Page of 9 雖然說這樣寫依然可以算出結果, 不過寫法本身是有問題的試想, = 這明明不是數, 為何可以寫出來運算呢? 同學在使用極限的四則運算之前請三思 (c) (7%) 看到指數型的極限, 通常的做法都是把底數換成 e, 因此原式 ep [ csc ln(cosh ) ] 0 [ ] =

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1 Page of 9. (0%) Find the its. ( ) (a) [ ( (b) + ln )] 0 微甲 0-04 班期中考解答 (c) 0 (cosh ) csc, where cosh = e + e (a) (7%) 這一題要怎麼做呢? 看到分子中間係數有個, 不難讓人想到可以把 + 分給左右兩邊因此 [( ) ( )] 原式 ( ) ( )( + + ) ( )( + + )( + + ) = = = = 其中中間用了兩次的分子有理化另外, 有許多人直接把 / 放到分母, 然後說這是不定型而使用羅畢達法則, 這是不對的! 理由是分子並不是 0, 而是一個無法計算的量 (b) (6%) 這一題要怎麼辦呢? 我們可以令 y = /, 則 [ 原式 y 0 + y + ] [ ] y + ln( y) ln( y) y y 0 + y ( ) 可以注意到這是一個不定型, 故根據羅畢達法則, 我們得到 ( ) = L + y y 0 + y y y 0 + y y 我們發現分子是趨近於的, 而分母是趨近 0 + 的, 因此 ( ) = 許多同學很喜歡使用 product rule, 而把原題目寫成兩個 it 的乘積, 這也是不對的 : [ ( 原式 + ln )] [ ( + ln )]

2 Page of 9 雖然說這樣寫依然可以算出結果, 不過寫法本身是有問題的試想, = 這明明不是數, 為何可以寫出來運算呢? 同學在使用極限的四則運算之前請三思 (c) (7%) 看到指數型的極限, 通常的做法都是把底數換成 e, 因此原式 ep [ csc ln(cosh ) ] 0 [ ] = ep 0 csc ln(cosh ) [ ] ln(cosh ) = ep 0 sin Limit 和 e 可以交換的原因是指數函數是連續的, 這麼一來只要處理其中的極限就好了! 而我們發現他是一個不定型, 因此可以用羅畢達法則得到 ln(cosh ) 0 sin L 0 sinh / cosh sin cos = 0 tanh sin L = 0 sech cos = = 9 因此這題的答案就是 e 9/ 另外, 如果同學對拆項法很熟的話, 也可以不要馬上用羅畢達做, 先拆項再羅畢達也是很漂亮的方法 : ln(cosh ) 0 sin [ ] ln(cosh ) cosh 0 sin cosh ( ) [ ] ( ) ln(cosh ) cosh 0 sin 0 cosh 0 = 0 cosh L sinh 0 L = 0 cosh = = 9 其中, 因為 cosh, 而我們又知道下列極限 y ln y y = 我們令 y = cosh, 就能知道我們要算的該極限值也是了 : ln(cosh ) 0 cosh =. (5%) Find the derivative of the functions. (You need not simplify your answer.) (a) f() = log ( ) (b) f() = sin ( cos ( tan ))

3 Page of 9 (a) (b) f = ln + 4 ln ( ) f = cos(tan )( sin(tan )) sec cos4 (tan ) { if 0. (5%) Let f() = if = 0. (a) Determine whether f() is continuous at 0. (b) Determine whether f() is differentiable at 0. (a) To determine whether f() is continuous at 0, that is to determine whether then we have to check whether similarly, Then consider 0 f() = f(0) 0 0 f() f() = f(0). + f() e ln = ep ln f() = ep ln( ) 0 0 ln ln is an indeterminate form, and then by L Hospital rule, we can get similarly, ln 0 + ln( ) 0 f() f() = = f(0) f() is continuous at = 0 f() e0 = = 0 f() e0 = 0 0

4 Page 4 of 9 (b) To determine whether f() is differentiable at 0, we only need to check the it eists or not, or to check whether f() f(0) 0 0 and both its eist. f() f(0) f() f(0) f() f(0) is an indeterminate form, and then by L Hospital rule, we can get similarly, (ln + ) 0 + = d d = d d e ln = e ln (ln + ) = (ln + ) 0 ( ) 0 0 ( ) (ln( ) + ) = d d ( ) = d d e ln( ) = e ln( ) (ln( ) + ) = ( ) (ln + ) f() f(0) f() f(0) 0 0 = and 0 f() f(0) 0 = does not eist, so f() is not differentiable at (5%) (a) Show that the function f() = + + is strictly increasing on R. (b) If g() is the inverse function to the function f() of part (a). Find g (5) and g (5). (a) y = + > 0, so y is strictly increasing (b) g (5) = 6 g (5) = 6 f (g())g () = or g (f())f () = or g ()g () + g () = f (g(5))g (5) = or g (f())f () = or g (5)g (5) + g (5) = ; g(5) =, g (5) = /6 f (g())g () + f (g())g () = 0 or g (f())f () + g (f())f () = 0 or 6g()g () + g ()g () + g () = f (g(5))g (5) + f (g(5))g (5) = 0 or g (f())f () + g (f())f () = 0 or 6g(5)g (5) + g (5)g (5) + g (5) = ; g (5) = /6 5. (0%) The minute hand on a watch is mm long and the hour hand is mm long. How fast is the distance between the tips of the hands changing at two o clock?

5 Page 5 of 9 Let the distance between the tips be X(t) and the angle be θ(t). X(t) = + cos θ(t) X(t) dx = ( sin θ(t))dθ dt dt dθ dt = π 60 π 60 rad/min By direct calculation we get θ(two o clock) = π X(two o clock) = 47 dx dt (two o clock) = 57π mm/min (5%) Two vertical poles P Q and ST are secured by a rope P RS as shown in the picture. S P Q R T Given that P Q = m, ST = m and QT = m, we want to find the position of R such that (a) the length of the rope P RS is maimized. (b) the angle θ = P RS is maimized. (a) (8%) Find the maimum, not the minimum, of the length of for the case R lie in the line QT STEP ONE Define the length function by Here we setting thus the domain of L() is P R + RS, L() = ( ) Q = (0, 0), R = (, 0), T = (, 0), P = (0, ), S = (, ) [0, ] STEP TWO Take the derivative correctly, the key point is chain rule. L () = + ( )( ) + + ( )

6 Page 6 of 9 STEP THREE(no point, but relate to final answer) Observe that you will get etreme value at crtical points of the length function or the boundary of the domain of L(), in this case, L() = ( ) [0, ] = 0, STEP FOUR Find all the crtical points of the length function, since L () eists on whloe domain, we just need to find what the solution of that is L () = 0 + ( )( ) + + ( ) = + ( ) ( ) + = ( ) A little but helpfully observation + > 0, + ( ) > 0 then ( + )( + ( ) ) > 0 This make everything be simple. Since we can drop something now!! + ( ) ( ) ( ) = 0 + ( ) ( ) + = 0 that is the solution does not chage, but the equation mucs easy to understand!! STEP FIVE Slove the equation correctly!! Observe that + ( ) = ( ) + + ( ) = ( ) + ( + ( ) ) = ( ) ( + ) For simplify, denoted that then Z = ( ) ( + ( ) ) = ( ) ( + ) ( + Z) = Z( + ) this notation make the computaion much esay ( + Z) = Z( + ) 9 + Z = Z + Z 9 = Z = ( ) 9 (4 4 + ) = 0 thus we only need to slove the much easy one equation 0 = = 4( )( + )

7 Page 7 of 9 therefore we get where are all the crtical points of the length function live in, that is =, but - not lie in our domain of the length function, so just forget about it... Thus STEP FOUR union STEP FIVE have three point STEP SIX List all possible position of that making the maimum value of the length function happen L() = ( ) By STEP THREE, STEP FOUR and STEP FIVE, we known = 0,, If we need to find the minimum value of the length, then = is the one that we want. Since by reflection on X-ais, we saw P S is the straight line through thus the length is the minimum value, globally!! P = (0, ) P = (0, ) R = (, 0) L(/) = 5 But we need to find the maimum value of the length function on the Domain, thus we still need to finish the comparison on STEP SEVEN Show that(or Only state it...): L(/) = 5, L(0) = +, L() = + 5 L(/) = 5 < L(0) = + < L() = + 5 By simple way: = 4 < 5 < 9 =, 5 4 = < 4 = 9 5 conclusion : 5... <.5 and Same method [L(/) = < 5 L() = > 5 = 9 < < 6 = 4, 6 = < 4 = 9

8 Page 8 of 9 conclusion : >.5 thus L(/) = < 4.5 < + < + 4 = 5 < = L() Finally state : L() is the maimum value of the length function. equivalent (b) (7%) θ = π arctan arctan To maimize θ, we need to minimize g() = arctan + arctan g () = = + ( ( ) ) ( ) ( ) ( ) Or some sentence g () = = 0 = 6 where 0 < < 6, g () < 0 and when 6 < <, g () > 0 Hence g() is minimized when = 6 which means θ is maimized at = 6 7. (0%) Let f() = ( ) 5 ( ) (a) What is the domain of f()? (b) Does f() have any vertical or horizontal asymptote? (c) Calculate (f() ) and find the slant asymptote of f(). ± (d) Find the intervals of increase or decrease. (e) Find the intervals of concavity and the inflection points. (f) Find the local maimum and minimum values. (g) Sketch the graph of f(). (a) domain of f is R but, (b) On the domain of f, f() = ( ) 4, f() = ±,so f does not have horizontal ( + ) ± asymptote, and f() = ±, so f has a vertical asymptote = (c) (f() ) ± ( ) 4 ( + ) ± ( + ) ± = 5, ( ) 4 ( + ) ( + ) [( ) 8 + ( ) 4 ( + ) + ( + ) ]

9 Page 9 of 9 so the slant asymptote is y = 5 (d) f () = 4 ( ) ( + ) ( ) 4 ( + ) ( + ) increasing on (, 5 ), (, ) and f is decreasing on ( 5, ), (, ) = ( ) ( + ) 4 ( + 5), f is (e) ln f () = ln + ln 4 ln + + ln + 5, f () f () = , then f () = ( ) (+) 7 [(+5)(+)+9( )(+) 4( )(+5)] = 9 6 ( ) ( + ) 7, f concave on (, ), (, ) and f concave down on (, ), f 9 does not have inflection point (f) critial point is = 5, and f ( 5 ) < 0, so f has local maimum f( 5 ) = 8 4 (g) (a) 分和 - 各分 (b) 分水平和鉛直漸近線各分 (c) 分算對 it 有分, 寫出斜漸近線有分 (d) 分算對 f 的一次微分有分, 寫出遞增遞減區間有分 (e)4 分算對 f 的二次微分有分寫出上凹下凹區間有分寫出沒有反曲點有分 (f) 分寫出 f() 在哪裡有局部極大值有分, 算出來有分未寫出 f() 沒有局部極小值不扣分 (g)4 分圖形若有畫出鉛直漸近線有分, 畫對斜漸近線有分 X= 畫出未定義空心點有分剩下的分就是圖形大致上的樣子

Untitled-3

Untitled-3 SEC.. Separable Equations In each of problems 1 through 8 solve the given differential equation : ü 1. y ' x y x y, y 0 fl y - x 0 fl y - x 0 fl y - x3 3 c, y 0 ü. y ' x ^ y 1 + x 3 x y 1 + x 3, y 0 fl