (ln 2) t/ 20 = ln 16 = ln 2 4 = 4 ln 2 t = 80 . So the bacteria culture reaches 2000 cells 80 minutes after 8pm, which is 9:20pm Here is another solution, which does not use part (a) ( 3pts bonus , whether or not the stan dard solution is present). By assumptions, the number of cells doubles every 20 minutes. Since 2000 / 125 = 16 = 2 4 , this (doubling) needs to happen precisely 4 times, which takes 20 4 = 80 minutes. So the bacteria culture reaches 2000 cells 80 minutes after 8pm, which is 9:20pm Note: In this case, the units pm or p.m. , with or without space after the numbers, are necessary. Problem 3 (15pts) The direction field for a differential equation is shown below....
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This note was uploaded on 01/25/2012 for the course MAT 127 taught by Professor Guanyushi during the Fall '07 term at SUNY Stony Brook.
 Fall '07
 GuanYuShi
 Calculus

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