MATH111-200530-PS03-Solutions

# MATH111-200530-PS03-Solutions - Math 111 Problem Set 3...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 non-zero 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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online