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exam1f09sol

# exam1f09sol - MAT 127 Calculus C Fall 2009 Solutions to...

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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 initial-value 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 0 = 2 check , y (0) = 2e 0 = 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.

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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 (ln2) 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 (ln2) t/ 20 = 2000 ⇐⇒ e (ln2) 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. y x 1 1 y ( 0 ) = 0 y ( 0 ) = - . 8 y ( 0 ) = 2 (a; 10pts) On the direction field, sketch and clearly label the graphs of the three solutions with the initial conditions y (0) = - . 8 , y (0) = 0 , and y (0) = 2 (each of these three conditions determines a solution to the differential equation). The first curve must pass through the point (0 , - . 8), the second through (0 , 0), and the third through (0,2). These curves must be tangent to the slope lines at those points and roughly ap- proximate the slopes otherwise. The sketch suggests that the middle curve is the line y = x .

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