Math 152 Exam 3, Fall 2005

Transcription

1 c IIT Dept. Applied Mathematics, December, 005 PRINT Last ame: KEY First ame: Sigature: Studet ID: Math 5 Exam 3, Fall 005 Istructios. For the multiple choice problems, there is o partial credit. For the workout problems, you must show the mathematical steps of the solutio i order to receive full credit; writig oly the aswer will receive o credit. Writte explaatios for each step are ot required (due to time costrait) but a icorrect solutio with a correct writte explaatio might receive partial credit. You do ot have to verify the precoditios of a covergece test uless it is explicitly requested. Coditios. No calculators, computers, otes, books, or scratch paper. By writig your ame o the exam you certify that all work is your ow, uder pealty of all remedies outlied i the IIT studet rules. Please do ot talk util you are away from the room. Time limit: 50 miutes (strict). NOTE: The topics may ot be i order either of icreasig difficulty or of the order they were covered i the course. POSSIBLY USEFUL FORMULAS sec x = ta x + M = x [ f ( ) ( x 0 +x + f x +x ) ( + + f x +x )] cos +cos x x = T = x [f(x 0) + f(x ) + + f(x ) + f(x )] dx = ta x + C +x ta x dx = l cos x + C P V = RT E M < K(b a)3 (K f (x)) 4 F = ρgad E T < K(b a)3 (K f (x)) R (x) M x (+)! a + (r r 0 ) = 0 S = x [f(x 3 0) + 4f(x ) + f(x ) + f(x ) + 4f(x ) + f(x )] E S < K(b a)5 (K f (4) (x)) 80 4 f(x) dx R + f(x) dx x dx = si x + C = +x ( ) x d dx (sec x) = x x s s a + x = x si x = si x cos x Vol = b π a [(f(x)) (g(x)) ] dx f(x) = f () (a) (x a)!

2 c IIT Dept. Applied Mathematics, December, 005 Part I. (0pts ea.) MULTIPLE CHOICE NO PARTIAL CREDIT NO CALCULATORS. Compute the limit lim a) 7 b) 9 c) 3 d) 4 e) 0 = = ( lim ( 8 ) / ) /4 sice the leadig powers of are both 5 = 3. Which statemet is true about the series ( ) l +? a) It is a coverget p-series whose th partial sum is s = l( + ). b) It is a diverget telescopig series whose th partial sum is s = l( + ). c) It is coverget because lim l(/( + )) = 0. d) It is diverget because lim /( + ) =. e) It is coverget because l(/( + )) < / ad is coverget. a) It is ot a p-series. c) The th term test does ot give covergece, oly divergece if the limit is ozero. d) This limit is i fact, but this is ot the limit of the th term. e) The compariso is ivalid diverges. To see that b) is true, ote that ( ) l + = (l l ) + (l l 3) + + (l l( + )) + Sice l = 0, the th term is l( + ), which goes to. 3. The sum of the series is a) b) 3 c) d) e) 3 The first term is /3. The ratio is (/9)/(/3) = /3. The formula for geometric series gives /3 /3 =.

3 c IIT Dept. Applied Mathematics, December, Select the correct completio of the statemet: The series a) coverges by the ratio test. b) diverges by the ratio test. c) coverges, by compariso with d) diverges, by compariso with e) coverges, by compariso with a) ad b) are false because the ratio test results i a limit of which is icoclusive. c) is false because / 3 is ot greater tha 3 /( 4 /). e) is false because /() is ot less tha 3 /( 4 /). d) is true sice? 3 4 / 4 /? 4, which is true for all. (? meas we are t sure if the iequality is true iitially, but goig backwards verifies it.) 5. Select the expressio which represets the area bouded by oe loop of the graph of the polar fuctio r = si(3θ). a) b) c) d) e) π/6 π/6 π/6 π/6 π/3 0 π/3 0 π/3 0 π si (3θ) dθ si(3θ) dθ si(3θ) dθ si (3θ) dθ si (3θ) dθ The formula for polar area is β α (f(θ)) dθ. Aswer d) is impossible sice it correspods to two loops of the graph: at all three values θ = 0, π/3, π/3, si(3θ) = 0. Aswer e) is the oly possible choice.

4 c IIT Dept. Applied Mathematics, December, Part II. SHOW WORK FOR FULL CREDIT NO CALCULATORS 6. (5 pts) Use oe of the itegral remaider estimates to estimate the value of withi 0 4. Clearly idetify the estimate, error, ad error boud. Show your work which guaratees that the estimate is withi this amout of the actual sum. actual estimate error error boud tolerace Way s s s s s s f(x) dx 0 4 R Way s s + f(x)dx+r + f(x)dx s est. s est. a 0 4 to Oly oe way is required for the solutio. First way. Pluggig i f(x) = x, set RHS of error boud tolerace ad solve for. x Therefore s 5000 is withi 0 4 of s. dx x 0 4 = = Now Way, agai with f(x) = x. The estimate is s 50 + which is withi 0 4 of s. a = f(x)dx + 5 f(x)dx = 50. = s 50 + = s 50 + x + 50 x , 5

5 c IIT Dept. Applied Mathematics, December, (5 pts) Assume that f(x) = si x has power series represetatio si x = ( ) x + ( + )! with radius of covergece R =. Fid a power series represetatio for g(x) = cos x ad determie, with justificatio, its radius of covergece. Way. Just take the derivative of both sides. Whe we take the derivative of a power series, the radius of covergece stays the same. d dx si x = d dx cos x = ( ) x + ( + )! d dx ( ) x + ( + )! = ( + )x ( ), ( + )! with radius of covergece R = as described above. Way. Note the idetity cos x = si(x + π/). So we ca just replace x with x + π/ i the power series for si x. (x + π/)+ ( ), ( + )! which coverges everywhere (R = ) sice the origial coverged everywhere ad we oly plugged i a shifted value. 8. (0 pts) Fid the ceter, radius ad the iterval of covergece of the power series (x 3) ( ). 5 The ceter of the power series is a = 3 from the way the series is preseted. Ratio test: lim a + a = lim (x 3) + (+) 5 + (x 3) 5 = lim (x 3) ( + )5 x 3 =, 5 ad the ratio test says the series coverges whe this quatity is < ad diverges whe > : x 3 5 < x 3 < 5 5 < x 3 < 5 < x < 8.

6 c IIT Dept. Applied Mathematics, December, The radius of covergece is half the distace betwee the edpoits of the iterval of covergece. Aother way to see that R = 5 it is that (, 8) = (3 5, 3 + 5). Fially we must check for covergece at the edpoits. At x = : ( ) ( 5) 5 = which diverges as a p-series with p =. (Or do itegral test.) At x = 8: ( ) (5) 5 = which coverges because it is a alteratig series. Therefore the iterval of covergece is (, 8]., ( ),