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SampleProblems (2)

# SampleProblems (2) - AUBG Spring 2009 MAT 103 Calculus I...

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AUBG Spring 2009, MAT 103 “Calculus I” Final Exam Sample Problems Problem 1. Evaluate the limit. (a) lim x →− 1 3 x + 1 x + 1 (b) lim x 0 sin 3 x sin 4 x (c) lim x →∞ x 2 + 1 x 4 + 1 + x 2 (d) lim x →∞ x ( radicalbig x 2 + 1 - x ) Hint: You may try to rationalize the numerator in part (d). Answers. (a) 1 / 3 (b) 3 / 4 (c) 1 / 2 (d) 1 / 2 Solutions. (a) lim x →− 1 3 x + 1 x + 1 = lim x →− 1 ( 3 x + 1)( 3 x 2 - 3 x + 1) ( x + 1)( 3 x 2 - 3 x + 1) = lim x →− 1 x + 1 ( x + 1)( 3 x 2 - 3 x + 1) = lim x →− 1 1 3 x 2 - 3 x + 1 = 1 3 p ( - 1) 2 - 3 - 1 + 1 = 1 3 (b) lim x 0 sin 3 x sin 4 x = lim x 0 (sin 3 x ) /x (sin 4 x ) /x = lim x 0 (sin 3 x ) /x lim x 0 (sin 4 x ) /x = 3 4 (c) lim x →∞ x 2 + 1 x 4 + 1 + x 2 = lim x →∞ x 2 x 4 + x 2 = lim x →∞ x 2 x 2 + x 2 = 1 2 (d) lim x →∞ x ( p x 2 + 1 - x ) = lim x →∞ x ( x 2 + 1 - x )( x 2 + 1 + x ) x 2 + 1 + x = lim x →∞ x x 2 + 1 + x = lim x →∞ x x 2 + x = lim x →∞ x x + x = 1 2 Problem 2. Express the limit as a derivative and evaluate. (a) lim h 0 1 10 1 + h - 1 h (b) lim h 0 cos h - 1 h (c) lim x π/ 4 tan x - 1 x - π/ 4 Answers. (a) - 1 / 10 (The derivative of f ( x ) = x 1 / 10 at x = 1.) (b) 0 (The derivative of f ( x ) = cos x at x = 0.) (c) 1 (The derivative of f ( x ) = tan x at x = π/ 4.) Solutions. (a) lim h 0 1 10 1 + h - 1 h = lim h 0 f ( a + h ) - f ( a ) h = f ( a ), where f ( x ) = 1 10 x = x 1 / 10 and a = 1. Since f ( x ) = - 1 10 x 11 / 10 , the limit is equal to - 1 10 1 11 / 10 = - 1 10 . (b) lim h 0 cos h - 1 h = lim h 0 f ( a + h ) - f ( a ) h = f ( a ), where f ( x ) = cos x and a = 0. Since f ( x ) = - sin x 2 cos x , the limit is equal to - sin 0 2 cos 0 = 0.

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(c) lim x π/ 4 tan x - 1 x - π/ 4 = lim x π/ 4 f ( x ) - f ( a ) x - a = f ( a ), where f ( x ) = tan x and a = π/ 4. Since f ( x ) = 1 2 tan x · 1 cos 2 x , the limit is equal to 1 2 p tan π/ 4 · 1 cos 2 π/ 4 = 1. Problem 3. Find y . (a) y = x 3 cos 4 x (b) y = x 3 + x x 2 + x + 1 (c) y = (1 + tan 2 x ) 5 (d) y = 3 radicalBig 1 + 1 + x Solutions. (a) y = 3 x 2 · cos 4 x + x 3 · 4 cos 3 x · ( - sin x ) = 3 x 2 cos 4 x - 4 x 3 cos 3 x sin x (b) y = 1 2 x 3 + x · (3 x 2 + 1) · ( x 2 + x + 1) - x 3 + x · (2 x + 1) ( x 2 + x + 1) 2 = - x 4 + x 3 - x + 1 2 x 3 + x ( x 2 + x + 1) 2 (c) y = 5(1 + tan 2 x ) 4 · 2 tan x · 1 cos 2 x = 10 tan x (1 + tan 2 x ) 4 cos 2 x (d) y = 1 3 (1 + 1 + x ) 2 / 3 · 1 2 1 + x = 1 6 1 + x ( 3 p 1 + 1 + x ) 2 Problem 4. Use implicit differentiation to find an equation for the tangent line to the curve at the given point. (a) ( x 3 + y 3 ) 3 = 2( x 2 + y 2 ) 2 at P (1 , 1); (b) sin x + cos y + tan y = 1 at P (0 , 0). Answers. (a) y - 1 = - ( x - 1) (b) y = - x Solutions. (a) 3 ( x 3 + y 3 ) 2 (3 x 2 + 3 y 2 y ) = 4 ( x 2 + y 2 ) (2 x + 2 yy ) 3 (1 3 + 1 3 ) 2 (3 · 1 2 + 3 · 1 2 · y ) = 4 (1 2 + 1 2 ) (2 · 1 + 2 · 1 · y ) 72(1 + y ) = 16(1 + y ) 1 + y = 0 y = - 1 (b) cos x - sin y · y + 1 cos 2 y · y = 0 cos 0 - sin 0 · y + 1 cos 2 0 · y = 0 1 - 0 · y + 1 1 2 · y = 0 y = - 1 Fig 1: Problem 5 – distance between the ships.
Problem 5. At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km/h and ship B is sailing north at 25 km/h. How fast is the the distance between the ships changing at 4:00 P.M.?

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SampleProblems (2) - AUBG Spring 2009 MAT 103 Calculus I...

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