Answer Key-Math 11- Optional Review Homework For Exam 2

Answer Key-Math 11- Optional Review Homework For Exam 2 1. Compute the derivative for each of the following functions: Please do not simplify your derivatives here. (I simlified some, only in the case that you want to check your answers) x3 + (a) f (x) = f ′ (x) = 1 x8 x2 2 x [3x2 3x4 − 8x−7 − 2x4 − 2x−7 x4 − 10x−7 − 8x−9 ] − (x3 + x−8 )(2x) = = = x4 x4 x4 1 − 10x−11 or more simply you can rewrite f (x) = x + x−10 and then f ′ (x) = 1 − 10x−11 .   1 (b) f (x) = sin2 3x − 2 x ′ f (x) = 2 sin(3x − x−2 ) · cos(3x − x−2 ) · (3 + 2x−3 )   3x 2 (c) f (x) = x tan sec x       3x (sec x)3 − (3x) sec x tan x 3x ′ 2 2 f (x) = x sec · + tan · 2x sec x sec2 x sec x x3 + (x5 + 3x)7 √ x2 − 7x √ 1 x2 − 7x(3x2 + 7(x5 + 3x)6 (5x4 + 3)) − (x3 + (x5 + 3x)7 ) · √ · (2x − 7) 2 2 x − 7x ′ f (x) = x2 − 7x 1   (e) f (x) = 1 1 − +x cos x x7   −1 1 1 First rewrite f (x) = cos − +x x x7      1 1 1 1 ′ −2 − + x · − sin − + x · (−1x−2 + 7x−8 + 1) f (x) = −1 · cos x x7 x x7   1 1 sin − + x · (−1x−2 + 7x−8 + 1) x x7   = 1 1 2 cos − +x x x7 p√ √ x + cos(x2 + x) (f) h(x) =    √ 1 1 1 2 ′ h (x) = p√ √ · √ − sin(x + x) 2x + √ 2 x 2 x + cos(x2 + x) 2 x √ 2 (g) g(y) = y cos y √ y 2 sin y 1 √ √ √ + 2y cos y g ′ (y) = y 2 (− sin y) · √ + cos y · 2y = − √ 2 y 2 y (d) f (x) = (h) f (t) = (3t − 1)7 (4t + 3)9 f ′ (t) = (3t − 1)7 · 9(4t + 3)8 (4) + (4t + 3)9 · 7(3t − 1)6 (3) = 3(3t − 1)6 (4t + 3)8 [12(3t − 1) + 7(4t + 3)] = 3(3t − 1)6 (4t + 3)8 [64t + 9] 1  5 √ (i) w(x) = x + x4 + 1  4   √ 1 3 ′ 4 1+ √ (4x ) w (x) = 5 x + x + 1 2 x4 + 1 p sin(3x2 + 17x) (j) f (x) = cos3 (4x) f ′ (x) = " # p 1 cos3 (4x) p · cos(3x2 + 17x)(6x + 17) − sin(3x2 + 17x)3 cos2 (4x)(− sin(4x))4 2 sin(3x2 + 17x) cos6 (4x) 2. Compute y ′ where xy + y 3 = 4x2 cos(x2 ). Differentiating both sides implicitly yields dy dy x + y + 3y 2 = 4x2 (− sin(x2 ))(2x) + cos(x2 )(8x) and then dx dx dy (x + 3y 2 ) = −8x3 sin(x2 ) + 8x cos(x2 ) − y so that dx −8x3 sin(x2 ) + 8x cos(x2 ) − y dy . = dx x + 3y 2 3. Evaluate each of the following limits: 6 sin(6x) 6 6 sin(6x) = lim = ·1= x→0 8x 8 x→0 6x 8 8 (a) lim 0 sin(6x) = = 0 cos x + x 1+0 3x − 8x2 1 x(3 − 8x) x 4x (c) lim = lim = lim · lim (3−8x) = lim · lim (3−8x) = x→0 sin(4x) x→0 sin(4x) x→0 sin(4x) x→0 4 x→0 sin(4x) x→0 3 1 ·3= 4 4
1 3 3x − 8x2 −8 3x − 8x2 x2
x −8 (d) lim = lim · = lim 1 9 = 7 5 x→∞ 7x2 + 5x − 9 x→∞ 7x2 + 5x − 9 x→∞ 7 + x − x2 x2
9 1 + x728 − 2010 9x18 + 7x3 − 2010 9x18 + 7x3 − 2010 31 x
x13 x31 = 0 (e) lim = lim · 1 = lim 2 5 x→∞ x→∞ 6x31 + 2x7 + 5 x→∞ 6x31 + 2x7 + 5 6 + x24 + x31 x31
1 x3 + x15 − 2010 x12 + x4 − 2010 x12 + x4 − 2010 x9
x9 (f) lim = lim · = ∞ = lim 1 1 x→∞ x→∞ x→∞ 6x9 + x2 6x9 + x2 6 + x9 x7 (b) lim x→0 1 . Discuss domain, intervals of increase or decrease, local −4 extreme value(s), concavity, inflection point(s), and any horizontal and vertical asymptotes. Use this information to give a detailed and labelled sketch of the curve. 4. Consider the function f (x) = x2 2 • Domain: f (x) has domain {x|x 6= ±2} • Vertical asymptotes: VA at x = ±2. • Horizontal asymptote: HA at y = 0 for this f since lim f (x) = 0. x→±∞ • First Derivative Information −2x We compute f ′ (x) = 2 and set it equal to 0 and solve for x to find critical numbers. (x − 4)2 The critical points occur where f ′ is undefined or zero. The latter happens when x = 0. The derivative is undefined when x = ±2, but those values are not in the domain of the original function. As a result, x = 0 is the critical number. Using sign testing/analysis for f ′ , f′ ⊕  f ր t −2 ⊕ t ⊖ 0 ր local max 2 t ց ⊖ - ց or our f ′ chart is x f ′ (x) f (x) (−∞, 0) ⊕ ր (0, ∞) ⊖ ց So f is increasing on (−∞, 0); and f is decreasing on (0, ∞). Moreover, f has a local max at 1 x = 0 with maximum value f (0) = − . 4 • Second Derivative Information 2(3x2 + 4) is never zero. Using sign testing/analysis for f ′′ around the Meanwhile, f ′′ = (x2 − 4)3 vertical asymptotes, f ′′  f or our f ′′ chart is ⊕ ⊖ t S −2 x f ′′ (x) f (x) T (−∞, −2) ⊕ S 2 t (−2, 2) ⊖ T ⊕ - S (2, ∞) ⊕ S So f is concave down on (−2, 2) and concave up on (−∞, −2) and (2, ∞) with no inflection points. • Piece the first and second derivative information together 3 t  f f f ր S −2  t 0 ր T  local max 2 ց T t  - ց S  • Sketch: 5. Consider the function f (x) = x2 (x − 1)2 . Discuss domain, intervals of increase or decrease, local extreme value(s), concavity, inflection point(s), and any horizontal and vertical asymptotes. Use this information to give a detailed and labelled sketch of the curve. • Domain: f (x) has domain (−∞, ∞) • Vertical Asymptotes: It is a polynomial, continuous everywhere, and so has no vertical asymptotes. • Horizontal Asymptotes: There are no horizontal asymptotes for this f since lim f (x) = ∞ x→∞ and lim f (x) = ∞. x→−∞ • First Derivative Information We compute f ′ (x) = x2 (2)(x − 1) + (x − 1)2 (2x) = 2x(x − 1)[x + (x − 1)] = 2x(x − 1)(2x − 1) and set it equal to 0 and solve for x to find critical numbers. The critical points occur where 1 f ′ is undefined (never here) or zero. The latter happens when x = 0, x = 1, x = As a result, 2 4 those numbers are the critical numbers. Using sign testing/analysis for f ′ , f′ ⊖  t ⊕ t ⊖ t ⊕ - 1 0 1 2 ց local ր ց local ր local min max min f or our f ′ chart is x (−∞, 0) ⊖ ց f ′ (x) f (x) (0, 1/2) ⊕ ր (1/2, 1) ⊖ ց (1, ∞) ⊕ ր So f is increasing on (0, 1/2) and (1, ∞), and f is dereasing on −∞, 0) and (1/2, 1). Moreover, f has local mins at x = 0 and x = 1, but a local max at x = 1/2. Here f (0) = 0 and f (1) = 0, 1 and f ( 12 ) = 16 • Second Derivative Information Meanwhile, f ′′ is always defined and continuous, and f ′′ = 12x2 − 12x + 2 = 0 only at our 1 1 possible inflection point x = ± √ . Using sign testing/analysis for f ′′ , 2 12 f ′′ ⊕  or our chart is x f ′′ (x) f (x) t  1 1 −∞, − √ 2 12 ⊕ S infl. point    ⊕ t √ 1/2 − 1/ 12 T S f f ′′ ⊖ - √ 1/2 + 1/ S 12 infl. point 1 1 1 1 −√ , +√ 2 12 2 12 ⊖ T    1 1 + √ ,∞ 2 12 ⊕ S    1 1 1 1 1 1 √ √ √ So f is concave down on − and concave up on −∞, − and , + 2 2 12 2 12 12    1 1 1 1 1 1 + √ , ∞ , with inflection points at x = ± √ . You can check that f ±√ = 2 2 2 12 12 12 1 . 36 • Piece the first and second derivative information together  t t 0 f ց S f f local min ր S t ր 5 T 1 2  infl.  pt. local max t ց T  t - 1 ց S ր S infl.  local  pt. min • Sketch: 6. A conical water tank (point facing down) with radius of 3 meters at the top and height of 8 meters is leaking water. At the moment when the water is 2 meters from the top of the tank water is leaking at a rate of 1 cubic meter per minute. How fast is the water level decreasing at that moment? The cross section (with water level drawn in) looks like: • Diagram 3  6 B  r wB - 8 6 aB t B h e B r B ? ? B • Variables Let r = radius of the water level at time t Let h = height of the water level at time t Let V = volume of the water in the tank at time t dh Find =? when h = 6 feet (2 from top) dt m3 dV = −1 and dt min • Equation relating the variables: 1 Volume= V = πr2 h 3 • Extra solvable information: Note that r is not mentioned in the problem’s info. But there is a relationship, via similar triangles, between r and h. We must have r h = 3 8 3 =⇒ r = h 8 After substituting into our previous equation, we get: 6  2 1 3 3 V = π h h = πh3 3 8 64 • Differentiate both sides w.r.t. time t.   dV 3 d 3 dh d 3 =⇒ (V ) = πh = π · 3h2 · dt dt 64 dt 64 dt =⇒ dV 9 dh = πh2 dt 64 dt • Substitute Key Moment Information (now and not before now!!!): dh 9 −1 = π(6)2 64 dt • Solve for the desired quantity: dh −64 −16 m = = dt 36π · 9 81π min 16 meters every • Answer the question that was asked: The height is decreasing at a rate of 81π second. 7