Department of Mathematical Sciences. Math 226 Calculus Spring 2016 Exam 2V2 DO NOT TURN OVER THIS PAGE UNTIL INSTRUCTED TO DO SO

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1 Department of Mathematical Sciences Math 6 Calculus Spring 6 Eam V DO NOT TURN OVER THIS PAGE UNTIL INSTRUCTED TO DO SO NAME (Printed): INSTRUCTOR: SECTION NO.: When instructed, turn over this cover page and begin the test. You will have 9 minutes to complete the test. If you have any questions, raise your hand and wait for the proctor to come to your seat. This test is 6 pages long and contains 7 problems, some with several parts. Write your work on the test papers. If you need etra space, use the backs of the pages and say so on the front. You must show all necessary work for each problem. Solutions presented with no supporting work may receive no credit. Numerical answers should be presented as eact mathematical epressions, simplified as appropriate, not by a decimal approimation, unless eplicitly required by the problem. YOU MAY NOT USE NOTES, CELL PHONES, CALCULATORS OR LAPTOPS AT ANY TIME DURING THE TEST PERIOD. Good luck! Here are some useful identities. sin () = cos(), cos () = + cos(), sec 3 ()d = [sec() tan() + ln sec() + tan() + C. Problem Points Credit Total

2 Math 6 Calculus Spring 6 Eam V () (35 Points) Evaluate the following integrals. (a) (7 Points) tan 3 () sec 3 () d (b) (8 Points) sin 4 (3) d

3 Math 6 Calculus Spring 6 Eam V () (Continued) Evaluate the following integrals (c) ( Points) ( + 3) d (d) (8 Points) / d

4 Math Calculus Spring 6 Eam V () (5 Points) Circle the correct form for the partial fraction decomposition of the rational function ( 4 )( + ) : a) A + B 4 + C + D ( + ) c) A + B + + e) A + B 4 + C ( ) + E + F ( + ) 3 C + + D ( + ) b) A + B + + d) A + B + C + D + + f) A + C ( + ) + D ( + ) 3 + E + F + F E + + B ( ) + C + + ( + ) D ( + ) + E ( + ) 3. ( d converge or diverge? Why? Evaluate if it con- + ) 5 (3) ( Points) Does verges. (4) ( Points) Does d converge or diverge? Why? Evaluate if it converges. 3

5 Math 6 Calculus Spring 6 Eam V (5) ( Points) Use the Comparison Theorem to determine whether the following improper integral converges or diverges. DO NOT COMPUTE THE EXACT VALUE OF THE INTEGRAL, but show all work needed for the Comparison Theorem. 4 d

6 Math 6 Calculus Spring 6 Eam V (6) ( Points) Let the curve C be y = f() = e for. (a) (5 Points) Set up the integral for the arc length of that curve. DO NOT TRY TO EVALUATE OR SIMPLIFY IT. (b) (5 Points) Find the area of the surface made by revolving that curve about the -ais. DO EVALUATE THIS INTEGRAL.

7 Math 6 Calculus Spring 6 Eam V (7) ( points) The curve defined by parametric equations = t and y = t 3 3t has been discussed in the tetbook. Answer each of the following questions about it. In parts (b), (c) and (d), just set up the appropriate integral but DO NOT EVALUATE OR SIMPLIFY IT. WRITE IT AS AN INTEGRAL INVOLVING EXACTLY ONE VARIABLE, t. (a) (3 Points) Find the equation of the tangent line to that curve at t =. SHOW YOUR WORK. (b) ( Points) Write the integral for the arclength of the part of the curve where 3 t 3, (c) ( Points) Set up the integral for the area between that curve and the -ais for t 3. (d) (3 Points) Set up the integral for the surface area when the curve for t is rotated about the y-ais.

8 Math 6 Calculus Spring 6 Eam V Solutions () (35 Points) Evaluate the following integrals. (a) (7 Points) tan 3 () sec 3 () d = tan () sec () tan() sec() d Let u = sec() so du = tan() sec()d, and tan () = sec () = u, so the integral becomes tan 3 () sec 3 () d = tan () sec () tan() sec() d = (u )u du = (u 4 u ) du = u5 5 u3 3 + C = sec5 () sec3 () + C. 5 3 ( ) cos(6) (b) (8 Points) sin 4 (3) d = d = ( cos(6) + cos (6) ) d = 4 4 sin(6) + ( ) + cos() d 4 = 4 sin(6) + ( + sin() ) = sin(6) + sin() + C (c) ( Points) ( + 3) (partial fractions) where d = ( A + B + C + D ) + 3 d = (A)( + 3) + B( + 3) + (C + D) = (A + C) 3 + (B + D) + (3A) + 3B gives the equations A + C =, B + D =, 3A = 3, 3B = 6 so A =, B =, C =, D =. Then the integral is ( + 3) d = d + d + + d + 3 d = ln + ln + 3 ( ) 3 tan + C 3 = ln + 3 ( ) 3 tan + C either epression is correct. 3

9 Math 6 Calculus Spring 6 Eam V Solutions (d) (8 Points) / d Since cos (u) = sin (u), use the trig substitution = sin(u) so = sin (u) = cos(u) and d = cos(u)du. The limits of integration also change: = u = and = u = π 6 from the usual degree triangle. Get = / π/6 π/6 π/6 d = sin (u) cos(u) du = sin (u) du = cos(u) ( cos(u)) du = [ u sin(u) ] π/6 = [ π 6 sin(π/3) ] π/6 cos(u) du = π 3 8. () (5 Points) Since ( 4 ) = ( + )( ) = ( + )( + )( ) and each of these factors is irreducible, the denominator factors into irreducibles as ( )(+) 3 ( +), so the correct form for the partial fraction is: b) A + B + + C ( + ) + D ( + ) 3 + E + F + ( d converge or diverge? Why? Evaluate if it con- + ) 5 (3) ( Points) Does verges. This integral is improper since the upper bound is infinity. With substitution u = +, we find the indefinite integral ( + ) 5 d = u 5 du = + C so the 8u4 improper integral is the converging limit t ( lim u 5 du = lim t t 8t 4 ) ( 8( 4 = lim ) t 8 ) 8t 4 = 8.

10 (4) ( Points) Does Math 6 Calculus Spring 6 Eam V Solutions d converge or diverge? Why? Evaluate if it converges. 3 This integral is improper since the denominator of the integrand is zero at =. We must break it up into 3 d + d. If either integral diverges, then so does the 3 original. The indefinite integral is + C. Both integrals diverge. The first improper integral diverges since it is defined to be t lim t ( d = lim 3 t t ) ( ) =. The second improper integral diverges since it is defined to be lim s + s ( d = lim 3 s + () ) (s) =. It is enough to show that one of these improper integrals diverges. (5) ( Points) Use the Comparison Theorem to determine whether the following improper integral converges or diverges. DO NOT COMPUTE THE EXACT VALUE OF THE INTEGRAL, but show all work needed for the Comparison Theorem. 4 d For we have < 4 < 4 so < 4 4 = so 4 > and 4 > = >. We know that d diverges for p, so p Theorem, the given integral diverges. d diverges. By the Comparison

11 Math 6 Calculus Spring 6 Eam V Solutions (6) ( Points) Let the curve C be y = f() = e for. (a) (5 Points) Set up the integral for the arc length of that curve. DO NOT TRY TO EVALUATE OR SIMPLIFY IT. We have f () = e so + (f ()) = + 4e 4 and the integral for the arclength is + (f ()) d = + 4e 4 d (b) (5 Points) Find the area of the surface made by revolving that curve about the -ais. That surface area equals πy + (f ()) d = π e + 4e 4 d. After the substitution u = e with du = 4e d, the limits of integration are from u = to u = e, and the above integral equals π e + u du. Then use the trig sub u = tan(θ) so du = sec (θ)dθ and the integration limits u e correspond to α = tan () θ tan (e ) = β. This gives (using the integration formula from the cover page of the eam) π β sec 3 (θ)dθ = π [sec(θ) tan(θ) + ln sec(θ) + tan(θ) ] α = π [sec(β) tan(β) + ln sec(β) + tan(β) sec(α) tan(α) ln sec(α) + tan(α) ]. 4 The triangle corresponding to = tan(α) has side opposite angle α, adjacent side and hypotenuse 5, so sec(α) = 5. The triangle corresponding to e = tan(β) has side e opposite angle β, adjacent side and hypotenuse + 4e 4, so sec(β) = + 4e 4. The final answer is then π 4 [e + 4e 4 + ln e + + 4e 4 5 ln + 5 ]. β α

12 Math 6 Calculus Spring 6 Eam V Solutions (7) ( points) The curve defined by parametric equations = t and y = t 3 3t has been discussed in the tetbook. Answer each of the following questions about it. In parts (b), (c) and (d), just set up the appropriate integral but DO NOT EVALUATE OR SIMPLIFY IT. WRITE IT AS AN INTEGRAL INVOLVING EXACTLY ONE VARIABLE, t. (a) (3 Points) Find the equation of the tangent line to that curve at t =. SHOW YOUR WORK. At t = we have the point (, y) = (4, ) and the slope of the tangent line is dy d = dy/dt d/dt = 3(t ) t 3(3) which equals ( ) = 9 at t = so the equation of the tangent line to 4 the curve at that point is y + = 9 4 ( 4). (b) ( Points) Write the integral for the arclength of the part of the curve where 3 t 3. The arclength integral is 3 3 (d/dt) + (dy/dt) dt = 3 3 (t) + (3t 3) dt. (c) ( Points) Set up the integral for the area between that curve and the -ais for t 3. The integral giving that area is 3 yd = 3 (t 3 3t) tdt. (d) (3 Points) Set up the integral for the surface area when the curve for t is rotated about the y-ais. The surface area of the curve rotated about the y-ais is π(t ) (t) + (3t 3) dt

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