Math 100 Dec 04 Answers - December 2004 Math 100 Exam...

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Unformatted text preview: December 2004 Math 100 Exam Solutions 1. Short - Answer Questions. (a) 2 . x—l , ($—1)(x+1) _ m+1 2 1 ————=1 ———:1 =—=~2. zmm2~3as+2 xLHi(x—1)(x—2) gigs—2 ~1 (b) 1 1 1. 132—1 _1. $20-?) WI. 1—? w 1—0 1 zir20$2_3x+2—x320 2 3 2 $320 3 2 ~1_0+0—~ m(1——+—2) ——+—2 J} (I? (I? {I} (c) Differentiate using the Chain Rule: 1 23: f'(-’13)= mm? 12$) : *W (d) Differentiate: y' = 1 - 6—2“: + m - (—2641) = (1 — 2:06—21, and therefore 3/” = —2 - 6—22: + (1 — 2$)(—26"2$) = (4:1: — 4)e‘2$. Since 6—21” > 0 for any 3:, the sign of y” is the same as the Sign of 4x — 4. Note that 4x—4=0when$=1. If m < 1, then 42: — 4 is negative. Thus y” is negative on the interval (—00, 1). If a: > 1, then 4% — 4 is positive. Thus y” is positive on the interval (1, +00). We conclude that y changes the concavity at a: = 1, thus (I: = 1 is the inflection point. e Divide 333 + x2 to m2 + 1 usin the lon division: 8 g x—l—l $2+1 3:34—12 933+J; (II—‘1' Therefore the slant asymptote is y = :1: + 1. (f) The equation of the tangent line is y — y(:r0) = m(a: — :50), where the slope m = 9/070)- Differentiate using the implicit differentiation: 31'2 + 3y2y’ = cosa: 1 In the above equation, let 1: = 0 and y = 1. Then 0 + 33/ = 1, so y’ = So the desired tangent line has equation 1 1 y — 1 = — 0)7 or equivalently y = 313 + 1. (g) Differentiate: y' = —— sinw. 1 1 Note that the slope of the line y = —; is m : —5 so we need — sin 3:0 2 —E, or 1 equivalently sin $0 = Thus 2:0 : (h) Take “In” on both sides of the given equation: lny :1n(a:1”) = (lnzr)(lnx) = (ln x)2 Differentiate both sides of the above equation with respect to m: y’ 1 _: 21 ._ y < m) 96 Therefore y, : 21mm .y: 2lnac 'mlnz. m a; (i) The linear approximation is : m m f(4) + f’(4)(w — 4) when (I: is close to 4. So for m = 4.1 we get: f(4.1) m f(4) + f’(4)(4.1 — 4) = 3 + «m - 0.1 = 3 + 5-0.1 = 3.5. (j) If f is differentiable for all x, then it is continuous for all 33. Thus f is continuous at x = 1 i.e. 11m me) = $1_igl+f($) = N) 33—d— So lim = lim IE3 : 1 x—d— z—’1* lim f(x)= lim (Am-PB) =A+B I—>l+ x—'1+ So we need A + B = 1. Next we differentiate f(3:): {33:2 ifac< 1 Mr): A ifar>1 The function has to be differentiable at a: z 17 i.e. f'_(1) : or equivalently A z 3. Therefore A = 3 and B = —2. (k) Let = tan‘1 x — $2. If 3:0 2 1 is the initial guess, the second estimate m1 is given by the Newton Method formula: 1 a;1=.’E0- {(560) z 1 ~ f,() f (330) f (1) Differentiate f’(:r) = 1 — 2x so f’(1) : l —- 2 2 —§ 1 +x2 7 2 2' Therefore 7T tan‘1 1 — 1 Z _ 1 7r 1 xlzl— —§ —1+—§-—-‘é+§. 2 2 . . . f"(Z) 2 (l) The max1murn error 18 given by: R1(a:) = 2 (m — a) , where a = 0, x = 0.1 and0<z<0.1. The second derivative is f”(z) : (310‘:—1 < clown—1 = 60 = 1. 1 1 Therefore |R1 < §(0.1)2 : §(0.01) = 0.005. (In) The Taylor series for cosx is: 1 $2 + $4 s = — — —— — co m 2 4 The Taylor series for sin(2w) is: 2 3 sin(2w) = 2x — ( m) 3! Thus the Taylor series for f is: x2 2:4 (2x)3 1 2 8 3 f(g;):(1_3+Z_...)+(2m_ 3! +... :1+2$_§x _6$ +.... (n)The Maclaurin series for f is Full - Solution Problems. 2. Let t be the time in minutes after turkey is put into the oven, any y = y(t) be the thermometer’s reading 2 the temperature of (inside of) turkey. By Newton’s Law of cooling: dy — = k — 2 dt (2/ 00) dy _ du _ __ 2 Z _ Z . T ‘ ' 1 _ . . - Let u v y 00, so dt dt 0 dt hus u satlsfies dt — ku, so the solutlon of this differential equation is u = C - 6]“. It is given that 71(0) 2 20 — 200 and u(30) = 30 — 200. From here: u(0) = —180 2 C - e“ = C, C' = —180. From the second condition u(30) : —170 : —180-e30k.. After solving the exponential nat'on we t k ~ 1 In 17 M _ . _. eq 1 ge 30 18 Therfore turkey is done when y(t) = 80, or equivalently when u : 80 — 200 = ~120. So, 1 17 2 —-1—t=l — <30 n18) n _301n§ _ ln% 3. Differentiate using the Product Rule: (m2 — 1)(2:L') — : 2:33 — 2a; — 2$3 —2a: / _ _ —— f (x) ‘ (x2 — 1)? (x2 — 1)? ($2 — 1)2 Since (a:2 — 1)2 is always strictly positive, the sign of f’ is the same as the Sign of —2x. Note that —2ac : 0 if and only if m = 0. Note also that f’ do not exist when a: = 1 and x = —1. Look first at the interval (—00, ——1). There, f’ is positive (use for example the ‘test point’ a: = —2). It follows that f(:r) is increasing in the interval (—00, ~1). Look next at (—1,0]. In this interval f’ is positive (use for example the ‘test point’ :c = —0.5). It follows that is increasing in the interval (—1,0]. Next look at the interval (0, 1). In this interval f’ is negative (use for example the ‘test point’ x = 0.5). It follows that is decreasing in the interval (0,1). Finally, look at (1,00). In this interval, f’(a:) is negative. (To show this, we can use a ‘test point’ like at z 2) . It follows that f is decreasing in the interval (1, We conclude that the point (0,0) is a localfi-i-ahfim. m whim/m (ii) Next we find the second derivative; differentiate using the Product Rule and the Chain Rule: ,, m2—122—2x-2232—1 29: 2952—2 —82 f($):_( )()($éa)1)4( )( )2 ( ($2_)1)3w Since 2+6m2 is always positive the Sign of f’(x) is the same as the sign of (a:2 — 1)3. Look first at the interval (—00, —1). There, ($2 —— 1)3 is positive, so f”(:r) is positive (use for example the ‘test point’ x : —2). . Next look at the interval (—1, 1). In this interval (x2 — 1)3 is negative , thus f”(m) is negative (use for example the ‘test point’ a: = 0). Finally, look at (1, 00). In this interval, f”(x) is positive. (To show this, we can use a ‘test point’ like :3 = 2) . Therefore function f (ac) changes its concavity in m = —1 and 1' = 1 but these points don’t belong to the domain. Thus f does not have any inflection point. (iii) Study the existence of horizontal asymptotes : 1' —1' x2 —1' $2 —1' 1 —1 1—13—1100 fl (1:2 — 1 — z—ir—noo 12(1 _ — z—{r—Iloo 1 — 512— — Similarly $3351me) = 1 So y = 1 is horizontal asymptote. Study the existence of vertical asymptotes : lim f(:1:): —oo; lim f(a:) = +00; z—v—IJF m—b—l— so as = —1 is vertical asymptote liIn = +00; lim 2 ——oo :c—>1+ z—al‘ so ac : 1 is vertical asymptote yA .V Lighl Let y = y(t) :the height of ball, and m = m(t)= the distance from the shadow to the ball’s landing spot. By similar triangles a:_x+3 ac y 8 8 Differentiate both sides with respect to t +§ 8. yi—“E-fit :ldg y2 8dt' d Note that when y = 4, 1% = #20. Substitute these values and :1: = 3 into the above equation to get: dz w 2 11m 16 8 dt d So, d—C: : —30. Thus the shadow is moving with 30 m/sec. (b) Let s = 5(t) be distance between the ball and its shadow. Since 82 = 3:2 + 312, we can differentiate both sides with respect to t: ds d2: dy 2—:2— 2— s a: +ydt dt dt d d ' Substitute IL‘ 2 3, y = 4 , d—g: = —20, E:— = —30: d3 32 + 423% = 3(—30) + 4(—20) : —170. d Therefore (1—: = —34. Thus, the distance between the ball and its shadow is changing at a rate of 34 m/sec. 5. Let a: be the width of pool and y be the length of the deep end, such that shallow end has length 2y. The volume of the shallow end is = 2113;. The volume of the deep end is y - m . 2 : 217;, so the total volume is 4mg 2 800, so my 2 200, or 2 yz—Oq,forx>0. :v The total surface area in contact with water is : S = my + 23:31 (bottom) + as + 211: (ends) + 2(2y + 23!) (sides) + :1: (middle) 1600 1600 :3$y+4x+8y:3-200+4a:+—x—=600+4m+ w. Differentiate both sides with respect to w. 1600 _ 42:2 — 1600 _ 4(332 — 400) 2 _ 2 2 Sl(1L')——— 4 I} (I) (E Find the critical points from S’ : 0. Since at > 0, the only critical point is m = 20. Let us look first at the interval (0,20). On this interval .5” < 0. It follows that S is decreasing in the interval (0,20). On the interval (20, +00), where S’ > 0, S(:r) is increasing. Therefore S takes its minimum value when a: = 20. 1 6. (a) Let f(:z:) = 1 2. By the definition of the derivative: — x 1 1 d 1 *, ~. f(m+h)—f(x) _. firm—fie 3(1—332) £136 —}lL1—I'I%) h a: _ 2 _ 2 thfl :3) (1 (w+h)) h—iO (1 — x2)(1 — (x + h)2)h 1—;r2—1+a:2+2$h—h2 : hall) (1 — m2)(1— x2 — 2xh+ h2)h _1im (2x+h)h _ h—»0 (1 — 9:2)(1 — m2 — 2Ih + h2)h _ 2z+0 2:1; (1—x2)(1—$2—0—0) _ (1—3132)2 (b) Since f is differentiable for all :r, so is g(x), hence g(x) is continuous for all $. Note that 9(0) = f(0)+f(1 — 0) = f(0) +f(1) and 9(1) = f(1) +f(1 — 1) = f(1) +f(0) = 9(0) Applying the Mean Value Theorem to g(m) on [0, 1], we conclude that there exists 1 — O 0 ac,0<c<1,suchthatg’(c)=gL£=I 1 _ 0 = 0, as required. ...
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This note was uploaded on 01/30/2011 for the course MATH 100 taught by Professor Lamb during the Fall '08 term at The University of British Columbia.

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Math 100 Dec 04 Answers - December 2004 Math 100 Exam...

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