Calculus I Practice Test 3

This is a practice test in differential calculus covering log and exponential derivatives, curve sketching, trig and inverse trig functions, and differential equations. The answers are on the 2nd page to allow you to correct your own work. A manual of complete solutions is available for $3 from Tammy the Tutor --see note at end of document. Time limit: 2 hours Instructions: No graphing calculators No notes or crib sheets allowed Show all work. Write neatly and big enough to see!! Numbers in parentheses ( ) are mark values. A/Differentiate : 5) y = e x ² – 3 + (7) x – 1/x – log 3 (x³ – 5x) 3) g (t) = ln² (3t + 1) 4) h (x) = csc (ln x) + arctan 5x – arccos (x³ + 7) 1) f (x) = ln 2) f (x) = ln (cos 2x + sin 2x) – 5 arcsec x² x2 − 1 x5 + 3 1 /4 (15) B/Use implicit or logarithmic differentiation to find y'. 1) y ² e 2x + xy³ = 1 ......................... 2) y = (x + 4) x ² – 1 (6) C/Find an equation for the tangent 1) to the curve y = e – x , perpendicular to 2x – y = 5. 2) to y = ln x at the point where x = 2. (7) D/1) Rewrite as an algebraic expression in x: (diagram) a) sec [arctan (x + 1)/5] b) sin {arccos ( –7/x)} (4) 2) Find equations for the tangent and normal lines to the graph of y = arcsin (x – 1) at point P (3/2, /6 ) . (4) 3) Sketch the graph of y = e – x ² . Clearly indicate y' and y'' and describe these properties: Intercepts, Asymptotes, Critical Points, Increasing and Decreasing intervals, Inflection points, Concavity, and Symmetry. (10) 4) Solve this differential equation for f (x) if f ΠΠ(x) = −3 x4 , f Π(1) = −1 f (1) = 12 (4) TOTAL (50) CALCULUS PRACTICE TEST # 3 Answers: A/Differentiate : 5) y Π = 2x e x2 − 3 + 1 + 1 x2 ( 7) x − 1/x (ln 7) − 3x2 − 5 (x3 − 5x) ln 3 4) h Π(x) = −csc ( ln x ) cot ( lnx ) x + 5 25x2 + 1 + 3x2 1 − (x3 + 7)2 3) g Π(t) = 6 [ ln (3t + 1) ] 3t + 1 2) f Π(x) = −2 sin 2x + 2 cos 2x cos 2x + sin 2x − 10 1) f x x4 − 1 Π(x) = 14 2x x2 − 1 − 5x4 x5 + 3 B/Use implicit or logarithmic differentiation to find y'. 1) y 2) Π = −2 y e 2 x − y 2 2 e 2 x + 3 x y yΠ = 2x ln ( x + 4 ) + x 2 − 1 x + 4 (x + 4) x 2 − 1 C/Find an equation for the tangent 1) The equation of the tangent is y = − 12 x + 12 ln 2 + 12 2) The equation of the tangent is y = 12 x + ln 2 − 1 D/1) Rewrite as an algebraic expression in x: a) sec arctan b) (x + 1) 5 = x2 + 2x + 26 5 sin arccos −7 x = x2 − 49 x 2)The equation of the tangent line is y = 23 ( x − 32 ) + 6 The equation of the normal line is y = − 3 2 ( x − 32 ) + 6 3) Max: (0, 1) Min: none Intercept: (0, 1) Inflection Points: x = ! 12 Asymptotes: y = 0 Concave Up: x < Concave Down: −12 , x > 12 −12 < x < 12 Increasing: x < 0 Decreasing: x > 0 Symmetry: to y-axis. 5 x 5 5 4 3 2 1 0 1 2 3 4 5 0.51 1.5 y = e – x ² . 4) f (x) = – ½ x – 2 – 2x + 3. CALCULUS PRACTICE TEST # 3 A solutions manual with complete solutions for these test questions is available for $3. Look for it in the Public Content of Tammy the Tutor at WiZiQ.com Do you need more help with these topics? Contact Tammy the Tutor for online tutorials. e-mail: mathster@the-mathroom.ca CALCULUS PRACTICE TEST # 3