Math 2153, Exam III, Apr. 17, 2008

Transcription

1 Math 53, Exam III, Apr. 7, 8 Name: Score: Each problem is worth 5 points. The total is 5 points. For series convergence or ivergence, please write own the name of the test you are using an etails of using the test. Otherwise no creit will be given.. Test the following series for convergence or ivergence. Then gives an estimate of R S S. 7 ln 7 ln ln 4 7 ln ln 6 Solution This is an alternating series an can be written as ( ) n+ 7 ln(n + ) n By using the alternating series test an notice that the absolute value of the general term satisfies 7 (a) is ecreasing as n increases; ln(n+) (b) lim n 7 ln(n+), we know the series is convergent. For alternating series, R a 7 ln.

2 . Determine whether sin(4n) n is absolutely convergent, conitionally convergent, 4 n or ivergent. Solution Notice that n sin(4n) 4 n an because n is a convergent geometric series. By using the comparison 4 n test, we know that n sin(4n) is convergent. Hence sin(4n) 4 n n is absolutely 4 n convergent by efinition. n 4 n 3. Determine whether ( n n + n n +) is absolutely convergent, conitionally convergent, or ivergent. Solution By using the root test ( n + lim n n n + ) n lim n Therefore the given series is absolutely convergent. n + n + lim + /n n + /n <.

3 4. Fin the raius of convergence an interval of convergence of ( ) n x n n. n+4 (You nee to specify whether the series converges or not at the enpoints of the interval.) Solution Using the Ratio test lim n lim n xn + 4 n + 5 lim + 4/n n x + 5/n x. ( ) n+ x n+ (n+)+4 ( ) n x n n+4 The power series converges when x < an iverges when x >. Hence the raius of convergence is. To fin the interval of convergence, we nee to know whether the series converges or not when x or x. On these two enpoints, the ratio test will give x an hence inconclusive. Other tests nee to be use in orer to raw the conclusion. (a) when x, the series becomes ( ) n n which is an alternating series. By n+4 using the alternating series test, it can be shown that the series is convergent; (b) when x, the series becomes with a ivergent p-series n n n+4, we know that n n. By using limit comparision test is also ivergent. n+4 Combine the above, we can see that the series converges on I (, ]. 3

4 5. Fin a power series representation of centere at for f(x) x. 9+x (You nee to give the general form of the series instea of writing only the first several terms. For example, n x n is a correct form while + x + 4x + 8x 3 + will not be accepte.) Solution We will use the formula x xn. f(x) x 9 + x x 9 + x 9 x 9 ( x) 9 x ( x 9 9 )n x 9 ( x 9 )n ( ) nxn+ 9 n+ 4

5 6. Fin the first 5 terms in the Taylor series representation centere at a for f(x) x. Solution n f (n) (x) f (n) () x / x / 4 x 3/ x 5/ x 7/ 5 6 So the first five terms of the Taylor series are f(a) + f (a)! (x a) + f (a)! (x a) + f (a) 3! (x a) 3 + f(4) (a) (x a) 4 4! + (x ) 4! (x ) ! (x ) ! (x )4 + + (x ) 8 (x ) + 6 (x )3 5 8 (x )4 + 5

6 7. Use Taylor series to evaluate the integral sin x x x Solution Since we know the Taylor series for sin x is xn+ ( )n, (n+)! xn+ sin x x x ( )n (n+)! x x ( ) n x n (n + )! x ( ) n x n (n + )! x ( ) n x n+ (n + )!(n + ) + C 6

7 8. Eliminate the parameter t to fin a Cartesian equation of the curve: { x ln(9t) y t Solution From the first equation, we have x ln(9t) 9t ex/ t ex/ 9 Substitute it into the secon equation gives e x/ y 9 Solution From the secon equation, we have Substitute it into the first equation gives t y x ln(9y ) 7

8 9. Fin an equation of the tangent line to the parametric curve at the point corresponing to t. { x e t y t ln(t 9 ) Solution First, we nee to fin the slope at t. The erivative y x is y y x t x t t 9 9t 8 e t 9 t t / e t t So at t, the slope is y x 9 t e 6 e Next, we nee to fin the xy coorinates at t. Finally, the tangent line is x t e, y t ln( 9 ) (y ) 6 e (x e) 8

9 . A cycloi is the curve efine by the path of a point on the ege of circular wheel as the wheel rolls along a straight line. If the rotating wheel has raius, the equation of the cycloi is { x (θ sin θ) Fin the area of the shae region. y ( cos θ) y q q p x Solution Area π π π π π π y x θ θ ( cos θ)( cos θ)θ 4( cos θ) θ (4 8 cos θ + 4 cos θ)θ (4 8 cos θ cos(θ) )θ (6 8 cos θ + cos(θ))θ (6θ 8 sin θ + sin(θ)) π π 9

10 Trigonometry csc θ sinθ tan θ sinθ cos θ cot θ tan θ sec θ cos θ cot θ cos θ sin θ sin θ + cos θ + tan θ sec θ + cot θ csc θ sin( θ) sinθ tan( θ) tan θ cos(π/ θ) sinθ sin(x + y) sinxcos y + cos xsiny sin(x y) sinxcos y cos xsiny tan(x + y) sinx sin xcos x tan x sin x tanx + tany tan xtan y tan x tan x cos x cos( θ) cos θ sin(π/ θ) cos θ tan(π/ θ) cotθ cos(x + y) cos xcos y sinxsiny cos(x y) cos xcos y + sinxsiny tan(x y) tan x tan y + tanxtan y cos x cos x sin x cos x cos x sin x + cos x sinacos B [sin(a B) + sin(a + B)] sinasinb [cos(a B) cos(a + B)] cos A cos B [cos(a B) + cos(a + B)] Differentiation rules (sin x) cos x x x (tanx) sec x (csc x) csc xcotx x x (ex ) e x x (ln x ) x (cos x) sinx x x (cot x) csc x (sec x) sec xtanx x x (ax ) a x lna x (log a x) xlna

11 x (sin x) x x (cos x) x x (tan x) + x x (cot x) + x x (csc x) x x x (sec x) x x Table of integrals u v u v e u u e u + C v u sinuu cos u + C sec u u tanu + C sec u tan u u sec u + C tan u u ln sec u + C sec u u ln sec u + tanu + C u a u sin u a + C u u + a ln(u + u + a ) + C u a u a ln u + a u a + C ue au u a (au )eau + C u ln u + C u a u u au lna + C cos u u sinu + C csc u u cot u + C csc u cot u u csc u + C cot u u ln sinu + C csc u u ln csc u cot u + C u a + u a tan u a + C u u u a a sec u a + C u u a ln u + u a + C u u a a ln u a u + a + C lnuu u lnu u + C Infinite sequences an series ar n a + ar + ar + n S n + n+ f(x)x S S n + n { a r for r < ivergent otherwise f(x)x

12 Taylor series f(x) f (n) (a) (x a) n n! f(a) + f (a)! (x a) + f (a)! Some Maclaurin series an interval of convergence (x a) + f (a) (x a) 3 + 3! x x n + x + x + x 3 + (, ) e x sin x cos x x n n! + x! + x! + x3 + (, ) 3! tan x ( ) n x n+ (n + )! x x3 3! + x5 5! x7 7! ( ) n xn (n)! x! + x4 4! x6 6! ( ) n xn+ n + x x3 3 + x5 5 x7 7 + (, ) + (, ) + [, ] Area, arc length, an surface area β ( ) x area y t α t β (x ) ( ) y arc length + t α t t β α surface area πy (x ) ( t + y t) t rotate aroun x-axis β πx (x ) ( α t + y t) t rotate aroun y-axis