Calculus 221 First Exam 50 Minutes Friday October 4 1996 I Find the limit or show that it does not exist Justify your answer t3 1 1 lim 2 t t 1 Answer 1 t 2 t3 1 t lim lim 1 t t t2 1 1 2 t t3 1 t 3 t2 1 2 lim Answer lim t 3 33 1 26 t3 1 2 2 t 1 3 1 8 t3 1 t 1 t2 1 3 lim Answer lim t 1 t 1 t2 t 1 t2 t 1 3 t3 1 lim lim 2 t 1 t 1 t 1 t 1 t 1 t 1 2 II Find the limit or show that it does not exist Justify your answer sin 3x 1 lim x x Answer Since 1 sin 3x 1 for all x we have for x 0 Hence lim x sin 3x 1 1 x x x sin 3x 0 by the Squeeze theorem x sin x2 x 0 x 2 lim Answer lim x 0 sin x2 sin x2 sin u sin x2 lim x lim lim x lim lim x 1 0 0 2 x 0 x 0 u 0 x 0 x 0 x x x2 u sin x sin 3 x 3 x 3 3 lim Answer Let f x sin x and a 3 Then f 0 x cos x and lim x 3 III sin x sin 3 f x f a 1 lim f 0 a cos 3 x a x 3 x a 2 1 Find f 0 x and f 00 x if f x sin x3 2 Answer f 0 x cos x3 2 3x2 f 00 x sin x3 2 3x2 2 cos x3 2 6x 2 Find g 0 3 if h0 9 17 and g x h x2 Answer g 0 x h0 x2 2x IV g 0 3 h0 9 6 102 Find the constant c which makes g continuous on x2 cx 20 g x if x 4 if x 4 Answer By the limit laws g x is continuous at any x 6 4 lim g x lim x2 16 x 4 lim g x lim cx 20 4c 20 x 4 x 4 x 4 The function g x is continuous when these are equal i e when c 1 Answer V Consider the curve y 2 xy x2 11 1 Find the equation of the tangent line to the curve at the point P 2 3 Answer Differentiate 2y dy dy y x 2x 0 dx dx Evaluate at the point P 2 3 6 Solve dy dx 3 2 x y 2 3 dy dx dy dx 4 0 x y 2 3 x y 2 3 1 8 This is the slope of the tangent line The point P 2 3 lies on the tangent line so the equation of the tangent line is y 3 2 Find x 2 8 d2 y at the point P 2 3 dx2 Answer Differentiate again 2 dy dx 2 y d2 y dx2 dy d2 y dy x 2 dx dx dx 2 0 Evaluate at x y 2 3 2 2 1 8 d2 y 3 dx2 x y 2 3 1 8 Solve d2 y dx2 x y 2 3 2 1 d2 y 2 8 dx2 1 2 8 1 8 6 2 2 0 x y 2 3 1 8 2 Answer VI Consider the function y f x whose graph is shown below Match the expression in the left column with the correct corresponding value in the right column f 0 0 6 f 0 0 9 0 f 0 1 3 f 0 1 732 0 6 2 1 5 1 0 5 0 0 5 1 1 5 2 2 1 5 1 0 5 0 0 5 1 2 1 5 Answer For positive x the slope of the tangent line is decreasing as x increases so f 0 0 f 0 0 9 f 0 1 f 0 1 732 The only possibility is f 0 0 3 f 0 0 9 0 6 f 0 1 0 f 0 1 732 6 Answer VII Prove the product rule f g 0 f 0 g f g 0 using the definition of the derivative high school algebra and the appropriate limit laws Justify each step Answer Let w x f x g x and assume that f and g are differentiable w0 x w x h w x h f x h g x h f x g x lim h 0 h f x h f x g x h g x lim g x h f x h 0 h h f x h f x g x h g x lim g x h f x lim lim h 0 h 0 h 0 h h f 0 x g x f x g 0 x lim 1 h 0 Reasons 1 Definition of w0 x 2 Definition of w x 3 High school algebra 2 3 4 5 4 Limit laws 5 Definition of f 0 x continuity of g x and definition of g 0 x In step 5 we used the fact that a differentiable function is continuous Here is the proof lim g x h g x lim g x h g x h 0 h 0 g x h g x h h g x h g x lim lim h h 0 h 0 h g 0 x 0 0 lim h 0 so lim g x h g x i e g is continuous h 0 Calculus 221 Second Exam Two Hours Thursday October 24 1996 I e5x 1 2xe2x 5x a Find f 0 x if f x Answer f 0 x b Find 5e5x 1 e2x e 1 e2x 2 2e d when sin 1 t2 dt 2 1 Answer Let y t so sin y Then d d dy 1 2t p 2t dt dy dt 1 t4 1 y2 II After 3 days a sample of radon 222 decayed to 58 of its original amount a What is the half life of radon 222 Answer Let Y be the amount of radon 222 at time t and Y0 be the amount when t 0 Thus Y Y0 ect for all time t When t 3 we have Y 0 58Y0 Hence ln 0 58 0 58Y0 Y0 ec3 Apply ln to find c c The half life is the time when 3 ln 0 5 3 ln 0 5 Y 0 5Y0 i e 0 5Y0 Y0 ec Apply ln to find c ln 0 58 b How long would it take for the sample to decrease to 10 of its original amount Answer The desired time t satisfies 0 10Y0 Y0 ect To find t cancel Y 0 and apply ln 0 10 3 ln 0 10 ln t c ln 0 58 III a Find the inverse function g y to the function f x 9 x Answer For y 0 we have y 9 x y 2 …
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