111_quiz3_spr08_key

111_quiz3_spr08_key - -3 5 Solution cot(arcsin-3 5 =-4...

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APMA 111 Quiz 3 February 5, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. [3 pts] A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420. Find an expression for the number of bacteria after t hours. Solution: The equation is y ( t ) = y (0) e rt = 100 e rt because the initial population is 100. We need to find r , so we use y (1) = 420: 420 = 100 e r · 1 21 / 5 = e r ln(21 / 5) = ln( e r ) r = ln(21 / 5) . Now plug in r to get the resulting equation: y ( t ) = 100 e t ln(21 / 5) = 100 e ln(21 / 5) t = 100(21 / 5) t . 2. (a) [1 pt] arcsin(sin 3 π 4 ) Solution: arcsin(sin 3 π 4 ) = arcsin(sin π 4 ) = π/ 4 (b) [1 pt] cos - 1 (cos 3 π 4 ) Solution: cos - 1 (cos 3 π 4 ) = 3 π/ 4 (c) [1 pt] cot(arcsin
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Unformatted text preview: -3 5 ) Solution: cot(arcsin-3 5 ) =-4 / 3 (see triangle below) 5 4-3 θ 3. [3 pts] lim x →∞ arctan( e x ) Solution: x → ∞ ⇒ e x → ∞ ⇒ arctan( e x ) → π/ 2 ⇒ lim x →∞ arctan( e x ) = π/ 2 Quiz 3, Page 2 of 2 February 5, 2008 4. [3 pts] Compute the derivative of y = tan-1 (cosh x ). Solution: y = 1 1 + (cosh x ) 2 · (cosh x ) = sinh x 1 + cosh 2 x (Chain Rule) 5. [3 pts] Z 1 1 / 2 dx √ 1-x 2 arcsin x Solution: Let u = arcsin x , then du = dx/ √ 1-x 2 , and Z 1 1 / 2 1 arcsin x dx √ 1-x 2 = Z π/ 2 π/ 6 du u = ln u ± ± ± ± ± π/ 2 π/ 6 = ln( π/ 2)-ln( π/ 6) = ln( π/ 2 π/ 6 ) = ln 3...
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This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.

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111_quiz3_spr08_key - -3 5 Solution cot(arcsin-3 5 =-4...

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