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# Math 100 Dec 05 Answers - Solutions for Math 100 180 Final...

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Unformatted text preview: Solutions for Math 100 / 180 Final Exam, December 2005 1. (a) rim z—3 _ hmM—‘P’l = lim : (i; ” 2 MW :c—»9 a; — 9 mag deg—92 I69 1 1 2 lim (f~3+_\/5) z—v9 2 _ § — 2 (b) Note that = ln(1000 tan—1(1‘)) = in 1000 + ln(tan_1(x)). Hence, f,( ) d(ln 1000+ ln(tan"1(z))) d(ln(tan_1(m))) 1 1 m _ — —————-——— = —— - dzc dz: tan—Kr) 1 + m2 (c) First differentiate the relation x3 + y3 = 3\$y with respect to an: d d 3m2 + 3y2—y 2 3g + 3IE—y dm dm dy _ 3:2 — y dx _ ac — y2 ' So we have, dy dm :ewg) _1 3/2,3/‘2 — — 3 - (%) - (1) From this we conclude that the line is given by (y — = —1(\$ — which simpliﬁes to - (d) The ﬁrst derivative of g(x) is g’(a:) = 2cos(e2m)e2z. From this we can ﬁnd that g”(m) is given by WQ) 2 —4 sin(e2“”)e4m + 4 cos(e%)e2—:l (e) Starting from cosine, we have 00 1.211 cos a: = (— 1)" 7;) (2n)! 00 22n I \$211 cos 2% = (— 1 " —— 1;) ) (2n)! using this result, we can write 1 1 1 °° 22%?" . 2 __ _ _ : _ __ _ _ n Sin .’L‘ — 2(1 cos(2w)) 2 2 7;} 1) (2n)! 1 0° n22n—lm2n Z 5 _ Z(—1) (2n): -(_1)325 2 Therefore, c6 : W, which simpliﬁes to c6 = 21—5— . (f) Noting that ln f(m) = xlnsi”, we write ’1: cos f()=lnsinx+x m sins: Therefore f’(x) = (sin x)\$[ln sin :1: + mcotm] (g) We know that g(1.1) m g(1)+g’(1) -0.1. Using g’(m) = f’(m3)3ac2 we ﬁnd that 9(1) 2 f(1)=1, and g’(1): f’(1)-3 = 3. We conclude that g(-1.1) m 1+ 3 - 0.1 = 1.3. (h) L“ = a: , f(16)+f’(16)-1=2+(%.%)= %. Using Taylor series expansion we have f (17) z 3 1332—7. Therefore, [L 12—7 3 '2—11 1 ﬂ A < 16 = = _ — WWI — (n+1)! 2! 3 (2) _ (j) Let f(w) = 20m — m3 — 24. The solution is approximated using f(w) _\$ 2025—39—24 a 24—2253 f’(a:) _ 20— 3x2 _ 20— 32:2 (i) For m E [16,18], we have lf”(x)| = l—%x‘% (25(56) = 96 Given 931 = 2 we ﬁnd 1132 = ¢>(\$1) = 45(2) = 1: m3 = mm) = ¢<1>= . (k) Since s’(t) = 11(t) = x/t, we have 5(10) m 3(9) + s’(9)(10 — 9) = 20 + 3 2 . 2. The angular velocity w 2 10 radian/min. The passenger’s velocity is v = 10w = 100 me— tre/min. The angle a indicated in the ﬁgure below has a cosine of cosa : V 1010—6 = 3. Hence vrising = v cosa = 80 metre/min . passenger ground 3. Let y(t) be the temperature of the beverage at time t. By Newton’s law of cooling: {3% = k(y — 4). Writing p = y — 4, we have if}? 2 kg, and therefore Mt) = 06’“, with unknown constants c and k. We know that ,u(15) = 26 — 4 = 6615’“ = 22, and M30) : 15 — 4 = ce30k = 11. Dividing ce30k by ce15k, we know that 815k : g = %, and so k = {Elna Substitution into M15) gives cemﬁ mi 2 celn% = %c = 22 giving c = 44. Now it’s easy to ﬁnd that M0) = 44, and therefore y(0) 2 MO) +4 =48. 5. (a) (i) (ii) f(m+h)—f(\$) h ‘/1—2(x+h)—\/1—2x h _ hm 1—2(a:+h)—~1+2x _ mow/1 —2h—2x+\/1—2x) —2 1 lim h—>O f’(m) = lim h—>0 II = lim——————————: __ h—>0\/1—2h—2m+v1—2II: x/l—2a: 3+12—z-2x The ﬁrst—order derivative of f is f’(a:) = critical points are :3 : ix/g . 2 W = (ﬁg—7, and it’s immediate that the / \ We conclude that x = — 3 is a local minimum , and a: 2 \/§ is a local maximum . H( ) _ _2x(3 + m2 _ (3 — \$2)2(3 +\$2)2x _ ~18a: + 2x3 _ 295(952 — 9) f a: _ (3+w2)4 — (3+w2)3 _ (3+m2)3 Therefore f’ — + — + ———+———~1———i— f _3 , 0 3 down up down up Since f does not grow arbitrarily large for any x and the degree of the numerator is smaller than the degree of the denominator, there is no vertical or slant asymptote. For the horizontal asymptote we ﬁnd limmaioo f(x) : 0, so there is a horizontal asymptote y = 0 . See ﬁgure 1 See illustration below. The distance between the two ships is given by D 2 ﬁx? + y2. The derivative of this with respect of time is given by d 13% + y 3% D d d E _ l2mﬁ+2y3¥ : dt 2 \$2+y2 d_m _ g1 _ _ g2 _ 20z—4OH dt —20, dt — 40, and so dt — D . We want to ﬁnd % = 0. This happens when 201 — 40y = 0, and therefore at a: 2 2y. We can express 1: : 20t, and y = 20 — 40t. Combining these expressions we need to satisfy From the speeds given we have IL‘ 2 2y 2015 2 2(20 — 40t) 100t = 40 —2 0 2 4 6 8 —4 Figure 1: Solution to question 5b. So, we conclude that the ships are nearest at t : 0.4h . 133% 40km/h 20km SZOkm/h Proceeding from the previous question, we see that at t = 0.4 we have as = 8 and y = 4. The distance between the ships at that point is \/ 82 + 42 = \/ 80 < V100 = 10, and therefore, , there is a time at which the people on the two ships can see each other. It is given that me) = { For f(2:) to be continuous at a: = 0, we must have limxm cos(aa:) + b : limxw 2 — 123, which gives 1 + b = 2 and therefore . In addition we need f;(0) = f’_(0) because of the differentiability requirement. It can be found that = limzaoﬁu —asin(a:c) = 0 and f’_(0) = liquoﬁ 3x2 2 0. As a result a can be any constant . Since the function 9(27) is twice differentiable everywhere, 9(1) is continuous and differentiable on [0, 2]. According to the intermediate—value theorem, there is a number k in [1, 2] such that g(k) = 0. Then, by Rolle’s theorem, we have at least one interior point c in [0, k] at which g’(c) = 0 since 9(0) = 93<0 \$20 2 — x3 cos(am) + b The End ...
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Math 100 Dec 05 Answers - Solutions for Math 100 180 Final...

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