A3_f2015_key.pdf - MATH 1500 D01/D02 Fall 2015 Assignment 3...

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MATH 1500 D01/D02 Fall 2015 Assignment 3 Total Marks: 60 Due Date: Oct 24, 2015 . SHOW ALL WORK to get full marks . Word problems should have sentence answers with units. This assignment covers sections 3.3, 3.4, 3.5, 3.6, 3.9. 1. Differentiate the following functions. DO NOT SIMPLIFY. (a) (3 points) y = 4 sin 2 x Solution: y 0 = 4 sin 2 x (ln 4) 2 sin x cos x (b) (4 points) y = cos p tan( π 2 x ) Solution: y 0 = - sin p tan( π 2 x ) 1 2 ( tan( π 2 x ) ) - 1 / 2 sec 2 ( π 2 x ) π 2 (c) (4 points) y = ( ex + e - x x ) 3 / 2 Solution: y 0 = 3 2 ( ex + e - x x ) 1 / 2 1 2 ( ex ) - 1 / 2 e - e - x x + e - x 1 (d) (3 points) f ( t ) = ln | sec( π 2 t - t 4 ) | Solution: f 0 ( t ) = 1 sec( π 2 t - t 4 ) sec( π 2 t - t 4 ) tan( π 2 t - t 4 ) ( π 2 - 4 t 3 ) (e) (4 points) y = log 3 (csc (5 x ) + ln ( x 2 + 1)) Solution: y 0 = 1 csc (5 x ) + ln ( x 2 + 1) 1 ln 3 - csc (5 x ) cot (5 x ) 5 + 1 x 2 + 1 2 x 2. Evaluate the following limits (a) (4 points) lim x 0 3 sin (7 x ) x - 3 x 5 1
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Solution: = lim x 0 3 sin (7 x ) x (1 - 3 x 4 ) = lim x 0 3 · 7 7 · sin (7 x ) x (1 - 3 x 4 ) = lim x 0 21 · sin (7 x ) 7 x · 1 (1 - 3 x 4 ) = 21 · lim x 0 sin (7 x ) 7 x lim x 0 1 (1 - 3 x 4 ) = 21 · 1 · lim x 0 1 (1 - 3 x 4 ) = 21 · 1 · 1 (1 - 0) = 21 (b) (4 points) lim θ 0 3 π sin θ θ - 2 sin θ Solution: = lim x 0 3 π sin θ θ - 2 sin θ ·
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