SOLUTIONS TO THE FINAL - PART 1 MATH 150 FALL 2016 KUNIYUKI PART 1: 135 POINTS, PART 2: 115 POINTS, TOTAL: 250 POINTS

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1 SOLUTIONS TO THE FINAL - PART MATH 5 FALL 6 KUNIYUKI PART : 5 POINTS, PART : 5 POINTS, TOTAL: 5 POINTS No notes, books, or calculators allowed. 5 points: 45 problems, pts. each. You do not have to algebraically simplify or bo in your answers, unless you are instructed to. Fill in all blanks after = signs. DERIVATIVES (66 POINTS TOTAL) = cos = cos + = + sin cos = cos sin, or cos 7 5 = D e + 4 = = 6 e + 4 tan = sec = 6e e + 4 ( cos( ) ) ( by Product Rule) sin( ) ( by Quotient Rule) 7, or 7 5 6, or 6 ( 5) ( 5) ( 5) 5 D e + 4 = 6 e by Gen. Power Rule ( ) cot = csc sec( ) = sec ( )tan( ) csc( ) = csc ( )cot sin = cos D ( by Gen. Trig Rule) = 4cos e = cos e = e = e = e

2 MORE! ( 5 ) = 5 ln 5 ( ) = ln = ln = ln ln 7 + = 7 + D 7 + = = log ( ) = D ln ln = ln ( ) = ln ln = ln sin ( ) = cos ( ) = tan ( ) = + sec ( ) = (Assume the usual range for sec ( ) in our class.) tan ( 7) = + 7 sinh = cosh cosh = sinh sech( ) = sech ( )tanh ( 7) = = 7 + 4

3 INDEFINITE INTEGRALS (4 POINTS TOTAL) d = + C d = ln + C e 7 d = e C = e C, or C 7e 7 6 d = 6 ln( 6) + C sin( ) d = cos( ) + C tan( ) d = ln cos( ) + C, or ln sec( ) + C cot( ) d = ln sin( ) + C sec( ) d = ln sec( ) + tan( ) + C csc( ) d = ln csc( ) cot( ) + C, or ln csc( ) + cot( ) + C cos( ) d = sin( ) + C sec ( ) d = tan( ) + C 6 + d = 6 tan 6 + C d = sin C cosh( ) d = sinh( ) + C WARNING: YOU'VE BEEN DEALING WITH INDEFINITE INTEGRALS. DID YOU FORGET SOMETHING? (+ C)

4 INVERSE TRIGONOMETRIC FUNCTIONS (6 POINTS TOTAL) lim tan = π (Drawing a graph may help.) If f ( ) = sin ( ), what is the range of f in interval form (the form with parentheses and/or brackets)? Range f = π, π. HYPERBOLIC FUNCTIONS (6 POINTS TOTAL) The definition of sinh( ) (as given in class) is: sinh( ) = e e Complete the following identity: cosh ( ) sinh (We mentioned this identity in class.) = TRIGONOMETRIC IDENTITIES (5 POINTS TOTAL) Complete each of the following identities, based on the type of identity given. tan ( ) += sec (Pythagorean Identity) cos( ) = cos( ) (Even/Odd Identity) sin( ) = sin( )cos( ) (Double-Angle Identity) cos( ) = cos ( ) sin ( ), or sin ( ), or cos ( ) (For cos sin ( ) = (Double-Angle Identity), I gave you three versions; you may pick any one.) cos( ) (Power-Reducing Identity)

5 SOLUTIONS TO THE FINAL - PART MATH 5 FALL 6 KUNIYUKI PART : 5 POINTS, PART : 5 POINTS, TOTAL: 5 POINTS A scientific calculator and an appropriate sheet of notes are allowed on this final part. ) Find the following limits. Each answer will be a real number,,, or DNE (Does Not Eist). Write or when appropriate. If a limit does not eist, and and are inappropriate, write DNE. Bo in your final answers. (4 points total) cos a) lim Show all work, as in class. (6 points) 4 b) lim + 4 cos lim =. Prove this using the Sandwich / Squeeze Theorem: cos Observe that 4 >, > ( we may assume this, since we let ). Divide all three parts by 4. As, 4 More precisely: lim 4 cos( ) 4 So, by the Sandwich/ Squeeze Theorem =, and lim 4 4 =. ( > ) cos Therefore, by the Sandwich / Squeeze Theorem, lim =. 4 lim + r + c) lim r r 5 + r = Show all work, as in class. (6 points) lim + ( + ) ( 5) + 7 Limit Form Answer only is fine. ( points) = r + lim =, because we are taking a long-run limit of a proper r ( r 5 + r) ( bottom-heavy ) rational epression as r. The degree of the denominator (, since the leading term would be r after epanding the square) is greater than the degree of the numerator ().

6 ) Let f ( ) = + 4,. Classify the discontinuity at =. Bo in one:, = ( points) Infinite discontinuity Jump discontinuity Removable discontinuity lim f = lim = + 4 = 7, which is f ( ). (The limit value and the were 7, then f would be continuous at. + 4 function value eist but are unequal.) If f =,. Do not use derivative short cuts we have used in class. ( points) ) Use the limit definition of the derivative to prove that Let f = f ( ) f + h f ( ) = lim h h = lim 5 + h + h h = lim h h + 7 h ( + h) h h + 5h h = lim h h ( ) h + 5h h h ( + 5h ) = lim = lim h h h = + 5( ) =. ( Q.E.D. ) h = lim h h ( + 5h ) ) Let f ( ) = log 4 ( ). Consider the graph of y = f ( ) in the usual y-plane. Find a Point-Slope Form of the equation of the tangent line to the graph at the point where = 6. Give eact values; you do not have to approimate. (8 points) = log 4 ( 6) =, so the point of interest is ( 6, )., the slope m of the tangent line to the graph at that point. f ( ) = log 4 ( ) = D ln( ) ln( 4) f 6 Find f 6 f ( 6) = = D ln( 4) ln 6ln 4 ln 4 6 ( ) =, or ( Change-of-Base Property of Logarithms) ln( 4 ) = ln 4 ( This is our desired slope, m. ) Find a Point-Slope Form of the equation of the tangent line. y y = m( ) y = 6ln( 4) ( 6 )

7 5) Assume that and y are differentiable functions of t. Evaluate D t ( e ) y when =, y = 5, d dt D t ( e ) y = e y =, and dy dt = e ( ) ( 5) D t = 4. (8 points) ( y) = ( 5) + ( ) ( 4) ey = d dt y + dy dt e5 6) Let f ( ) = 4 +. (4 points total) a) Find the two critical numbers of f. = 4 4 = ( + ) f ( by the Product Rule) = e 5 = e 5, which is never undefined ( DNE ) but is at = and =. These numbers are in Dom( f ), which is, so the critical numbers are: and. Note: We could use the Quadratic Formula (QF) with a =, b = 4, and c = 4. = b ± = + 4 b 4ac a ± ( 4) 4( ) ( 4) ( ) = 4 = 6 = or = 4 = = = 4 ± = 4 ± 64 6 = 4 ± 8 6 = ± 4 b) Consider the graph of y = f ( ), although you do not have to draw it. Use the First Derivative Test to classify the point at = as a local maimum point, a local minimum point, or neither. f is continuous on, so the First Derivative Test ( st DT) should apply wherever we have critical numbers (CNs). Both f and f are continuous on, so use just the CNs as fenceposts on the real number line where f could change sign. f sign (see below) f Classify point at CN ( st DT) f (not needed) / Test = Test = = ( + ) ( ) = ( + ) ( ) = = ( + ) ( + ) = + f f Also, the graph of y = f -intercepts at, + L.Min. Pt. or Evaluate f at and directly. ( ) is an upward-opening parabola with two distinct and (, ). The multiplicities of the zeros of f are both odd (), so signs alternate in our windows. Answer: Local Minimum Point

8 c) Use the Second Derivative Test to classify the point at the other critical number as a local maimum point or a local minimum point. f =, so we may apply the Second Derivative Test for =. = 4 4 ( ) = 6 4 f f f = 6 4 = 8 f < Therefore, the point at = is a Local Maimum Point. Think: Concave down ( ) at (actually, on a neighborhood of) =. 7) Evaluate the following integrals. ( points total) a) 6 e d. Give an eact answer. ( points) Let u = or / du = / d du = d Can use: d = du Method (Change the limits of integration.) = u = = u = = 6 u = 6 = 4 u = 4 6 e d = = e u du 4 6 = e u e d 6 ( by Compensation) = e d 4 = ( e 4 e ), or e ( e ) Method (Work out the corresponding indefinite integral first.) e d e = d by Compensation = e u du = e u + C = e + C = e Now, apply the FTC directly using our antiderivative (where C = ). 6 e d = e 6 = e = ( e 4 e ), or e ( e ) 6 = e 6 e d

9 b) d ( points) Hint: Consider the Chapter 8 material on inverse trigonometric functions! Use the template: u + 8 du du, or = u 8+ u tan + C. Let u = u = d = 7 du = d Can use: d = du 8+ d = d 4 = 7 du = 7 u 8+ u tan + C = 7 8 tan + C Alternative Method (using basic template u + du, or d = 7 8+ d = 7 d = = 7 8 = 7 8 d + ( ) ( by Compensation) du = tan u + u d + C ): Let u =, or u = = 4 8 du = d du = d ( or use Compensation) + d = 7 8 du = 7 + u 8 tan u + C = 7 8 tan + C

10 8) The velocity function for a particle moving along a coordinate line for t > is given by v( t) = t t, where t is time measured in seconds and velocity is 4 given in meters per second. The particle s position is measured in meters. Find s( t ), the corresponding position function [rule], if s( ) = (meters). ( points) v( t) dt = dt t t 4 dt = t 4 t / s( t) = t t/ / + C = t t/ + C Find C. We know: s s = (meters). = ( ) ( ) / + C = + C = + C C = s( t) = t t/ +, or t t or t t / t (in meters) t t ( 4 t ) t, or, t ) The region R is bounded by the -ais, the y-ais, and the graphs of y = cos and = π in the usual y-plane. Sketch and shade in the region R. Find the 4 volume of the solid generated by revolving R about the -ais. Evaluate your integral completely. Give an eact answer in simplest form with appropriate units. Distances and lengths are measured in meters. Hint: Use a Power- Reducing Identity. (8 points) Let f ( ) = cos. Then, f is nonnegative and continuous on the -interval, π 4.

11 The region R and the equation y = cos( ) suggest a d scan and the Disk Method. V, the volume of the solid, is given by: V = = π π radius = π cos d = π cos d + cos = π + sin( ) ( by a Power-Reducing Identity) = π d ( by "Guess-and-check," or using u = ) = π π 4 + sin π 4 + sin( ) = π π 4 + sin π + sin ( ) = π π 4 + ( ) = π π 4 + = π π + 4 = π π + 8 ) Rewrite tan sin 5 (7 points) Let θ = sin 5 + cos d as an algebraic epression in, where < < 5. ( θ acute) sin( θ ) = 5 m tan( θ ) = opp. adj. = 5 ß found by the Pythagorean Theorem Note: Rationalizing the denominator is usually unnecessary if the radicand involved is variable. ( ) ) Find D r sin sinh r. (5 points) Use the template: D r sin ( u) = D u r D r sin sinh r = cosh r = sinh r D r ( u). = sinh r ( ) ( sinh( r) ) ( r). Note: ( sinh r) is not equivalent to cosh r sinh cosh ( r ), but + sinh r is.