D sin x. (By Product Rule of Diff n.) ( ) D 2x ( ) 2. 10x4, or 24x 2 4x 7 ( ) ln x. ln x. , or. ( by Gen.

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1 SOLUTIONS TO THE FINAL - PART MATH 50 SPRING 07 KUNIYUKI PART : 35 POINTS, PART : 5 POINTS, TOTAL: 50 POINTS No notes, books, or calculators allowed. 35 points: 45 problems, 3 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) ( ) e = e e 4 sin( ) = 4 sin + 4 D sin (By Product Rule of Diff n.) = D = ( ln( ) ) 4 + = 4 ln tan = sec cot = csc ( ) = 4 3 sin( ) + 4 cos( ) by Quotient Rule , or ( 5 + 8) = 4 ln ( ) ( ) sec( ) = sec ( )tan + + csc( ) = csc ( )cot cos 3 4 = sin 3 4 = 3sin D ln + 3 4ln, or + 3, or D 3 4 = sin( 3 4) ( by Generalized Trig Rule) ( 5 + 4) 6 5 ( by Gen. Power Rule) 3 ( e ) 7 = e 7 ( 7) = e 7 7 = 7e 7, or 7 e 7

2 MORE! ( 9 ) = 9 ln = ln = + 74 ln( 7 ) 43 + ln( 6 +) = 6 + D 6 + = = log 3 ( ) = D ln ln 3 = ( ) = D ln( 3) ln ln( 3 ) = ln 3 sin ( ) = cos ( ) = tan ( ) = + sec ( ) = (Assume the usual range for sec ( ) in our class.) tan e = + e ( e ) = + e e = e + e sinh = cosh cosh = sinh sech( ) = sech ( )tanh

3 INDEFINITE INTEGRALS (4 POINTS TOTAL) 5 d = C d = ln + C e 3 d = e C 8 d = 8 ln( 8) + C cos( ) d = sin( ) + 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 sin( 7) d = cos( 7) + C 7 csc( )cot( ) d = csc( ) + C d = sin C 49 + d = 7 tan 7 + C cosh( ) d = sinh( ) + C WARNING: YOU'VE BEEN DEALING WITH INDEFINITE INTEGRALS. DID YOU FORGET SOMETHING? (I m referring to + C. )

4 INVERSE TRIGONOMETRIC FUNCTIONS (6 POINTS TOTAL) If f ( ) = cos ( ), what is the range of f in interval form (the form with parentheses and/or brackets)? Range f = 0, π lim ( sin ) = π (Drawing a graph may help.) HYPERBOLIC FUNCTIONS (6 POINTS TOTAL) The definition of cosh( ) (as given in class) is: cosh( ) = 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. + cot ( ) = csc (Pythagorean Identity) sin( ) = sin( ) (Even/Odd Identity) sin( ) = sin( )cos( ) (Double-Angle Identity) cos( ) = cos ( ) sin ( ), or sin ( ), or cos ( ) (For cos cos ( ) = (Double-Angle Identity), I gave you three versions; you may pick any one.) + cos( ) (Power-Reducing Identity)

5 SOLUTIONS TO THE FINAL - PART MATH 50 SPRING 07 KUNIYUKI PART : 35 POINTS, PART : 5 POINTS, TOTAL: 50 POINTS ) 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. (6 points total) a) lim r r 4 7 8r 4 + r b) lim t Answer only is fine. ( points) Answer:. We take the ratio of the leading coefficients of the polynomials in 8 the numerator and the denominator. This is because those polynomials have the same degree (4), and we are taking a long-run limit as r. t 5 + 3t 7 t 6 t Answer only is fine. ( points) Answer: 0, because we are taking a long-run limit of a proper ( bottomheavy ) rational epression as t. The degree of the denominator (6) is greater than the degree of the numerator (5). c) lim d) lim 0 + = lim 5 ( 5) ( + 3) 0 8 cos 3 Show all work, as in class. (6 points) Limit Form = 0 Show all work, as in class. (6 points) Answer: 0. Prove this using the Sandwich / Squeeze Theorem: cos 0 3 Observe that > 0, 0. Multiply all three parts by. As 0, cos So, 0 by the Sandwich/ Squeeze Theorem More precisely: lim 0 = 0, and lim = 0. 0 Therefore, by the Sandwich / Squeeze Theorem, lim 0 cos 3 = 0.

6 ) Use the limit definition of the derivative to prove that = for all real 0. Do not use derivative short cuts we have used in class. ( points) Let f ( ) =. f = lim = h 0 + h h + h + 0 f ( ) f + h h h h + h = ( Q.E.D. ) + h h ( ) h h ( ) + h + h ( + h) h ( + h) + h 3) Consider the given equation 4y y + 5e y = + 5e. Assume that it determines an implicit differentiable function f such that y = f ( ). Find dy (you may use the y notation, instead). ( points) d 8y y y + 8y y + 3 y y y + 5e y Product Rule to D y + 5 e y y = 0 34 y + 5 e y y = 0 8y y + 3 y y + 5e y y = 0 Isolate the terms with y on one side. Factor out y on that side. Divide to solve for y. = ( + 5e ) 8y y y + 5e y y = 3 y = 3 y y 8y e y y = 3 y 8y e y

7 4) Consider the graph of the equation in Problem 3), 4y y + 5e y = + 5e, in the usual y-plane. Find a Point-Slope Form for the equation of the tangent line to the graph at the point (, ). Give an eact answer; do not approimate. You may use your work from Problem 3). (7 points) (, ) satisfies the given equation, so the point, and we may use the y formula from Problem 3). Evaluate y at (, y) =, : y, Point-Slope Form for the tangent line at, y y = m 3 ( ) + 3( ) 4 + 5e = 8 : 4 y = ( ( ) ), or y = 9 + 5e e + lies on the graph of the equation, = 5) Let f ( ) = Show all work, as in class. (7 points total) e 0.49 a) Give the -interval(s) on which f is increasing. Write your answer in interval form. Do not worry about the issue of parentheses vs. brackets. (0 points) = 3 + = 3( + 4) f is never undefined ( DNE ). f ( ) = 0 at only = 0 and = 4. Dom( f ) =, so 0 and 4 are critical numbers (CNs) in Dom( f ). f f and f are everywhere continuous on, so we use just the CNs as fenceposts where f could change sign. Test = 5 4 Test = 0 Test = f sign + + f = ( 3) ( ) ( + 4) ( 5) = ( + ) ( ) ( ) = + ( ) = ( + ) ( ) ( + ) = = ( + ) ( + ) ( + ) = + f f f f The multiplicities of the zeros of f are both odd (), so we get alternating signs in our windows. Also, the graph of y = f with two distinct -intercepts at 4, 0 signs. ( ) is an upward-opening parabola and 0, 0 ; this eplains the first and last f is increasing on: (, 4, 0, ). (Brackets due to one-sided continuity.)

8 b) Give the -interval(s) on which the graph of y = f ( ) is concave up. Write your answer in interval form. Do not worry about the issue of parentheses vs. brackets. (7 points) = 6 + = 6( + ) f is never undefined ( DNE ). f ( ) = 0 at only =. Dom( f ) =, so is a PIN (Possible Inflection Number) in Dom( f ). f f, f, and f are everywhere continuous on, so we use just the PIN as a fencepost where f could change sign. Test = 3 Test = 0 f sign + f graph f f Also, the graph of y = f CD ( ) CU ( ) ( ) = ( 6) ( + ) ( 3) = ( + ) ( ) = = ( + ) ( + ) = + ( ) is a rising line with -intercept at (, 0). f 0 The graph of y = f ( ) is concave up on:, ). (Bracket due to one-sided continuity.) 6) Evaluate the following integrals. (7 points total) dθ a) sin θ (7 points) Use a Power-Reducing Identity (PRI). cos θ sin ( θ ) dθ = dθ = = θ sin( θ ) = θ sin( θ ) + C, or 4 cos θ dθ + C ( By "Guess-and-check," or using u = θ. ) θ sin( θ ) + C 4

9 b) d (0 points) Hint: Consider the Chapter 8 material on inverse trigonometric functions! Use the template: du, or 5 u du u = sin 5 u 5 + C. Let u = 3 u = 9 4 du = 6 d du = d ( or use Compensation) 6 d = ( Compensation) = du u d = u 6 sin 5 + C = 3 6 sin 5 + C Alternate Method (using more basic template): du du, or = sin u u u d = d = d = 5 d Let u = 3 5 = 3 5 u = 94 5 du = 6 5 d 5 du = d ( or use Compensation) = d = du 6 = u 6 sin u 7) Rewrite cos tan 7 Let θ = tan 7 tan θ ( Compensation) + C = 6 sin C as an algebraic epression in. (7 points) = 7, so cos tan Use the Pythagorean Theorem to find the hypotenuse. 7 = cos θ = adj. + C hyp. = (We usually don t rationalize a denominator where a radicand is variable.)

10 8) Find D w sech 5 e w. (7 points) D w sech 5 e w = D sech e w 5 w = 5 sech e w 4 Dw sech e w = 5 sech( e ) w 4 sech ( e w )tanh( e ) w D w ew = 5 sech( e ) w 4 sech ( e w )tanh( e ) w ew = 5ew sech ( 5 e w )tanh e w 9) Distances and lengths are measured in meters. ( points total) a) Find the area of the shaded region R. Evaluate your integral completely. Give an eact answer in simplest form with appropriate units, and also approimate your answer to four significant digits. (8 points) The area of the region R is given by: (WARNING: Use radian measure!) + d = tan ( ) = tan ( tan ) = tan π 4 m 0.38 m b) Find the volume of the solid generated by revolving the shaded region R about the y-ais. Evaluate your integral completely. Give an eact answer in simplest form with appropriate units, and also approimate your answer to four significant digits. (3 points) The region R and the given equation (solved for y in terms of ) suggest a d scan and the Cylindrical Shells (Cylinder) Method. V, the volume of the solid, is given by: V = π radius r( ) height h ( ) d = π + d Let u = + V = π du = d Can use: d = du Change the limits of integration: = u = + = u = = u = + ( ) = 5 u = d = π u du = π ln u = π ln 5 = π ln 5 ln ln( ). 5, or π ln 5 m3.879 m 3