1 dx (5%) andˆ x dx converges. x2 +1 a

Transcription

1 微甲 - 班期末考解答和評分標準. (%) (a) (7%) Find the indefinite integrals of secθ dθ.) d (5%) and + d (%). (You may use the integral formula + (b) (%) Find the value of the constant a for which the improper integral ( + a ) d converges. + (c) (3%) Evaluate the improper integral for this a. (a)(7pts) [part (5pts)] + d let tanθ dsec θ dθ (pts) + d sec θ secθ dθ secθ dθ ln secθ+tanθ +c ln ++ +c [part (pts)] + d ln + +c (b)(pts) lim ( + a + )d t ( + a + )d consider the indefinite integral ( + a + )d ln ++ aln + +c Page of

2 Let I I lim t ( + a + )d ( + a + )d ( + a + )d lim ( ln( t ++t) ln aln(t+)+aln ) t ++t lim ln( (t+) a )+aln (pts) Thus we only need to consider lim ln let L lim ln I t ++t (t+) a ( + a )dln L+ln + (case) a > L I ln L +aln dverges t ++t (t+) a (case) a < then L, and I ln L +aln dverges (case)and(case) (pts) (case3) a L lim ln t ++t (t+) Iln L+ln ln+ln ln Thus I converges when a (pts) (c)(3pts) When a L lim ln t ++t (t+) - (pts) Iln L+ln ln+ln ln (pts) Page of

3 . (%) Find the following indefinite integrals: 3t +t+ (a) (%) t 3 dt. +t (b) (%) cos d (a) 3t +t+ t 3 A +t t + Bt+C t ( point) + A,B,C ( point) t + t+ t + dt ln t ln(t +)+tan t+c( pointt for each) (b) Let u, d udu. cos d u cos udu( points) udsin (usinu sin udu)( points) (usinu+cosu)+c ( sin +cos )+c( point for each) Page 3 of

4 3. (%) (a) (%) Solve the initial-value problem: (b) (%) Find lim (t). d dt (a )(b ), where a > b >, for (t) with (). (a) d (a )(b ) dt d (a )(b ) dt ( b a a )d t+c b ( ln a +ln b ) t+c b a a b ln a b t+c () C a b ln a b a b ln a b ln a b (a b)t+ln a b a b a b e(a b)t abe (a b)t ae (a b)t ab b (t) ab(e(a b)t ) ae (a b)t b ab(e (a b)t ) ab(a b)e (a b)t (b) lim (t) lim lim b ae (a b)t b a(a b)e (a b)t 評分標準 (a) 變數分離得 分積分出來得 分把 C 求出來得 分帶回 C 並解出 得 分 (b) 沒過程不給分 Page of

5 . (%) (a) (%) Solve the initial-value problem: y y sin, with y(). (b) (%) Find lim + y(). (a) The linear differential equation is y y sin. We multiply the integrating factor e d e ln ( 積分因子有算出來得 5 分, 有錯整題最多給 分 ) on both sides of the differential equation and get y y sin d ( ) d y sin y sind cos+c. So y() ( cos+c). ( 寫到這裡可得 8 分 ) The initial condition y() implies (+C) and we get C, ( 解對積分常數再得 分 ) so the solution of the differential equation is (b) The limit is y() ( cos ). ( 最後完整的函數寫出再得 分 ) y() ( cos ) cos lim + lim + lim. + 注意到, 若用以上觀察, 寫 或 不存在 或 發散 都可以算對, 得 分 若是用羅必達法則計算該極限, 最後的答案只能寫, 而不能寫 不存在 或 發散 因為使用羅必達法則後, 若分子 分母各自微分後的極限 不存在 或 發散, 並不能對原極限下結論 ; 於是判定你對羅必達法則使用錯誤或是沒有正確理解, 無法給分 課本第 3 頁指出, 使用羅必達後, 唯有 極限存在 或是 ± 這種狀況可以反推原極限的結果 Page 5 of

6 5. (%) Evaluate lim ( tant) /t dt. First, note that lim ( tant) t dt is of the form. We can use l Hospital s rule to evaluate the limit. lim ( tant) t dt lim( tan) By fundamental theorem of calculus. (5 pts) Then we take log to evaluate the last limit. ln( tan) lim is of the form sec lim by l Hospital s rule tan Thus we have lim ( tant) t dt e e (5 pts) Note lim where f() FTC:If f() ( tant) t dt ( tant) t dt f () g(t)dt for [,a] then f () g() for (,a) but not for,a Page of

7 . (%) (a) (%) Show that the area of an ellipse with the semi-major ais of length a and the semi-minor ais of length b is ab. See Figure (a). (b) (8%) A toothpaste tube is modeled in Figure (b). y y b a z (a) z (b) z Figure : (a) The area of an ellipse is ab. (b) Find the volume of the modeled toothpaste tube. One side is flat and is located at,y,z. The other side is a circle with radius, so the equation of the circle is +y,z. Each cross-section for < z < is an ellipse with the semi-major ais of length a and the semi-minor ais of length b, where a +( ) z and b z. Find the volume of the modeled toothpaste tube with z. (a) Ellipse equation: Aera a ab ab b a + y y ±b b b ad (%) a a a d a (%) cos θdθ (let asinθ, d acosθdθ) (%) (+cosθ)dθ (%) ab(θ+ sinθ) ab (%) (b) Volume ( +( ) z ) zdz (%) z +( ) z dz ( z + ( ) z 3 ) (%) (%) Page 7 of

8 7. (%) (a) (3%) Find all intersection points of the two curves r sinθ and r cosθ in their polar equations. Figure : (a) Find all intersection points of the two curves. (b)find the area of the shaded region. (b) (9%) Find the area of the shaded region in Figure. (a). If we solve the equations r sinθ and r cosθ, we get { r sinθ r cosθ sin θ cosθ sin θ θ, 5, 7, r,,, We have found two points of intersection: (r,θ) (, ) (pt) (r,θ) (, 5 ) (pt) However, if we plug θ into r sinθ and θ / into r cosθ,we can find one more point of intersection: r (pt) (b). A ( ( sinθ) dθ + ( cosθ)dθ + cos θdθ (θ sinθ ) + 3 (pt) + sinθ cos θdθ (pts) ) (pts) Page 8 of

9 8. (%) The front door of a school bus is designed as in Figure 3. It is a folding door with AB BO. The door is opened or closed by rotating OB about z-ais, while A is moving along y-ais. y z School Bus A γ(t) y A γ(t) B O Figure 3: (a) Find the enclosed area by the curve γ(t) and two aes. (b) Find the length of the curve γ(t). The region swept out by the bus door on the y-plane is enclosed by the two aes and the curve γ(t) parametrized by (sin 3 t,cos 3 t) if t γ(t) ((t),y(t)) ( sint, ) cost if t, where t denotes the angle between the positive y-ais and OB. (a) (%) Find the area of this region. (b) (%) Find the length of the curve γ(t). O t B C The curve consists of two parts: a half of a branch of an astroid (for t ), and one-eighth of a circle (for t ). (a) () Formulation: A yd 3cos tsin tdt+ y(t) (t)dt+ y(t) (t)dt cos tdt Or, noting that as y goes from to, t varies from A dy (t)y (t)dt+ (t)y (t)dt 3cos tsin tdt+ sin tdt to, cos 3 t 3sin tcostdt+ cost costdt Alternatively, if polar coordinates are to be used (letting the polar ais be in the +y direction and θ be in the clockwise direction), θ t for the circle part, while θ tan ( sin3 t cos 3 t ) tan (tan 3 t), such that A r dθ r (t) dθ dt dt+ (sin t+cos t) 3tan tsec t +tan dt+ t 3 sin tcos tdt+ dt r (t)dt dt Page 9 of

10 () Integration and evaluation: cos tsin tdt ( +cost )( sin t)dt ( cost +sin tcost)dt 8 (t sint+ 3 sin3 t) cos tdt ( +cost )dt t+ sint A yd 3 3cos tsin tdt+ cos tdt ( + 3 )+ ( 8 ) ( 3 + )+( 3 ) 5 (For your reference, the other methods yields respectively.) yd ( 3 ) + ( 3 + ) 5 and r dθ Grading policy: for the astroid part: 3% for formulation, 3% for integration and % for evaluation; % in total for the one-eighth circle part. (b) () For the astroid part: L ds ( d dt ) +( dy dt ) dt (3sin tcost) +( 3cos tsint) dt (3%) 3 sintcost sin t+cos tdt 3 [ sin t] (%) 3 () For the circle part: L ( cost) +( sint) dt Or directly L 8 ( ) 8 dt 8 since it is one-eighth of a circle. (%) Thus the total length is L L +L Page of