. (% Solve the initial value problem: 微甲 7- 班期末考解答和評分標準 y ey, y(. + With y(, we have e tan + C C. C e (% The solution is e y e tan y ey + (separable e y y (% + e y tan + C (% for each sie. (a (% Solve the initial value problem: ( + y + (4 + y ( +, y(, >. (b (% Fin lim y( an lim y(. + (a Divie the equation by ( + We get y + 4 + ( + y ( + Multiply integration factor I( on both sie, then Iy + I 4 + ( + y I ( + Solve that I I 4 + e ( +, we let I 4+ (+ 4 + Where ( + ( + + ln + ln ( + + C, when > Hence I( e C ( +, Let C I( ( + Iy + I 4 + ( + y Iy + I y (Iy I + ( + Iy ( + + + D ( + y + + D Bring y( into the equation, we fin D Hence y + 6 + ( + (b lim y lim ( + 6 + 4( + 4 Because lim + 6 +, an lim ( + + + So lim y + + 評分標準如下, (a 解出 I( +5 分解出 y 的不定積分 + 分代入初始條件求出完整解答 + 分 Page of 7
計算錯誤該部分加分折半 (b (a 解出的 y 錯誤不管答案一律 分 ( 有些同學計算錯誤會影響答案, 有些不會, 覺得不應該同樣因為計算錯誤而有差別 此外對一題 分, 兩題 分. (a (7% Fin the length of the curve in polar coorinates: r + sin θ, θ π. (b (7% Fin the area enclose by the curve given in (a. (a The length of the curve L is L r θ cos θ + sin θ θ + sin θ π π π π π r + ( points ( r θ ( points θ + sin θ + cos θ + sin θ θ + sin θ + sin θ + cos θ + sin θ + sin θ + sin θ θ θ π ( points θ (b Because r for θ π (or ue to observation from the figure below, the area A of the enclose region is A π π r θ (4 points ( + sin θ θ [θ ] π cos θ ( π + π ( points Page of 7
4. (a (7% Fin the length of the parametric curve C: ln(sec t + tan t sin t, y cos t, t π. (b (7% Rotate the curve C about -ais. Fin the surface area. (c (7% Fin the volume of the soli boune by the surface given in (b an the planes, ln(+ (a.. ((t sec(t cos(t, t (y(t sin(t.... ( points t π s π tan(tt... ( points ln cos(t π/... ( points ln(... ( points, where s [ ] [ ] t ((t + t (y(t t tan(tt. (b. π πy(ts π πcon(ttan(tt... ( points πcos(t π/... ( points π... ( points (c. πy (t((t, where ((t [sec(t cos(t]t. π π πcon (t[sec(t cos(t]t... ( points πsin (tcon(tt π sin (t π/... ( points 8 π... ( points 5. (% Fin ( + +. Let ( + + a + b + c +, where a, b, c, are etermine. ( point Then solve a,b + +,c,. ( points, if solve partially point so we have ( + + + + + + Page of 7
ln + ( points + + + + + + + + + ln( + + ( points ( + + ( arctan + ( points so Answer is ln + ln( + + arctan + + C If you on t wirte +C, you will lose point. 6. (% Fin sec. Solution. 後面的三項積分個別計分, 第一項積出 sin sec sec sec + sec + sec + sec ( 分 ( sec + cos tan ( 分 sin + cos sin sin sin sin + sin sin sin + csc cot + C. sin 再得 4 分, 其它兩項及常數 C 一共佔 分 Solution. Let t tan ( 佔 分, then we have cos t + t ( 佔 分, sin t + t ( 沒有用到, an + t t. ( 佔 分 So sec t cos cos ( +t 分 +t t t +t ( t t t (7 分 + t tan t + C (9 分 tan tan ( tan t t ( + t t (5 分 + C cot + C. ( 分 前面兩種解法是大部分人採用的積分策略, 所以在此列出詳細配分標準, 以下還有數種解法, 處理方式都很漂亮, 在此也完整呈現供大家參考及學習, 配分以個案處理, 但原則上和前兩種配分方式雷同 Solution. sec cos sin cos sin csc ( cot + C. Solution 4. sec sec sec + sec + sec + sec + sec tan. Page 4 of 7
Since an we get Solution 5. Since an sec sec tan tan sec tan tan sec tan sec cot sec tan + sec cot csc sec cos tan + sin 4 sec tan + sec sin 4 sin tan sin + C, tan sec cot (csc cot + C, sec sec tan sin cot + C. sec sec + sec + sec + sec + sec tan. sec tan cos tan sin cot csc csc + C, cos csc cos cot cos cot + cot cos cos cot cot cos sin cos cot cos cos cot ( + cos cos sin cot + C cos cot sin cos + C we get sec csc cos cot sin cos + C. Solution 6. Let t cot, then we have So cos t t, sin, an + t + t + t t. sec cos cos ( + + t cot + tan ( cot t +t +t t t t + t t +t t t + tan t + C + C. Solution 7. Let t sec +tan. Since sec tan (sec +tan (sec tan, we have t an sec t + t, tan t, t t +. t sec tan Page 5 of 7
(b Discuss the asymptotic behavior lim y( an lim y(. + Problem. (%+% (a Fin the length of the polar curve: r + sin θ, θ π. So (b Fin the area enclose by the curve given in (a. Problem 4. (%+%+% sec t + t t + t 4t (a Fin the length of the parametric curve C: ln(sec t + (ttan t (t sin t, + y cos t t, t π. ( (t (b Rotate the curve C about the -ais. Fin the surface area. t t + t tan t + C (c Fin the volume of the soli boune by the surface given in (b an the planes, ln( +. sec + tan tan (sec + tan + C. Problem 5. (% Fin ( + +. 7. Consier the intersection Problemof6. three (% Fin circular cyliners with raius R an all aes of cyliners lie in a plane with polar equations θ, θ π, an θ π sec. Problem 7. (%+%. Consier The cross-section the intersection is of a heagon, three circular an cyliners the shape with raius of ther soli an all looks like the union of two umbrellas. aes of cyliners lie in a plane with polar equations θ, θ π, an θ π. The cross-section is a heagon, an the shape of the soli looks like the union of two umbrellas. θ π θ π R θ y (a (b Figure : Intersection of three cyliners. (a Vertical view. (b The soli. (a (6% Fin the area (a Fin of the cross area ofsection the crossof section the soli of the when soli when the the height is isy, y [, R]. R]. (b (6% Fin the volume (b Finof the the volume soli. of the soli. (aobserve the figures below. (aobserve the figures below. Problem 8. (% Evaluate the improper integral e. θ π 6 y R When the height is y, the area of the cross section(heagon is ecie by When the height is y, the area of the cross section(heagon is ecie by R y.(% R y. Hence, Hence, Area 6 Area of Area 6 Area of equilateral triangles 6 equilateral triangles 6 (R y (% (R y R R R V A(yy (R y y (R y y R 8 (b V A(yy(% R R (R y y (R y y R (% 8 R (% 8. (% Evaluate the improper integral e. Page 6 of 7
t t t e lim e lim e lim u e u u ( 分 lim ( t u e u lim [ u e u] ut t e u u (6 分 u ( lim t e t [e u] ut u lim (7 分, ( t e t e t + where we use the l Hospital Rule to get fin the limit: t lim t e t lim e t (,L lim t t e t lim. e t 這題重點在於是否具有 瑕積分的觀念, 而判定的依據是以有寫出 極限 為準 ; 縱使前面所有的式子都是寫 是上 下限寫, 但最後一定要用極限判斷 lim t e t, 就認定你知道 瑕積分的意義 ; 這個極限值不是顯然 的, 一定要討論, 佔 分, 也就是說, 最後有完整討論這個極限就可以拿到滿分 分 另一個極限式 lim e t 可接受直接寫答案, 沒有解釋不會扣分 或 Page 7 of 7