practice_final_fa05_solutions - Practice Final Exam Math...

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Practice Final Exam, Math 191, Fall 2005 No calculators. Show your working. Clearly mark each answer. 1. (a) Method I. The arc length is integraldisplay 2 0 radicalBigg 1 + parenleftbigg dy dx parenrightbigg 2 dx = integraldisplay 2 0 radicalBigg 1 + parenleftbigg 2 2 x 2 2 x x 2 parenrightbigg 2 dx = integraldisplay 2 0 radicalbigg 1 + (1 x ) 2 2 x x 2 dx = integraldisplay 2 0 radicalbigg 1 2 x x 2 dx = integraldisplay 2 0 1 radicalbig 1 ( x 1) 2 dx = integraldisplay 1 1 1 1 u 2 du = bracketleftbig sin 1 u bracketrightbig 1 1 = π 2 parenleftBig π 2 parenrightBig = π Method II. y = 2 x x 2 = radicalbig 1 ( x 1) 2 , 0 x 2 traces out the half of the radius 1 circle centred at (1 , 0) , that lies above the x -axis, and this has length π . (b) See page 536 of the textbook for the sketch. The volume is integraldisplay −∞ π y 2 dx = integraldisplay −∞ π sech 2 x dx = lim b + integraldisplay 0 π sech 2 x dx + lim a →−∞ integraldisplay 0 a π sech 2 x dx = lim b + [ π tanh x ] b 0 + lim a →−∞ [ π tanh x ] 0 a = π lim b + e b e b e b + e b 0 + 0 π lim a →−∞ e a e a e a + e a = π lim b + 1 e 2 b 1 + e 2 b 0 + 0 π lim a →−∞ e 2 a 1 e 2 a + 1 = π. 1 π. ( 1) = 2 π (c) The surface area is integraldisplay 2 0 2 πy radicalBigg 1 + parenleftbigg dy dx parenrightbigg 2 dx = 2 π integraldisplay 2 0 cosh x radicalbig 1 + sinh 2 x dx = 2 π integraldisplay 2 0 cosh 2 x dx = 2 π integraldisplay 2 0 parenleftbigg e x + e x 2 parenrightbigg 2 dx = 2 π integraldisplay 2 0 e 2 x + 2 + e 2 x 4 dx = 2 π bracketleftbigg e 2 x + 4 x e 2 x 8 bracketrightbigg 2 0 = 2 π e 4 + 8 e 4 8 = π e 4 + 8 e 4 4 (Alternatively, use the identity 2 cosh 2 x = 1 + cosh 2 x in this calculation.) 2. (a) The degree 1 Taylor polynomial of f ( x ) = x 2 3 + integraldisplay x 2 2 sec( t 2) dt at x = 2 (otherwise

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