06 微甲 0-04 06-0 班期中考解答和評分標準. ( poits) Determie whether the series is absolutely coverget, coditioally coverget, or diverget. Please state the tests which you use. (a) ( poits) (b) ( poits) (c) ( poits) = = = ( ) l(!) 3 l ( ) ( 3 si ( 3 )) ( ) 3 ( + ) Solutio: Questio (a) Solutio pts Correct Icorrect = ( ) l(!) 3 l Cosider that Ad series = l(!) 3 l l 3 l =. is coverget by p-series for p = >. Therefore we ca apply Limit Compariso Test to determie ( ) l(!) is abso- 3 l lutely coverget. = l(!) Those who thought = 3 l is diverget for ay reaso ad the prove that ( ) l(!) = 3 l is coverget by Alteratig Series Test with a correct process get poits. (a) is goig to be coverget by Itegral Test. = (b) l(!) = l + l + + l < l + l + + l = l (c) l(!) l( ) = l (d) l + = l x dx < l(!) < (e) Stirlig Formula:! π ( e ) = + l x dx = ( + ) l( + ) l(!) l + l π (a) Ratio Test: lim a + = leads o coclusio. a (b) L Hôspital Rule: Differetiate l(!) leads mistakes. (c) Test for Divergece: The limit of a as is zero. So we ca ot use it to coclude the series is diverget. (d) Limit Compariso with leads ad o coclusio. 3 (e) = = π 6 = x dx l (f) lim l + l l + + l = 0 + 0 + + is wrog. l (b) Let a = 3 si 3, for ay positive iteger. The, a 0. (For x 0, si(x) = si(x) si(0) = x cos(ξ) x, for some ξ (0, x), by Mea Value Page of 9
Theorem. Hece, 3 si 3 0.) This problem ca be decomposed ito two parts. Covergece of. Divergece of Covergece of ( ) a = a = = ( ) a (3 poits) ( poits) There are two kids of gradig, depedig o what kid of method oe used.. Directly applyig Alteratig Series Test: ˆ ( poit) a is decreasig : Let f(x) = 3 si x 3. f (x) = x 3 x 4/3 ( cos(x)) 0, for x > 0. Hece, f(x) is decreasig as x icreases (whe x > 0). Sice a = f() for all, a is decreasig as icreases. ˆ ( poit) lim a = 0 : lim a = lim ( 3 si( 3 )) = lim 3 lim si( 3 ) = 0 0 = 0, sice si(x) is cotiuous with respect to x. Hece, by Alteratig Series Test, ( ) a = ( ) ( 3 si 3 ) is coverget.. Cosiderig covergece of = = ( ) 3 ad = = ( ) si( 3 ): ( ) ˆ ( poit) Covergece of 3 : = 3 0, lim 3 = 0 ad 3 is decreasig as icreases. Therefore, by Alteratig Series Test, 3 is coverget. ( ) = ˆ ( poit) Covergece of = ( ) si( 3 ): si( 3 ) 0, lim si( 3 ) = 0 ad si( 3 ) is decreasig as icreases. Therefore, by Alteratig Series Test, Divergece of a = = We will use Compariso Test to demostrate that ( ) si( 3 ) is coverget. = a is diverget. Note that, if oe oly use upper boud or egative lower boud of a to get the divergece of will get 0 poit i this part. = a, he/she Page of 9
ˆ ( poits) x si(x) lim x x 3 6 > 0. Therefore, ˆ ( poit) Compare = = = lim x cos(x) 3x a with some appropriate series: = lim x si(x) 6x a ad = Divergece of the appropriate series: = 6 lim both coverges or both diverges. a / = lim By p-series test or itegral test or compariso test, we kow that ad hece, so is = a. That is, = (c) Let a = ( ) 3 ( + ), for all positive iteger. ˆ ( poits) lim a = lim ( + ) = e ˆ ( poits) e > (strictly larger tha ) ˆ ( poit) By Root Test, = a = ( 3 si 3 ) is diverget. ( ) 3 = ( + ) is diverget. = 3 si 3 (/ 3 = ) 3 is diverget, Page 3 of 9
(0-0 班 ) Determie whether the series is absolutely coverget, coditioally coverget, or diverget. Please state the tests which you use. (a) ( poits) ( ) l(!) 3 l (b) ( poits) (c) ( poits) Solutio: (a) = ( ) l = = 4 3 ( ) + 3 (b) Observe that / is decreasig to zero, ad l() is icreasig to ifiity. So / l() is descreasig to zero. Hece by alteratig series test, is coverget. ( %) O the other had, observe that Sice = l() = / l() = el()/ l() ( ) / l() is diverget by itegral test, the series is also diverget by compariso test. (3 %) Thereofore, = ( ) / l() (c) Let a = 4 3. Observe that ( ) + 3 lim a + a Hece by ratio test, ( + l() ) = l() l() l(). / = l() is coditioally coverget. ( + ) = lim ( + ( 3) + + ( 3) 4+ ( + ) 3 ) + 4 3 = ( + ) + ( 3/) = lim ( + ( 3) ( 3/) 4 ( + )3 /4 ) 3 /4 = 4 = 4. ( %) a is absolutely coverget. Page 4 of 9
. (0 poits) Fid the radius of covergece ad the iterval of covergece of the power series (x ). = (l ) 3 4 Solutio: Write a = (x ), by ratio text, we have (l ) 3/4 lim a + a = lim (x ) + + ) (l( ) = x l 3/4 Hece, we have x <, or x < (4%). Now we check the covergece of edpoits: ˆ Whe x = 0: ( ) Now a = (l ). Note that lim 3/4 (l ) = 0(%) ad 3/4 (l ) 3/4 decreasig.(%) By Leibitz test, it is coverget.(%) is obviously ˆ Whe x = : Now a = (l ). Write f(x) =, the f is obviously positive (for x > ), 3/4 x(l x) 3/4 decreasig(%), ad cotiuous. By itegral test, we have dx x(l x) 3/4 = 4(l x)/4 = (%) Therefore, it is diverget.(%) Hece, the radius of covergece is ad the covergece iterval is [0, ) Page of 9
(0-0 班 ) (a) ( poits) Fid the costat p such that lim + + + (b) ( poits) Fid the iterval of covergece of the power series p is a fiite ozero costat. = ( 3x) + + +. Solutio: (a) Let L = lim So + + + + p Hece by squeeze theorem,. Observe that dx + + + + x + p Thus the costat p = /. ( %) + + +. p p If p > /, the L = 0, If p = /, the L =, If p < /, the L =. (b) Let f (x) = + + + ( 3x). The lim f +(x) f (x) = lim = lim + + + + + + + + + + = 3x = 3x. x dx. (3 %) 3x + + + + + + 3x So by ratio test, if 3x <, or equiveltly, 0 < x < 3, the f (x) is coverget (3 %), ad if x < 0, or 3 < x, the f (x) is diverget. Now we are goig to check x = 0 ad x = 3. Observe that = = f (0) = = = + + + lim. By (a), sice / + + + f (0) is diverget by limit compariso test. ( %) = =, Page 6 of 9
Observe that = f ( 3 ) = ( ) + + +. So it is clear that = f ( ) is cover- 3 get by alteratig series test. ( %) Therefore, the iterval of covergece of = f (x) is (0, 3 ]. = Page 7 of 9
3. (0 poits) Let F (x) = 0 x l ( + t ) dt. (a) (6 poits) Fid the Maclauri series of F (x) ad its radius of covergece. (b) (4 poits) Estimate F (0 ) up to a error withi 0 7. Solutio: (a) Let F (x) = l( + x ). 0 Sice l(+x) = ( ) x = x l( + t )dt ad by Fudametal Theorem of Calculus we have F (x) = = ( ) x, substitute x F (x) is itegrate F (x) term by term F (x) = Next, use ratio test to fid radius of covergece. lim a + a = lim with x ad we have l(+ x ) = = ( ) x + = ( + ) ( ) x +3 + ( + )( + 3) ( + ) = ( ) x x+ ( ) ( x ) I order to make this alterative series coverge, x eed to be less tha. We have x <, hece the radius of covergece is = (b) ( 0 )+ Suppose b = ( + ) ad M = bouded by b +. Note that ( ) ( = 0 )+ ( + ) b = ( 0 ) 3 3 = = 0.00067 > 0 7 6000 b = ( 0 ) 4 = 4000000 =. 0 > 0 7 b 3 = ( 0 ) 7 8 3 7 = 68 0 7 =.9 0 0 < 0 7, we kow that the error of M (x) is Hece the summatio of first two term of F (0 ) is sufficiet to make the error less tha 0 7. F (0 ) 6000 4000000 0.000664 GRADING CRITERIA (a) Fidig the Maclauri series ad radius of covergece are 3 poits respectively. Write dow the basic formula of Maclauri series will get poit, aswer correct will get poits. Page 8 of 9
For radius of covergece, use ratio test to fid the aswer will get poit, aswer correct will get poits. You will loss poit for each calculatio error. (b) If you try to estimate the error F (0 ), you will get poits eve if the fial aswer is wrog. If the aswer is correct, you will get aother poits. You will loss poit for each calculatio error. Page 9 of 9
4. (8 poits) (a) (4 poits) Idetify the power series =0 (b) (4 poits) Fid the sum 3 9 3 + 4 3 89 3 + ( ) + x + as a elemetary fuctio. + Solutio: (a) Method. ( ) x+ =0 ( ) (x)+ =0 ta x = + (%) + = ta x (%) Method. (b) for x < ( ( ) (x)+ =0 ( ) (x)+ =0 + + ) = ( ) (x) = = =0 + 4x (%) + 4x dx = ta x + C (%) The series equals 0 whe x = 0, so C = 0. (%) Therefore, ( ) (x)+ =0 + = ta x 3 9 3 + 4 3 89 3 +... = 3 3( 3) + 3 ( 3) 7( 3) +... 7 = =0 ( ) + ( 3 ) + (%) = ta 3 (%) = π 6 (%) Page 0 of 9
. ( poits) Let r(t) = (si t t cos t)i + (cos t + t si t)j + t k, 0 t π, be a vector fuctio that parametrizes a curve i space. (a) (3 poits) Fid the arc legth of the curve. (b) (6 poits) At what poit o the curve is the osculatig plae parallel to the plae x+ 3y z = 0? (c) (3 poits) Fid the curvature of the curve. Solutio: (a) r (t) = (cos t cos t + t si t)i + ( si t + si t + t cos t)j + tk = t si ti + t cos tj + tk r (t) = t si t + t cos t + (t) = t ( poits) π Arc legth L = r π (t) dt = tdt = 0 0 π ( poit) (b) Osculatig plae is spaed by the taget ad ormal vector of the curve r(t), so we eed to fid T(t) ad N(t). r (t) = (t si t, t cos t, t) ad r (t) = t T(t) = r (t) r (t) = (si t, cos t, ) T (t) = (cos t, si t, 0) N(t) = (cos t, si t, 0) Normal vector of osculatig plae = (, 3, ) parallel to T N = ( si t, cos t, ) t = π 3π 3 6, that is, the osculatig plae at (, + π, π ) is parallel to the plae 36 x + 3y + z = 0. ( poits for each T, T, poit for each N, Poit) (c) (Method I)By (b) we have T (t) =, ( poit) Curvature κ = T (t) = = ( poits) r (t) t t (Method II) By (a) we have r (t) = t si ti + t cos tj + tk r (t) = (si t + t cos t)i + (cos t t si t)j + k ( poit) r (t) r (t) = t si ti + t cos tj t k r (t) r (t) = t ( poit) κ = r (t) r (t) t r (t) 3 = ( t) 3 = ( poit) t Page of 9
(sol II) (i) r (t) = t si ti + t cos tj + tk r (t) = t (ii) r (t) = (si t + t cos t)i + (cos t t si t)j + k (iii) r (t) r (t) = t si ti + t cos tj t k r (t) r (t) = t (iv) T(t) = r (t) r (t) = (si t, cos t, ) (v) T (t) = (cos t, si t, 0) T (t) = (vi) N(t) = T (t) T = (cos t, si t, 0) (t) (vii) B(t) = T(t) N(t) = ( si t, cos t, ) (a) Arc legth L = 0 π r (t) dt = 0 π tdt = π (b) Normal vector of osculatig plae = (, 3, ) parallel to B(t) = ( si t, cos t, ) t = π 3π 3 6, that is, the osculatig plae at (, + π, π ) is parallel to the plae 36 x + 3y + z = 0. (c) κ = T (t) r (t) = t = t or κ = r (t) r (t) r (t) 3 = t ( t) 3 = t Page of 9
6. (0 poits) Let surface S be give by S = {(x, y, z) R 3 si(xyz) = x + y + 3z}. (a) (4 poits) O the surface, compute z y ad x x. (b) ( poits) Fid a equatio of the taget plae to the surface S at (,, 0). (c) (4 poits) Suppose, whe restricted to the surface S, a differetiable fuctio f attais a local maximum value at the poit (,, 0) with f(,, 0) = 0 ad f x (,, 0) =. Let (x 0, y 0, z 0 ) be a poit which is close to the poit (,, 0) ad lies o aother surface si(xyz) = z + y + 3z + 0. Use the liear approximatio to estimate f(x 0, y 0, z 0 ). Solutio: Defie g(x, y, z) = si(xyz) x y 3z. (a) Treatig z implicitly as a fuctio of x ad y, by chai rule we ca differetiate the equatio g(x, y, z) = 0 as follows: We obtai Similarly, g x = g x x x + g y y x + g z z x = g x + g z z x = 0. z x = g x yz cos(xyz) = g z xy cos(xyz) 3. y x = g x yz cos(xyz) = g y xz cos(xyz). (b) The taget plae to the surface S at (,, 0) is g(,, 0) < x, y ( ), z 0 > = (x ) (y + ) z = 0, or x + y + z = 0. (c) Sice f(,, 0) is a local maximum value, by the method of Lagrage multiplier, there is a umber λ such that f(,, 0) = λ g(,, 0). From the x-expoet of the equatio ad the fact that f x (x, y, z) = we fid that λ = ad thus f(,, 0) = g(,, 0). It follows from the liear approximatio of g at the poit (,, 0) that 0 = g(x 0, y 0, z 0 ) g(,, 0) g(,, 0) < x 0, y 0 +, z 0 > Therefore, the liear approximatio of f at (,, 0) yields Markig Scheme f(x 0, y 0, z 0 ) f(,, 0) + f(,, 0) < x 0, y 0 +, z 0 > = 0 g(,, 0) < x 0, y 0 +, z 0 > 0 0 = 9.98. (a) poit for each derivatio usig chai rule or direct use of formula; poit for each correct aswer. (b) poit for the formula of the taget plae ad poit for the correct equatio. (c) poit for usig Lagrage s method; 0. poit for the correct λ. poit for each approximatio of f ad g; 0. poit for the correct estimate. Page 3 of 9
x 3 + y 3 if (x, y) (0, 0). 7. (3 poits) Let f(x, y) = x + y 0 if (x, y) = (0, 0). (a) (3 poits) Is f(x, y) cotiuous at (0, 0)? Justify your aswer. (b) ( poits) Fid the gradiet vector f(0, 0). (c) (4 poits) Is f x (x, y) cotiuous at (0, 0)? Justify your aswer. (d) (4 poits) Fid the maximum ad miimum directioal derivatives of f at the poit (0, 0) amog the directios of all the uit vectors u. Solutio: (a) x = r cos θ, y = r si θ f(x, y) = r3 (cos 3 θ + si 3 θ) r = r cos 3 θ + si 3 θ r cos 3 θ +r si 3 θ r So, f(x, y) 0 = f(0, 0) as r 0 is as (x, y) (0, 0) Therefore, f is cotiuous at (0,0). Gradig Policy: () 3 poits for correct proof. () No partical poits. h 3 h 0 f(h, 0) (b)f x (0, 0) = lim = lim h 0 h h h = f(0, h) h f y (0, 0) = lim = lim 3 h 0 h h 0 h h = f(0, 0) = i + j Gradig Policy: ()Correct limits for poit. ()Correct aswer for poit. (c)away from (0, 0), f x = 3x (x + y ) (x 3 + y 3 )(x) (x + y ) = x4 + 3x y xy 3 (x + y ) Let x = r cos θ, y = r si θ f x = cos 4 θ + 3 cos θ si θ cos θ si 3 θ f x is D θ=0. θ = 0 f(x, y) =, ad θ = π f(x, y) = f x (x, y) is NOT cotiuous at (0,0). Other method: f x = x4 + 3x y xy 3 (x + y ) Let y = mx f x = m + m Page 4 of 9
So, the limit as (x, y) (0, 0) alog the differet lies y = mx is differet for differet m. f x (x, y) is NOT cotiuous. Gradig Policy: ()Correct f x for poits. ()Correct limit at (0, 0) for poits. (d) u = (cos θ, si θ) f(r cos θ, r si θ) D u f(0, 0) = lim r 0 r r 3 (cos 3 θ+si 3 θ) r r = lim r 0 = cos 3 θ + si 3 θ Let g(θ) = cos 3 θ + si 3 θ g (θ) = 3 si θ cos θ + 3 cos θ si θ If g (θ) = 0, the θ = 0 or π 4 or π or π 4 Maximum is g(0) = g( π ) =. ad so o. Miimum is g(π) = g( 3π ) =. Gradig Policy: ()Fid the directioal derivative i directio of u = (cos θ, si θ) for poits. ()Correct maximum ad miimum arguets for poit. (3)Correct aswer for poit. Page of 9
8. ( poits) Let f(x, y) = + 3x x 3 + 3y y 3. (a) (6 poits) Fid the local maximum ad miimum values ad saddle poit(s) of f(x, y). (b) (6 poits) Fid the extreme values of f(x, y) o the regio D bouded by the triagle with vertices (, ), (, ) ad (, ). Solutio: (a) f x = 6x 6x = 6x ( x) f y = 3 3y = 3 ( y) ( + y) f x = 0 solve imply critical poits are :(, ), (, ), (0, ), (0, ) f y = 0 f xy = f yx = 0 f xx = 6 x f yy = 6y D = f xx f yy = 7xy 36y at poit (, ) D (, ) = 36 > 0, f xx (, ) = 6 < 0,local maximumf (, ) = 4 at poit (, ) D (, ) = 36 < 0,saddle poit at poit (0, ) D (0, ) = 36 < 0,saddle poit at poit (0, ) D (0, ) = 36 > 0, f xx (0, ) = 6 > 0,local miimumf (0, ) = (b) for y=, x f (x, ) = x 3 + 3x deote g (x) g (x) = 6x (x ),solve g (x) = 0, x = 0, f (, ) = 7, f (0, ) =, f (, ) = 0, f (, ) = for x=, y f (, y) = y 3 + 3y 3 deote g (y) g (y) = 3 (y ) (y + ),solve g (y) = 0, y =, f (, ) =, f (, ) =, f (, ) =, f (, ) = for x+y=0, x f (x, x) = x 3 + 3x 3x + deote g 3 (x) g 3 (x) = 3 (x ),solve g 3 (x) = 0, x = f (, ) = 7, f (, ) = 0, f (, ) = maximum = 7, at (, ) Comparig above poit ad critical poits we get, miimum =, at (, ), (, ) [Gradig] (a) ( poits) f x, f y, f xx, f yy ad fid 4 critical poits (4 poits) correct determie each critical poit is locatio maximum,locatio miimum,saddle poit if you do t write the local maximum ad local miimum values, you will lose poit (b) if you oly cosider poits i the iterior of the triagle ad extremum o boudary, you will get at most 4 poits if you oly cosider poits i the iterior of the triagle ad corers of the triagle, you will get at most 3 poits Page 6 of 9
if you cosider all poits but does t dissusio, you will get at most poits if you oly cosider poits i the iterior of the triagle, you will get at most poit Page 7 of 9
9. (0 poits) By the Extreme Value Theorem, a cotiuous fuctio o a sphere attais both absolute maximum ad miimum values. Fid the extreme values of f(x, y, z) = l(x+)+l(y+)+l(z+) o the sphere x + y + z = 3. Solutio: Step. Let g(x, y, z) = x + y + z = 3. Accordig to the method of Lagrage multipliers, we solve the equatio f = λ g ad g(x, y, z) = 3. This gives (3pts) x + = λ x y + = λ y z + = λ z x + y + z = 3 Step. Note that λ 0 because λ = 0 implies Thus we have From x(x + ) = y(y + ), we have which gives = 0, which is impossible. x + = x(x + ) = y(y + ) = z(z + ) λ 0 = x y + x y = (x y)(x + y + ) y = x or y = x Similarly, from x(x + ) = z(z + ), we have z = x or z = x Case. y = x ad z = x From x + y + z = 3, we have 3x = 3 ad the x =,. Thus we have two poits (,, ), (,, ). Case. y = x ad z = x From x + y + z = 3, we have 3x + 4x + 4 = 3 ad the x = 3,. Thus we have two poits ( 3, 3, ), (,, ). 3 Case3. y = x ad z = x From x + y + z = 3, we have 3x + 4x + 4 = 3 ad the x = 3,. Thus we have two poits ( 3, 3, ), (,, ). 3 Page 8 of 9
Case4. y = x ad z = x From x + y + z = 3, we have 3x + 8x + 8 = 3 ad the x = 3,. Thus we have two poits ( 3, 3, ), (,, ). 3 Hece f has possible extreme values at the poits (,, ), (,, ), ( 3, 3, 3 ), ( 3, 3, 3 ) ad ( 3, 3, 3 ). (pts) Step3. We compare the values of f(x, y, z) at these poits: ˆ f(,, ) = l 7 ˆ f(,, ) = 0 ˆ f( 3, 3, 3 ) = f( 3, 3, 3 ) = f( 3, 3, ) = l 3 7 Therefore the maximum value of f o the sphere x + y + z = 3 is l 7 ad the miimum value is l 7. (pts) Page 9 of 9