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Math16B - Spring 08 Final Solutions

# Math16B - Spring 08 Final Solutions - SOLUTIONS TO MATH 16B...

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SOLUTIONS TO MATH 16B FINAL EXAM OF SPRING 2008 1. Let f ( x ) = x 3 - x - 1 so that f 0 ( x ) = 3 x 2 - 1. Start with x 0 = 1, then x 1 = 1 - f (1) /f 0 (1) = 1 - ( - 1) / (2) = 3 / 2. x 2 = (3 / 2) - f (3 / 2) /f 0 (3 / 2) = (3 / 2) - 7 8 / 23 4 = 31 / 23. 2. Since ( y (0)) 2 = 25 6 = 0, we may divide both sides of the equation by y 2 and then integrate to obtain 1 5 - 1 y ( T ) = - 1 y ( T ) - - 1 y (0) = - 1 y ( t ) | t = T t =0 = Z T 0 y 0 ( t ) y ( t ) 2 dt = Z T 0 (1 + tan( t )) dt = ( t + ln(cos( t ))) T 0 = T + ln(cos( T )) - 0 - ln(cos(0)) = T + ln(cos( T )) So, 1 y ( T ) = 1 - 5 T - 5 ln(cos( T )) 5 Thus, y ( T ) = 5 1 - 5 T - 5 ln(cos( T )) 3. We know the Taylor series for e x is given by e x = X n =0 1 n ! x n Replacing x with x 3 , we have e x 3 = X n =0 1 n ! x 3 n Subtracting 1, we have e x 3 - 1 = X n =1 1 n ! x 3 n Dividing by x we have e x 3 - 1 x = X n =1 1 n ! x 3 n - 1 1

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2 SOLUTIONS TO MATH 16B FINAL EXAM OF SPRING 2008 Integrating, we have Z e x 3 - 1 x dx = ( X n =1 1 (3 n )( n !) x 3 n ) + C Hence, Z 3 2 0 e x 3 - 1 x dx = X n =1 1 (3 n )( n !) ( 3 2 ) n 4. Let P ( t ) be the population in thousands t hours after noon. We know P (0) = 2 P (1) = 3 P 0 ( t ) = kP ( t ) 2 where k is a constant to be determined.
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Math16B - Spring 08 Final Solutions - SOLUTIONS TO MATH 16B...

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