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Unformatted text preview: AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I Practice Final Exam Solution Notes 1. Solve for x : 10 3 2 x = 3 (1 . 12) x Taking the natural log of both sides, we get ln 10 + 2 x ln 3 = ln 3 + x ln 1 . 12 x (2 ln 3 ln 1 . 12) = ln 3 ln 10 x = ln 3 ln 10 2 ln 3 ln 1 . 12 = ln(3 / 10) ln(9 / 1 . 12) 2. Determine which function has a larger value as x : (a). f ( x ) = 0 . 0005 x 3 or g ( x ) = 52 , 000 2 x f ( x ) grows like x 3 (we can ignore the constant factor). g ( x ) grows like 2 x , and exponential function (again, we can ignore the constant factor). Thus, g ( x ) has a larger value as x . Another way to say this is that lim x f ( x ) g ( x ) = 0 , so that g dominates f as x , which we compute using lHopitals rule (since it has form ): lim x . 0005 x 3 52 , 000 2 x = lim x . 0005 3 x 2 52 , 000 2 x ln 2 = lim x . 0005 6 x 52 , 000 2 x ln 2 ln 2 = lim x . 0005 6 52 , 000 2 x ln 2 ln 2 ln 2 = 0 . (b). f ( x ) = log x 7 or g ( x ) = 6 log 2 x 2 f ( x ) grows like log x (since log x 7 = 7 log x ), while g ( x ) grows like log 2 x (since log 2 x 2 = 2 log 2 x ), so g ( x ) dominates f ( x ) as x . 3. Find the equation of the line that goes through the point (8,3) and is perpendicular to the plot of the equation 2 x +1 y 1 = 2. We rewrite 2 x +1 y 1 = 2: 2 y 2 = 2 x + 1, so that it is the equation of the line y = x + 3 2 . The slope of this line is 1; thus, the slope of a line perpendicular to it is 1 1 = 1. Thus, the equation of the line, L , we desire is y = x + b . We find b by requiring that (8,3) lies on line L : 3 = 8 + b , so b = 11. Thus, line L has equation y = x + 11. 4. The quantity of moisture in a loaf of bread decreases with time as it sits on the table. Suppose that the moisture, M ( t ), at time t minutes after being placed on the table decreases according to the function M ( t ) = Qe kt . If 10% of the moisture is gone at the end of 1 hour, (a). What percentage of the original moisture is present after 30 minutes? We are told that M (60) = ( . 9) M (0), which implies that Qe 60 k = ( . 9) Q 60 k = ln( . 9) k = ln( . 9) 60 Now, we want to find M (30) M (0) 100 = Qe 30 k Q 100 = 100 e 30 ln( . 9) 60 = 100 e 1 2 ln( . 9) = 100( . 9) 1 / 2 (b). How long will it take until the moisture is reduced to 60% of its original quantity? We want to find t so that M ( t ) = 0 . 6 M (0) = 0 . 6 Q . Thus, we want t so that Qe kt = 0 . 6 Q , i.e., so that e kt = 0 . 6. Thus, we want t so that kt = ln(0 . 6), i.e., t = ln(0 . 6) k = ln(0 . 6) ln(0 . 9) 60 = 60 ln(0 . 6) ln(0 . 9) 5. Evaluate the following expressions: (a). cos(sin 1 ( 5 11 )) = Draw a right triangle and let be one of the two angles that is not / 2. Then, = sin 1 ( 5 11 ) if the side opposite to is of length 5 and the hypotenuse is of length 11. Then, the side adjacent to is of length...
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Zhang

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