MATH111-200630-PS03-Solutions

# MATH111-200630-PS03-Solutions - MATH 111 Problem Set 3...

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Unformatted text preview: MATH 111 Problem Set 3 Solutions DRAFT Edward Doolittle October 16, 2006 1. (a) In a right triangle with angle θ sin 1 4 5 , the opposite side O and hypoteneuse H are 4 and 5 units respectively. Then the adjacent side A 52 42 9 3 units. Then sec sin 1 4 5 sec θ H A 5 3. See Figure 1(a). (Question: is A 3 possible? Why or why not?) (a) 3, 4, 5 right triangle (b) 1, 2x, 1 4x2 right triangle Figure 1: Two right triangles (b) In a right triangle with angle θ tan 1 2x , the opposite side O and adjacent side A are 2x and 1 units respectively. Then the hypoteneuse is H 2x 2 12 1 4x2 units, and sin tan 1 2x sin θ 2. See Figure 1(b). (Question: what happens if x is negative? Can the adjacent side be O H 2x 1 4x 1 instead of 1?) 2. (a) The limit is of the form 0 0. Applying L’Hˆ pital’s rule, o lim et 2t 0 tant lim 0 et t t ln 2 2t sec2 t 1 ln2 1 which is 0 3069 to four decimal places. (b) Factoring the denominator, The ﬁrst factor is of the form 0 0. Applying L’Hˆ pital’s rule until it is no longer of that form, o lim lim lim lim x x x 0 x 0 1 X The second factor is of the form 1 0, but we must be careful before we conclude that it is ∞. As x tan2 x x2 0 so 1 tan2 x x2 tends to ∞. You should also try this problem with x3 instead of x5 in the denominator. V § U sin x x 0 x3 cos x 1 0 3x2 sinx 6x § § U § T T x 0 x3 tan2 x x5 x W lim lim V T U sin x x sin x x 1 3 2 x x2 0 x tan cos x 6 1 6 ¤ ¢ ¤ ¢ ¨¤¤ ¢ ¡ ¢ ¤¨¤ £ ¢ ¡ ¢ 'FE § ¤ ¢ § T ¥ H G B@ 4 2 A9FE DCA9876531 T ¥ T ¤ ¢¤ ¡ ¢ P A9F DE SQPH I G R § ¥ § ¥£ ¤ ¦¢ ¡ § U § § ¤ T £ U § U " £ U £ ¢£ U 0 (    )'&%\$#"!  © & T ¥£ £ § T X T V £ £ 0, x ∞ lim 1 5x sin 1 x x ∞ lim sin 1 x 1 5x 1 x lim cos 1 x 1 x2 ∞ 5 1 5x 2 Now we try to evaluate the limit using limit laws and algebra: lim cos 1 x 1 x2 ∞ 5 1 5x 2 x x ∞ lim cos 1 x (d) This limit is of the form 1∞ . Let L be the desired limit. Taking logarithms, This new limit is of the form ∞ 0. Moving x to the denominator (where it becomes x 1 ) and applying L’Hˆ pital’s rule, o Multiplying the second factor and the denominator through by ln L x ∞ Therefore L 3. (a) By the chain rule, (b) Again by the chain rule, h z cot 1 ez d z e dz (You can stop there if you like, but a little more algebra gives an interesting result: h z In fact, h z ez 1 e2z e z π 2, a constant. Question: is the function k x (c) By the chain rule, g t cos 1 3 Differentiating again, by the chain rule, g t 1 3 2t 2 3 2 Further simpliﬁcation is only useful if you are taking a third derivative or doing something else with the function. dV¨¤¤ § ¢ ¤ § ¢ § ¢ W c ¡ ¤ ¤ § ¢ § ¢ § ¤ ¤ § ¢ § ¢ W c ¡ ¤ ¤ § ¢ § ¢ § ¤ ¢ Y Y `V¤ ¡ § ¢ ¤ ¡ ¢ T § ¤ ¢ T § ¡ ¤ ¡ ¢ ¤ ¡ ¢ T V § ¥ W ¤ ¢ § W ¤ ¢Y ¤ ¢Y V V ¤ T ¢W ¡¤ T T ¢ ¤ ¡ T ¢W ¡¤ ¡ T ¡ T ¢ U § U U § V ¤ ¡ § ¡ § ¢ W ¡ ¡ ¤ ¡ T ¡ T ¢ ¤ ¡ T ¡¡ T ¢ ln L x ∞ lim ln 1 2x x 1 3x 2 1 x ∞ lim 1 2x 1 3x 2 1 2x 2 6x x 2 x2 , lim 1 2x 1 3x 2 1 2 6x 1 1 0 0 1 2 0 2 e2 7 3891 to four decimal places. d x e dx f x arcsin ex 1 1 ex 2 ex ex 1 e2x cot 1 e z d e dz z 1 1 ez ez 2 1 1 e z 2 1 e 2z e2z e2z ez 1 e2z cot ez e2z 1 0 1 x cot 1 1 x constant? Careful!) 2t 2 1 d 1 dt 3 2t 2 2 V c ¡¤ ¤ § ¢ § ¢ ¤ § ¢ § 2 3 2t 2 21 3 2t 2 1 2 1 3 2t 2 3 2 23 ¡ ¤ £ ¢ ¡ a¤ ¢ ¡ ¤ ¢ T V T T T § W ¡ T T ¡ ¤ § ¢ W ¤ § ¢ bY¤ ¡ ¢ ¤ ¢ Y U ln L ln lim 1 x ∞ § ¤ §¤ £ § U ¤ § § UW ¨¢¢ ¢ ¡¤ £ § ¢ U W ¤ £ ¢ § x lim 1 x2 ∞ 5 1 5x 2 1 lim x ∞ 1 5x 5x2 x ∞ lim x ln 1 V T T U T T 2 x 3 x2 x V ¤ £ ¡ ¤§ W ¤§ £¢ ¢ ¢ 2 x ∞ U ¤¡ ¤ £ § ¢ U ¤ £ ¢ ¤ § ¢ U ¢ W T § ¤ ¢Y W (c) This limit is of the form ∞ 0. Dividing through by 1 § 5x and applying L’Hˆ pital’s rule, o ¤ U V ¤ £ ¡ § ¢ W ¤§ £¢ ¢ ¤ ¢ ¤Y ¡ ¢ ¤ ¢ Y £ ¤ ¢ U lim 1 x 5 5 2 5 2 x 3 x2 3 e z 2t 2 xy 1 2 By the product rule, 1 Solving for y , xy 4. (a) Dividing through by 9, Let u t 3; then du dt 3, dt 3 (After reading chapter 8.3, it is more natural to make the substitution t (b) This is a case where we have to be careful of the sign of the square root. Recall that value of u. Therefore the integrand is 1 cos2 x sin x 2 On the interval in question the above function corresponds to the Heaviside step function H x x x (see Figure 2). Informally, since H is an odd function (H x H x ), its integral over any interval symmetric about the origin must be 0. A more formal way of evaluating the integral is to break up the domain of integration into pieces: π 4 (Actually, sin x sin x is a ‘square wave’. Graph it on the larger interval called a square wave.) 16 The ﬁrst term on the left side of (1) can be evaluated by making the substitution u du 2 x dx. Then x2 16, du T x2 16 dx 3 Tg g x 1 2 du u 1 ln u 2 C 1 ln x2 2 16 C T V f¤ T ¢ T T T e T x2 x2 dx x2 16 dx V T x 4 dx 16 x e e T e (c) The integral can be rewritten 4 p p § V £ T £ § c ¡ c e T § § ¥ c ¡e π 4 sin x dx 1 cos2 x 0 π 4 sin x dx 1 cos2 x π 4 0 π 4 0 sin x dx 1 cos2 x 0 π 4 1 dx 1 dx π 4 π 4 0 3π x 3π to see why it is g g£ ¤ ¢ § ¥ c e T g e ¥ sin x sin x sin x sin x ¤ ¢ § ¤ §¢ p V q p q § ii T§ g § ¥ c ¡ c e § ¥ T t2 9 dt g gh ¥ V Tf¤ £ T f¤ ¢ T ¢ e V T ¤ £¢ T e e T ¢£ Y ¢£ § Y §Y T 2x2 yy 2xy2 y 2xy2 x 1 y 1 x2 y2 3 t2 9 dt 1 3 t 3 3du, and 1 3 3du u2 1 arctan u V ¤ §T¤ T § x y 1 x2 y2 y 1 x2 y2 x2 y2 2x2y 1 2 1 dt C arctan t 3 1 1 YV T 1 y x2 y2 2xy2 2x2 yy dV¤ C 3 tan θ.) u2 u , the absolute π 4 T Y¤ T ¢ ¤ ¢ T 1 xy d xy dx d 2 2 x y dx ¢ ¤ ¢ By the chain rule, dV¤ 0 x x 0 π 4 d tan dx 1 d 1 dx T ¢ ¤ ¢ ¡ (d) By implicit differentiation, x2 y2 £ e e g g£ Y £ (1) 2x dx, £ H x 15 1 05 0 05 1 15 2 15 1 05 0 05 1 15 Figure 2: Graph of the Heaviside step function H x The second term on the left side of (1) can be evaluated by dividing through by 16 and making the substitution v x 4, dv dx 4, dx 4dv: x2 16 dx x 4 2 1 dx ln (d) Let u x3 . Then du 3x2 dx so x2 dx dx 2 3 du 3 and 1 x6 5. (a) Multiplying the given limit through by the conjugate radical and then dividing through by x, x ∞ lim x2 2x x x ∞ lim x2 2x x x2 2x x x ∞ lim x2 2x x x ∞ lim 1 (b) Taking out a factor of x, putting it in the denominator, and then applying L’Hˆ pital’s rule, o x ∞ lim x2 2x x x ∞ lim 1 2 x 1 x 0 0 1 x ∞ lim 1 2 1 2 x 1 1 x2 2 2 x2 x ∞ lim 1 2 x 6. This limit is of the form 0 0 because of Calculus, x tan t 2 dt lim 0 x 0 x3 f t dt 0. Applying L’Hˆ pital’s rule and the Fundamental Theorem o lim sec2 x2 2x 0 6x lim sec2 x2 0 3 1 3 x lim tan x2 0 3x2 x x 7. The limit is of the form 1∞ . Taking logarithms of both sides, x ∞ 4 § ln lim ln e i x x a a x V T £ T U ¤ ¢ U ¤ ¢ U ¤ ¢ U ¤ ¢ r £ ¤¢ r U ¤ § T V c ¡ ¤ £ T ¢ U ¤ £ § c ¤£ £ § T ¤ £ U § £ £ T ¢ ¡ ¢ ¢ T T U T T § ¤ T ¥ ¢ U T T ¥¥ W ¤ § T x2 du 1 u2 2 arcsin u 3 C 2 arcsin x3 3 2x x x2 2x x2 V C¤ ¢ T T f¤ ¢ 2x2 T T x x2 4 dx 16 x2 16 arctan x 4 V f¤ £ ¢ T Altogether, T T § ¥ § ¥ e e £ T T e T ¤ £¢ T 4 1 4 1 dv v2 1 arctan v C arctan x 4 C V Tf¤ £ ¢ C V ¤¢ C 2 2 x 1 1 1 2 2 T f¤ ¢ ¤¢ V e U V § e ¢ § £ U ¤ § T V § § V § £e V V § V § ¢ ¢ U U 1 i.e., x ∞ Dividing through by x and applying logarithm laws and L’Hˆ pital’s rule, o x ∞ 1 x ∞ x 1 x ∞ x x ∞ x2 so the equation becomes 2a 1, i.e., a 1 2. 5 § lim lim lim lim ln x a x x a ln x U ¡ ¡ § § s U ¤ § ¢ ¡§ ¤ T ¢ a ln x a 1 x a 1 x a 2 § lim x ln V 1 x x T a a U £ U ¨¤¤ § ¢ `£¤¡ T Q¢¢ 2ax2 a2 2a U ...
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## This note was uploaded on 01/12/2010 for the course MATH 111 taught by Professor Doolittle during the Fall '06 term at Berkeley.

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