Math 211 Solutions To Sample Exam 2 Problems

Transcription

1 Solutios to Sample Exam Problems Math Math Solutios To Sample Exam Problems. Solve each of the followig differetial equatios / iitial value problems. (a) (+x) dy dx 4y (b) y dy dx x x 6 Solutio: (a) Usig separatio of variables, we have (+x) dy 4y (+x)dy 4ydx dx so our fial aswer is y D(+x) 4. (b) Usig separatio of variables, we have y dy dx x x 6 y y so our fial aswer is y ± x 6+C. ydy 4y dy +x dx l y l +x +C 4 l y 4l +x +4C y e 4C e 4l +x y D(+x) 4, xdx x 6 du/ u (where u x 6, du xdx) u / du y u+c x 6+C,. Test each of the followig series for covergece. Justify each aswer with the appropriate supportig work ad explaatios. ( (a) e (e) ) ( ) ( ) (i) (m) 4 (b) ( ) (f) cos( ) (j) ( ) () + (c) 4 ()! (g) e + e (k) 9 (o) (l) (d) Solutio: ( 5) 6 (h) + (l)! (p) ( ) l + (a) Sice e ( e) ( e) + ( e) + ( e) +, we see that this series is a geometric series with r /e. Therefore, sice < /e <, the series coverges by the Geometric Series Test. (Note: The Ratio Test would also work o this series.)

2 Solutios to Sample Exam Problems Math (b) First, ote that ( ), so we choose to compare the give series with the series. Sice ( ) for all, ad sice is a diverget p-series (p ), the give series diverges by the Compariso Test. (c) Sice this series has a factorial i it, we will apply the ratio test. We have 4 + ((+))! ()! 4 4 (+)(+). Therefore, sice a +/a <, the give series coverges by the Ratio Test. (d) We first simplify the th term of the series a bit: ( 5) 6 6 ( 5) ( ) 6 ( ) 5. 9 We ca ow see that this is a geometric series. Therefore, sice the give series is geometric with r 5/9, it coverges by the geometric series test. (Note: The Ratio Test would also work o this series.) (e) Sice this series is geometric with r /4, it coverges by the geometric series test. (Note: The Ratio Test would also work o this series.) (f) First, ote that cos( ) for all. Therefore, we have cos( ), ad sice is a coverget p-series (p ), we coclude that cos( ) coverges by the Compariso Test. It follows that the give series cos( ) (g) Let f(x) ex + e x, ad ote that by L Hospital s Rule, e + Therefore, e (h) First, ote that e x + x e x coverges by the Absolute Covergece Test. e x x ex. x, so the give series diverges by the Test for Divergece. +, so we choose to compare the give series with the series. Sice +, ad sice is a coverget p-series (p ), the give series coverges by the Compariso Test. (i) Let b /. The we have b + b for all, + ad b (/ ). Therefore, sice b + b for all, ad sice b, the give series coverges by the Alteratig Series Test. (j) Sice the series has a i it, we will try the Ratio Test: ( ) + + ( ) + Sice L >, we coclude that the give series diverges by the Ratio Test. +

3 Solutios to Sample Exam Problems Math (k) Sice 9 the give series diverges by the Test for Divergece. 9, (l) Because the series cotais a factorial, we will apply the Ratio Test. We have a + a (+) (+)!! +. Therefore, sice a +/a <, the give series coverges by the Ratio Test. (m) Sice ( ) / is a coverget p-series (p /), we coclude that the give series coverges by the Absolute Covergece Test. () First, ote that + 7/ 5/, so we choose to compare the give series with the series. Sice 5/ for all, ad + 5/ sice is a coverget p-series (p 5/), the give series coverges by the Compariso Test. 5/ (o) For this series, applyig the Ratio Test yields L ad is therefore icoclusive, ad all attempts to compare it to a related p-series or geometric series fail. Therefore, sice f(x) /(x(lx) ) is cotiuous, positive, ad decreasig o [, ), we will apply the Itegral Test. We have dx x(lx) ] t [ (lx) dx (substitutig u lx, du (/x)dx) x(lx) [ lt + ] l l, idicatig that the above itegral coverges to /(l ). Therefore, sice the itegral coverges, the give series also coverges by the Itegral Test. (p) Sice ( ) ( l l + the give series diverges by the Test for Divergece. ) l + ( ), x(lx) dx. For which of the 6 series i the previous problem is it possible to fid the exact value of the sum of the series? Do it. Solutio: Fidig the sum is possible for all of the series which are coverget geometric series. This applies to oly three of the series: e, (r /e, a /e), ( 5) 6, (r 5/9, a 5/54), ad (, 4) (r /4, a /4). We calculate the sum of these series below: e /e (/e) e ( 5) 6 ( 4) 5/54 +(5/9) 5 84 /4 +(/4) 5

4 Solutios to Sample Exam Problems Math 4 4. Recall that a radioactive substace decays proportioal to the amout of the substace that is preset at ay give time. (a) Fid the half-life of the radioactive substace Vikiium, which is observed to decay to 5% of its origial mass i years. (b) How much of a -gram sample of Vikiium would remai after 5 years? Solutio: (a) (Kow: y(t) y e kt, Give: y().5y ) Usig the give ad kow iformatio, we have y().5y y e k.5y e k.5 k l(.5) k l(.5)/.5 Therefore, y(t) y e.5t. We are beig asked to fid the half life of the substace, which is equivalet to fidig t such that y(t).5y. Calculatig, we have y e.5t.5y.5t l(.5) t l(.5)/(.5) 6.6. Therefore, the half life of the substace is approximately 6.6 years. (b) Now, we re beig give that y. Therefore, our formula for the amout of the substace preset after t years becomes y(t) e.5t, ad we have y(5) e (.5)(5).8. Therefore, there are about.8 grams of the substace preset after 5 years. 5. Determie whether the followig itegrals coverge or diverge. Evaluate those that do coverge. (a) xe x dx (b) x dx Solutio: (a) We have Therefore, the itegral xe x dx [ x xe x dx 4e x ] t 4e + t 4e t 4e + 8e t 4e. xe x dx coverges to /(4e ). (Formula 96) (by L Hospital s rule) (b) Note that the fuctio we are itegratig has a discotiuity at x. Therefore, ( ) dx x t dx+ x t + t x dx. We must determie whether the two improper itegrals o the right had side of the above equatio coverge or diverge. Sice dx t x t (l t l), the first itegral o the right had side of (*) diverges. Therefore, the give itegral diverges. dx also x

5 Solutios to Sample Exam Problems Math 5 6. For each of the followig power series, fid the radius of covergece ad the iterval of covergece. x (x) (a) (b) (c) (x )! Solutio: (a) First, we apply the Ratio Test. We have a + a x + (+)+ x x + x. Therefore, the give power series coverges by the Ratio Test if x <, which meas that it coverges if x <, so we ca coclude that the radius of covergece is R. Now, we eed to test the edpoits, x ad x. If x, our series becomes ( ) ( ), which coverges by the Alteratig Series Test sice /(+) / for all, ad sice (/). If x, the our series becomes, which is a diverget p-series. Therefore, the iterval of covergece of our power series is [, ). (b) First, we apply the Ratio Test. We have a + a (x) + (+)!! (x) x + for all x. Therefore, sice a +/a < for all x, the power series coverges for all x by the Ratio Test, ad we coclude that the radius of covergece is R ad that the iterval of covergece is (, ). (c) First, we apply the Ratio Test. We have a + a (+)(x ) + (x ) + x x. Therefore, by the Ratio Test, this power series coverges if x <. Also, x < < x < < x <, so we coclude that that radius of covergece is R ad that x ad x are our edpoits. We will ow test these two edpoits for covergece or divergece. If x, the our series becomes (x ) ( ) , so because the sequece,,, 4, 5, does ot coverge to, the series ( ) diverges by the Test for Divergece. If x, the our series becomes (x ), so because, the series diverges by the Test for Divergece. Sice the power series diverges at both of its edpoits, we coclude that the iterval of covergece is (, ).

6 Solutios to Sample Exam Problems Math 6 7. For each of the followig choices of a, decide whether the sequece {a } coverges ad whether the series a coverges. Justify your coclusios with appropriate supportig work ad explaatios. (a) a + (b) a e (c) a ( ) Solutio: (a) For the sequece {a }, we have so the sequece {a } coverges to /. +, + O the other had, because /, the series a diverges by the Test for Divergece. (b) Let f(x) xe x. The we have x xe x x Therefore, the sequece {e } coverges to. x e x H x e x. To decide whether the series a coverges, ote that a + a (+)e (+) [( )( + e e e + )] e. Therefore, sice a +/a /e <, the series e coverges by the Ratio Test. (c) For the sequece {a }, we have ( ). ( ) Therefore, by Theorem 4 i Sectio 8., we coclude that that the sequece {a } coverges to. For the series a, we ote that ( )., which is equivalet to sayig Therefore, sice a is a coverget p-series (p ), we coclude that a coverges by the Absolute Covergece Test. 8. Evaluate each of the followig its: (a) (t ) Solutio: (b) (t ) (c) t +(t ) (d) t +(t ) (a) As t approaches ifiity, so does t. Therefore our aswer is (t ). (b) We have (t ) t because, as t approaches ifiity, the t i the deomiator of the fractio /t approaches ifiity while the umerator remais costat. Therefore, our aswer is (t ).

7 Solutios to Sample Exam Problems Math 7 (c) As t approaches, so does t. Therefore, our aswer is (t ). t + (d) We have (t ) t + t + t because, as t approaches from the right had side, the t i the deomiator of the fractio /t approaches through positive umbers while the umerator remais costat. Therefore, our aswer is t + (t ). 9. (a) Explai why the Compariso Test for series caot be used to test the series or divergece. (b) Use the Absolute Covergece Test to show that the series i part (a) coverges. Solutio: si for covergece + (a) I order to apply the Compariso Test, the terms i the series must all be greater tha or equal to zero. Therefore, sice si ca be either positive or egative, we caot apply the Compariso Test to this series. (b) Sice si has values betwee ad for all, it follows that si for all. Therefore, si + + for all, ad sice is a coverget p-series (p ), the series si + coverges by the Compariso si Test. Thus, the series coverges by the Absolute Covergece Test. +