Solutions to quizzes Math Spring 2007

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Isabella Leonard
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1 to quizzes Math 4- Sprig 7 Name: Sectio:. Quiz a) x + x dx b) l x dx a) x + dx x x / + x / dx (/3)x 3/ + x / + c. b) Set u l x, dv dx. The du /x ad v x. By Itegratio by Parts, x(/x)dx x l x x + c. l x dx x l x. Quiz a) lim x x x Hit : Compute the limit of l of the fuctio. b) x dx Hit : A trig sub formula is eeded. a) Set l y l x l x l x x. The lim l y lim x x x x lim y lim x x el y e lim x l y e. b) Set x si θ, θ π/. The dx cos θdθ. dx x si x + c. Now dx x π/. lim R lim x /x by L Hopital s rule. y el y, so cos θ dx x cos θ dθ R dx x dθ θ + c. Thus, lim R (si R si ) 3. Quiz 3 a) Show the sequece {! }. Hit: Show it is decreasig ad bouded below.

2 to quizzes Math 4- Solutio Sice {! } > for all, the sequece is bouded below by. Next we show that the sequece is mootoe decreasig. The ratio a + a (+)! (+) (+)! ( + )! ( + ) (+)! ( + ) < for all. Hece the sequece is mootoe decreasig ad it is covergig by the theorem. b) Fid the limit of ( e 3 )k. Hit: Use the geometric series property. k Solutio First ote that this is a geometric series with the iitial term of the series ( e 3 ) ad the ratio ( e 3 )k+ ( e 3 )k e 3. Sice e 3 <, the geometric series coverges to e e. 4. Quiz 4 a) Determie if the series k coverges or diverges. k k + b) Determie if the series k 3 k + k coverges or diverges. a) Cosider the fuctio f(x). The fuctio is positive ad decreasig i (, ). This ca be see x by f (x) R <, (, ). Now f(x) dx lim dx lim x 3 R x R x R. Hece, by the itegral test, the series coverges. b) Cosider k. This series coverges by the itegral test. Now compare this series with the series i the k problem. To this ed take the ratio k 3 + k lim k k 3 k + lim coverges. k k+ k 3 k+ k k+ k 3 k+ k k (k + ) ad it s limit as k. lim lim k k k 3 k + + k k 5/. By the limit compariso test, the series i the problem + k 3

3 to quizzes Math Quiz 5 a) Fid if ( ) coverges. Justify your aswer. b) Test if ( )! coverges. Justify your aswer. a) This is a alteratig series with a > for all. Now a + a + < for all. Hece + the sequece is decreasig. Also lim. By the alteratig series test, the series coverges. b) Set a! >. a + a! Now lim < lim the series coverges. (+)! (+) +! ( + ) < for all. Hece {a } is a decreasig sequece....! < lim. Hece, lim. By the alteratig series test, 6. Quiz 6 a) Use the root test to see if the series Solutio ( k + 3k + k k + )k coverges or ot. k + 3k + lim k (k k k + )k /k + 3k + lim k k k k + lim + 3k + )( ) k (k k (k k + )( ) k series is absolutely coverget. <. By the root test, the b) Fid the radius of covergece of the power series or coditioally. Solutio (x ). Also state where it coverges absolutely lim + (x )+ lim (x (x ) x. By the ratio test, the series is absolutely coverget if ) + x < or < x < 3. The radius of covergece is. Now whe x, x, 3. At x the series is ( ). By the alteratig series test, we see this alteratig series coverges (coditioally). At x 3 the series becomes, which is the harmoic series ad it diverges. Thus the iterval of covergece is

4 to quizzes Math 4-4 [, 3). The series coverges absolutely i (, 3) ad coverges coditioally at x. Elsewhere the series is diverget. 7. Quiz 7 Give + x ( ) k x k, x <, fid the power series represetatios of k a) b) (x + ), + x, c) ta x. Hit: Use term by term differetiatio ad itegratio. a) First ote that the give series is a geometric series of ratio x ad iitial value. Hece it absolutely coverges to the fuctio + x i x <. Next ote that (x + ) is the derivative of the fuctio + x. Hece, by term by term differetiatio oe gets (x + ) ( ) k kx k ( ) k kx k x <. b) Substitute x for x i k + x ( ) k x k, x <. The oe gets power series represetatio. This covergece is absolute i x <. c) Note that ta x k k + x ( ) k x k, x < as a k + x dx. By term by term itegratio oe gets ta x ( ) k x k dx ( ) k k + xk+ + c, where c is a costat. To determie the value of c, substitute x i ta x ( ) k k + xk+ +c. Sice the left had side ad the right had side of the above equatio must agree k k at x, we coclude c. Hece, ta x k ( ) k k + xk+ as a power series represetatio. k

5 to quizzes Math Quiz 8 a) Usig a kow Taylor series, fid the limit lim x cos x x 4. b) Usig a kow Taylor series, give a approximatio of the Taylor series. a) cos x ( ) k (k)! xk. Hece, cos x k k ( ) k (k)! (x ) k cos x The lim x x 4 lim{ x + ( ) k (k)! x4(k ) }. b) si x k si x dx. Use up to the third o zero term of x ( ) k (k)! x4k. ( ) k (k + )! xk+. Usig the first three o-zero terms of the series, k si x dx is ap- x proximately equal to 3! x + 5! x4 dx (x 3! 3 x3 + 5! k 5 x5 ) 3! 3 + 5! quiz 9 a) Fid the slope of the taget lie to the curve defied by x cos t, y si 3t at (, ). b) Fid the x, y coordiates of all poits at which the curve x cos t, y 7 si t has a horizotal taget. a) dx d cos t dy d si 3t dy si t. 3 cos 3t. Thus, dx 3 cos 3t. At (, ), x cos t si t, y si 3t. The first equatio gives t π. Thus t π. the secod equatio gives 3t π. So t (/3)π. Combiig them, we see t 3π satisfies both equatios. At t 3π, dx d cos t si t ad dy d si 3t 3 cos 3t ±. Thus the curve has the vertical taget at at (, ). b) dx ad dx d cos t si t ad dy d7 si t 7 cos t. To be horizotal, we have dy d7 si t 7 cos t d cos t si t. The first equatio gives t π ± (/)π. At all these t values, dx d cos t si t ±. Now x cos(π ± (/)π) ad y 7 si(π ± (/)π) ±7. Hece, the curve has a horizotal taget at (, ±7)..

6 to quizzes Math 4-6 Quiz Give a ellipse (x ) + y 3, compute the area of the regio eclosed by the ellipse usig oe of the area formulas. Solutio Sice the ellipse x + y ad the origial ellipse eclose the regios which have the same area. We will 3 compute the area usig the ellipse x + y. Prametrize the ellipse by x cos x, y 3 si x; t π. 3 with this parametrizatio, the ellipse is parametrized couterclockwise as t icreases. By the area formula, the area is give by π x(t)y (t) cos t) 3(t + (/) si t) π 6π. π ( cos t)(3 cos t) 6 π (cos t)(cos t) 6 π (/)( + Remark Of course oe ca also use the origial ellipse with parametrizatio x + cos t, y 3 si t.. Quiz Show your work. Usig a graphic calculator without explaatio will ot be acceptable. a) Compute the slope of the cardioid r si θ at a geeral θ. b) Fid the poits i the rectagular coordiates (x, y) where the graph has a horizotal taget. a) x r cos θ ( si θ) cos θ, y r si θ ( si θ) si θ. Hece, dx dθ si θ)( si θ) ( si θ si θ ) ad dy dθ Hece, dy cos θ( si θ) dx si θ si θ. ( cos θ)(cos θ) + ( ( cos θ)(si θ) + ( si θ)(cos θ) cos θ( si θ). b) Horizotal implies dy dx dy but. gives cos θ( si θ). Hece, cos θ or dθ dθ dθ si θ. From the former oe gets θ π/, 3π/, 5π/ etc. From the latter, oe gets θ π/6, 5π/6 etc. At θ π/, dx dx (π/6). At θ 3π/, (3π/6) 4. Thus θ 3π/ gives a horizotal taget. dθ dθ dx At θ π/6 ad 5π/6,. Hece, these θ s also give a horizotal taget. At θ 3π/, dθ 3 x, y 4. At θ π/6, x, y 3. At θ 5π/6, x, y. Hece, the polar curve has a 3 horizotal taget at (, 4), (±, )..

7 to quizzes Math 4-7 Quiz Fid the area of the regio bouded by oe leaf of the polar curve r si θ. Solutio Set r si θ to fid where the curve hits the origi. This happes whe θ π; hece, at θ, π/, π, 3π/, π ad so o. The first time the curve comes back to the origi is at θ π/. Hece by the formula the area of the leaf bouded by the curve from θ to θ π/ is give by π/ π/ (si θ) dθ 4 ( cos 4θ)dθ 4 (θ si 4θ) π/ π Quiz 3 a) Draw the cofiguratio idicatig X + Y ad X Y, where X (, ) ad Y (, ). Show all the pertiet coordiates. b) Express X Y for X (, ) ad Y (, ). c) Compute the legth (magitude) of X Y. a) Ufortuately drawig the diagram is out of my computer skill. Let me just state the followig. Create the parallelogram geerated by X, Y. The diagoals of the parallelogram are the the desired vectors X + Y ad X Y. Probably you have see this diagram too may times ad you are sick of it. b) X Y ( ) +. c) X Y ( ( ), ) (, ). Hece, X Y + ( ) Quiz 4 Let X (,, ), Y (,, ). a) Fid the projectio of Y oto X ( P roj X Y ). b) Fid the orthogoal (vertical) projectio of P roj X Y. c) Show they are perpedicular. a) The compoet Comp X Y X Y X (,, ) (,, ). (,, ) (,, ). P roj X Y (Comp X Y ) X X

8 to quizzes Math 4-8 b) The orthogoal compoet equals Y P roj X Y (,, ) (,, ) (,, ). c) (,, ) (,, ). Hece they are perpedicular. Good Luck!!

Review Problems Math 122 Midterm Exam Midterm covers App. G, B, H1, H2, Sec , 8.9,

Review Problems Math 122 Midterm Exam Midterm covers App. G, B, H1, H2, Sec , 8.9, Review Problems Math Midterm Exam Midterm covers App. G, B, H, H, Sec 8. - 8.7, 8.9, 9.-9.7 Review the Cocept Check problems: Page 6/ -, Page 690/- 0 PART I: True-False Problems Ch. 8. Page 6 True-False