Unformatted text preview: Math 111 Problem Set 3 Solutions
Edward Doolittle October 26, 2005
1. (a) In a right triangle with angle 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 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 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 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. o (a) The limit is of the form 0 0. Applying L'H^ pital's rule, 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) (This problem had a typo which made it hard. x5 , not x4 , should have appeared in the denominator. Try it! Also try it with x3 instead of x4 in the denominator.) Anyway, factoring the denominator,
x 0 x3 tan2 x x4 x The first factor is of the form 0 0. Applying L'H^ pital's rule until it is no longer of that form, o lim lim lim lim x 0 x 0 x 0 x 0 The second factor is of the form 1 0, but we must be careful before we conclude that it is . As x 0 , tan2 x x 0 so 1 tan2 x x tends to . However, as x 0 , the situation is not so clear. Roughly 1 a ` X sin x x x3 cos x 1 3x2 sinx 6x T T ` W W W Y lim lim X T sin x x sin x x 1 2x 3 0 x tan x cos x 6 1 6 'FE T H G [email protected] 4 2 A9FE DCA9876531 T U T V P A9F DE SQPH I G R T W W W " W W 0 ( )'&%$#"! & T X b T (1) speaking, tan x x for small x so tan2 x x2 which is insignificant compared to x for small x, so tan2 x x x 0 for small negative x, and therefore 1 tan2 x x tends to . (Can you make the preceding argument more rigorous?) So the first factor of (1) tends to a finite nonzero value and the second factor does not exist, so the limit as a whole does not exist. cos 1 x 1 x2 5 1 5x 2 x lim 1 5x sin 1 x x lim sin 1 x 1 5x 1 x lim 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, ln L ln lim 1
x This new limit is of the form 0. Dividing through by x and applying L'H^ pital's rule, o ln L
x lim ln 1 2x x 1 1 Multiplying the second factor and the denominator through by ln L
x lim 1 2x 1 Therefore L 3. e2 7 3891 to four decimal places. arcsin ex d x e dx (a) By the chain rule, f x (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 1 e
2z e2z 2, a constant. Question: is k x cot 1 x cot 1 1 x constant? Careful!) (c) By the chain rule, g t cos
1 Differentiating again, by the chain rule, g t 1 1 2 3 2t
2 3 2 Further simplification is only useful if you are taking a third derivative or doing something else with the function. 2 X iQ Y p Y p g g X p U Y hg g 3 2t 2 2 1 3 2t 2 1 d 1 dt 3 2t 2 1 1 2 3 2t 2 X T T ez 1 e2z W WY W Y x lim 1 x2 5 1 5x 2 1 lim x 1 5x 5x2 X i T T T X Y U Y g g X X T Y T T T Y T T W W W X Y T T T Y T W f f T T e W e x lim x ln 1 3x 2 x lim 1 2x 1 3x x2 , 3x 2 1 2 6x 1 1 1 1 ex 2 ex cot 1 e z d e dz z 1 1 ez ez 2 e z e2z e2z ez 1 e2z X f T T e 2 x 3 x2 x 2 0 X Y 2 x T T Y T T W W W Y T g (c) This limit is of the form 0. Dividing through by 1 5x and applying L'H^ pital's rule, o lim 1 x 5 5 2 2 x 3 x2 1 2x 2 6x 3 x 2 0 1 2 0 2 ex 1 e2x 1 1 e z 2 e z ez 1 0 3 2t 2 1 2 3 2 23 2t 2 c T
5 T c W X Y hg g c W d xy 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 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: 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 (2) can be evaluated by making the substitution u du 2 x dx. Then x2 16, du T x2 16 dx 3 Tr r x 1 2 du u 1 ln u 2 C 1 ln x2 2 16 C T X T T T T q T x2 x2 dx x2 16 dx X T x 4 dx 16 x q q T q (c) The integral can be rewritten 4 w w X T p p q T p q 4 sin x dx 1 cos2 x 0 4 sin x dx 1 cos2 x 4 0 4 0 sin x dx 1 cos2 x 0 4 1 dx 1 dx 4 4 0 3 x 3 to see why it is r r p q T q sin x sin x sin x
2 w X d w d vv T u r r t p p q T t2 9 dt r rs T X T T q X T T q q T g T g g 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 sin x sin x X T T x y 1 x2 y2 y 1 x2 y2 x2 y2 2x2y 1 2 1 dt C arctan t 3 1 1 gX T T g T T
1
2 1 xy d xy dx d 2 2 x y dx 2xy2 1 y x2 y2 %X 2x2 yy C 3 tan .) u2 u , the absolute 4 By the chain rule, %X
0 x x 0 4 d tan dx 1 d 1 dx T (d) By implicit differentiation, x2 y2 q q r r g (2) 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 (2) can be evaluated by dividing through by 16 and making the substitution v x 4, dv dx 4, dx 4dv: 16 x 4 1 Altogether, x x2 4 dx 16 ln x2 (d) Let u x3 . Then du 3x2 dx so x2 dx dx 2 3 du 3 and du 1 u2 2x2 1 x6 2 arcsin u 3 5. (a) Multiplying the given limit through by the conjugate radical and then dividing through by x,
x lim (b) Taking out a factor of x, putting it in the denominator, and then applying L'H^ pital's rule, o
x 6. This limit is of the form 0 0 because of Calculus, x tan t 2 dt lim 0 x 0 x3 7. The limit is of the form 1 . Taking logarithms of both sides, ln lim
x T T T T T T U W X T W T T Y T U W T U W x2 2x x x lim x2 2x x x2 x2 2x 2x lim W W W y W y X p T W p T W T U W T U W x2 2x x x lim 1 2 x 1 x
0 0 1 x lim 1 2 1 f t dt 0. Applying L'H^ 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 ln e 4 v f f T e W e x x a a x T x2 X x T T q q T X T T T U T q T T dx 2 dx 16 x x 4 1 4 1 dv v2 1 arctan v C arctan x 4 arctan x 4 C C 2 arcsin x3 3 x lim x2 2x x2 x x2 2x x lim 2 x 1 1 x2 2 2 x2 x lim 1 x X T C X C 1 2 2 x 1 1 2 x
1 2 2 T X q X q X X q X X X 1 i.e.,
x lim x ln Dividing through by x and applying logarithm laws and L'H^ 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 a ln x a 1 x a 1 x a 2 W a W T W Q T Q X f T e
x x a a 1 W 2ax2 a2 2a W ...
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 Summer '05
 edwarddoolittle
 Calculus, Derivative, lim

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