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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—z2x 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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